NCERT Class 12 Determinants MCQs with Solutions

📘 Concept & Theory

Concept/Theory

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For a square matrix \(A\), the adjoint (or adjugate) of \(A\) is obtained by taking the transpose of its cofactor matrix.

If the cofactor matrix of \(A\) is denoted by \(C\), then

\[ \operatorname{adj}(A)=C^{T} \]

For a \(2\times2\) matrix

\[ A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}, \]

its adjoint is given directly by

\[ \operatorname{adj}(A)= \begin{bmatrix} d & -b\\ -c & a \end{bmatrix}. \]

Thus, for a \(2\times2\) matrix, the diagonal entries are interchanged and the signs of the two off-diagonal entries are changed.

🗺️ Solution Roadmap

Step-by-step Plan

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Write the given matrix \(A\).
Identify the four entries \(a,b,c,d\).
Use the \(2\times2\) adjoint formula.
Interchange the diagonal entries \(a\) and \(d\).
Change the signs of the off-diagonal entries \(b\) and \(c\).
Write the resulting matrix as \(\operatorname{adj}(A)\).

✏️ Solution

Complete Solution

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Step-by-step Solution · 11 steps

Given —
\[A=\begin{bmatrix}1 & 2\\3 & 4\end{bmatrix}\]
Comparing it with the general \(2\times2\) matrix
\[A=\begin{bmatrix}a & b\\c & d\end{bmatrix}\]
we have
\[a=1,\quad b=2,\quad c=3,\quad d=4\]
For a \(2\times2\) matrix, the adjoint is
\[\operatorname{adj}(A)=\begin{bmatrix}d & -b\\-c & a\end{bmatrix}\]
Substituting \(a=1,\ b=2,\ c=3,\ d=4\), we get
\[\operatorname{adj}(A)=\begin{bmatrix}4 & -2\\-3 & 1\end{bmatrix}\]
Hence,
\[\boxed{\bbox[5pt]{\operatorname{adj}(A)=\begin{bmatrix}4 & -2\\-3 & 1\end{bmatrix}}}\]
Verification Using Cofactors
The above result can also be verified directly from the definition of adjoint. The cofactor of an element \(a_{ij}\) is
\[C_{ij}=(-1)^{i+j}M_{ij}\]
where \(M_{ij}\) is the corresponding minor.
For
\[A=\begin{bmatrix}1 & 2\\3 & 4\end{bmatrix}\]
the cofactors are
\[C_{11}=4,\quad C_{12}=-3,\quad C_{21}=-2,\quad C_{22}=1\]
Therefore, the cofactor matrix is
\[C=\begin{bmatrix}4 & -3\\-2 & 1\end{bmatrix}\]
Since the adjoint is the transpose of the cofactor matrix,
\[\operatorname{adj}(A)=C^{T}=\begin{bmatrix}4 & -2\\-3 & 1\end{bmatrix}\]
This confirms the result obtained using the direct \(2\times2\) formula.

🎯 Exam Significance

Exam Significance

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The concept of adjoint is fundamental to the Determinants chapter and is directly connected with the inverse of a matrix. In board examinations, questions may require students to find the adjoint, calculate the inverse using the adjoint, or verify identities involving \(A\operatorname{adj}(A)\).

For a \(2\times2\) matrix, remembering the direct formula can save considerable time, but students should also understand that the formula comes from the more fundamental rule

\[ \operatorname{adj}(A)=\left[\text{cofactor matrix of }A\right]^T. \]

Significance for Competitive Entrance Examinations

For competitive examinations, this result is useful because adjoint-related questions are often combined with determinants, inverse matrices, matrix multiplication, and properties such as

\[ A\operatorname{adj}(A)=\operatorname{adj}(A)A=|A|I. \]

For the present matrix,

\[ |A|=(1)(4)-(2)(3)=4-6=-2. \]

Therefore,

\[ A\operatorname{adj}(A) = -2I \]

This property provides a quick way to verify an adjoint and is particularly useful in time-bound objective questions.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 6 points

For a \(2\times2\) matrix \(A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\), use \(\operatorname{adj}(A)=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\).
Interchange the two diagonal elements.
Change the signs of the two off-diagonal elements.
The fundamental definition is \(\operatorname{adj}(A)=C^T\), where \(C\) is the cofactor matrix.
Always distinguish between the cofactor matrix and the adjoint; the latter is the transpose of the former.
The identity \(A\operatorname{adj}(A)=|A|I\) is an important verification tool.

📘 Concept & Theory

Concept/Theory

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For a square matrix \(A\), the adjoint is defined as the transpose of its cofactor matrix.

\[ \operatorname{adj}(A)=C^T, \]

where \(C\) is the cofactor matrix of \(A\).

The cofactor \(A_{ij}\) corresponding to the element \(a_{ij}\) is given by

\[ A_{ij}=(-1)^{i+j}M_{ij}, \]

where \(M_{ij}\) is the minor obtained by deleting the \(i\)-th row and \(j\)-th column.

The signs of the cofactors follow the pattern

\[ \begin{bmatrix} + & - & +\\ - & + & -\\ + & - & + \end{bmatrix} \]

🗺️ Solution Roadmap

Step-by-step Plan

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Write the given \(3\times3\) matrix \(A\).
Find each minor \(M_{ij}\) by deleting the corresponding row and column.
Apply \(A_{ij}=(-1)^{i+j}M_{ij}\) to obtain all nine cofactors.
Arrange the nine cofactors in the cofactor matrix.
Transpose the cofactor matrix to obtain \(\operatorname{adj}(A)\).

✏️ Solution

Complete Solution

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Step-by-step Solution · 26 steps

Given —
\[A=\begin{bmatrix}1 & -1 & 2\\2 & 3 & 5\\-2 & 0 & 1\end{bmatrix}\]
Find the cofactors of the first row
For \(A_{11}\), delete the first row and first column:
\[ M_{11} = \begin{vmatrix} 3 & 5\\ 0 & 1 \end{vmatrix} =(3)(1)-(5)(0)=3\]
Therefore,
\[A_{11}=(-1)^{1+1}M_{11}=1(3)=3\]
For \(A_{12}\), delete the first row and second column:
\[M_{12} = \begin{vmatrix} 2 & 5\\ -2 & 1 \end{vmatrix} =(2)(1)-(5)(-2) =2+10 =12\]
Therefore,
\[A_{12}=(-1)^{1+2}M_{12}=-1(12)=-12\]
For \(A_{13}\), delete the first row and third column:
\[ M_{13} = \begin{vmatrix} 2 & 3\\ -2 & 0 \end{vmatrix} =(2)(0)-(3)(-2) =0+6 =6\]
Therefore,
\[A_{13}=(-1)^{1+3}M_{13}=1(6)=6\]
Hence, the first row of the cofactor matrix is
\[ \begin{bmatrix} A_{11} & A_{12} & A_{13} \end{bmatrix} = \begin{bmatrix} 3 & -12 & 6 \end{bmatrix} \]
Find the cofactors of the second row
For \(A_{21}\), delete the second row and first column:
\[ M_{21} = \begin{vmatrix} -1 & 2\\ 0 & 1 \end{vmatrix} =(-1)(1)-(2)(0) =-1 \]
Therefore,
\[ A_{21}=(-1)^{2+1}M_{21} =(-1)(-1) =1 \]
For \(A_{22}\), delete the second row and second column:
\[ M_{22} = \begin{vmatrix} 1 & 2\\ -2 & 1 \end{vmatrix} =(1)(1)-(2)(-2) =1+4 =5 \]
Therefore,
\[ A_{22}=(-1)^{2+2}M_{22} =1(5) =5 \]
For \(A_{23}\), delete the second row and third column:
\[ M_{23} = \begin{vmatrix} 1 & -1\\ -2 & 0 \end{vmatrix} =(1)(0)-(-1)(-2) =0-2 =-2 \]
Therefore,
\[ A_{23}=(-1)^{2+3}M_{23} =(-1)(-2) =2 \]
Hence, the second row of the cofactor matrix is
\[ \begin{bmatrix} A_{21} & A_{22} & A_{23} \end{bmatrix} = \begin{bmatrix} 1 & 5 & 2 \end{bmatrix} \]
Find the cofactors of the third row
For \(A_{31}\), delete the third row and first column:
\[ M_{31} = \begin{vmatrix} -1 & 2\\ 3 & 5 \end{vmatrix} =(-1)(5)-(2)(3) =-5-6 =-11 \]
Therefore,
\[ A_{31}=(-1)^{3+1}M_{31} =1(-11) =-11 \]
For \(A_{32}\), delete the third row and second column:
\[M_{32} = \begin{vmatrix} 1 & 2\\ 2 & 5 \end{vmatrix} =(1)(5)-(2)(2) =5-4 =1\]
Therefore,
\[A_{32}=(-1)^{3+2}M_{32}=-1(1)=-1\]
For \(A_{33}\), delete the third row and third column:
\[ M_{33} = \begin{vmatrix} 1 & -1\\ 2 & 3 \end{vmatrix} =(1)(3)-(-1)(2) =3+2 =5 \]
Therefore,
\[A_{33}=(-1)^{3+3}M_{33}=1(5)=5\]
Hence, the third row of the cofactor matrix is
\[ \begin{bmatrix} A_{31} & A_{32} & A_{33} \end{bmatrix} = \begin{bmatrix} -11 & -1 & 5 \end{bmatrix}. \]
Form the Cofactor Matrix
Collecting all nine cofactors, we obtain
\[ C= \begin{bmatrix} A_{11} & A_{12} & A_{13}\\ A_{21} & A_{22} & A_{23}\\ A_{31} & A_{32} & A_{33} \end{bmatrix} = \begin{bmatrix} 3 & -12 & 6\\ 1 & 5 & 2\\ -11 & -1 & 5 \end{bmatrix} \]
Transpose the Cofactor Matrix
The adjoint of \(A\) is the transpose of the cofactor matrix:
\[\operatorname{adj}(A)=C^T\]
Therefore,
\[ \operatorname{adj}(A) = \begin{bmatrix} A_{11} & A_{21} & A_{31}\\ A_{12} & A_{22} & A_{32}\\ A_{13} & A_{23} & A_{33} \end{bmatrix} \]
Substituting the values of the cofactors,
\[ \operatorname{adj}(A) = \begin{bmatrix} 3 & 1 & -11\\ -12 & 5 & -1\\ 6 & 2 & 5 \end{bmatrix} \]
Hence, the required adjoint is
\[ \boxed{\bbox[5pt]{ \operatorname{adj}(A) = \begin{bmatrix} 3 & 1 & -11\\ -12 & 5 & -1\\ 6 & 2 & 5 \end{bmatrix}}} \]

🎯 Exam Significance

Exam Significance

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Finding the adjoint of a \(3\times3\) matrix is an important skill in the NCERT Class 12 Mathematics chapter on Determinants. The question tests whether a student can correctly calculate minors, apply cofactor signs, construct the cofactor matrix, and transpose it.

A common source of errors is stopping after forming the cofactor matrix. Remember that the adjoint is not the cofactor matrix itself; it is its transpose.

The result is also directly used in the formula for the inverse:

\[ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A), \qquad |A|\neq0. \]

Since \(|A|=27\neq0\), this matrix is invertible and

\[ A^{-1} = \frac{1}{27} \begin{bmatrix} 3 & 1 & -11\\ -12 & 5 & -1\\ 6 & 2 & 5 \end{bmatrix}. \]

Significance for Competitive Entrance Examinations

In competitive examinations, adjoint problems often appear as part of questions involving determinants, inverse matrices, matrix equations, and matrix identities. Speed is important, but accuracy in the cofactor signs is even more important.

The identity

\[ A\operatorname{adj}(A)=|A|I \]

is particularly useful for checking an answer or simplifying expressions involving a matrix and its adjoint.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 8 points

The cofactor of \(a_{ij}\) is \(A_{ij}=(-1)^{i+j}M_{ij}\).
The cofactor signs follow the pattern \(+\), \(−\), \(+\); \(−\), \(+\), \(−\); \(+\), \(−\), \(+\).
First calculate all nine cofactors carefully before forming the adjoint.
The cofactor matrix is \(C=\begin{bmatrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{bmatrix}\).
The adjoint is the transpose of the cofactor matrix: \(\operatorname{adj}(A)=C^T\).
Do not confuse the cofactor matrix with the adjoint matrix.
The identity \(A\operatorname{adj}(A)=|A|I\) provides an effective verification method.
Here, \(|A|=27\), so \(A\) is non-singular and its inverse exists.

📘 Concept & Theory

Concept/Theory

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For every square matrix \(A\), the fundamental adjoint identity is

\[ A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I, \]

where \(|A|\) is the determinant of \(A\) and \(I\) is the identity matrix of the same order as \(A\).

For a \(2\times2\) matrix

\[ A= \begin{bmatrix} a & b\\ c & d \end{bmatrix}, \]

the adjoint is

\[ \operatorname{adj}(A)= \begin{bmatrix} d & -b\\ -c & a \end{bmatrix}. \]

The verification requires three calculations:

Calculate \(A(\operatorname{adj}A)\).
Calculate \((\operatorname{adj}A)A\).
Calculate \(|A|I\).

If all three results are identical, the required identity is verified.

🗺️ Solution Roadmap

Step-by-step Plan

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Write the given matrix \(A\).
Find \(\operatorname{adj}(A)\) using the \(2\times2\) adjoint formula.
Multiply \(A\) by \(\operatorname{adj}(A)\).
Multiply \(\operatorname{adj}(A)\) by \(A\) in the reverse order.
Calculate the determinant \(|A|\).
Multiply \(|A|\) by the identity matrix \(I\).
Compare the three resulting matrices.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 18 steps

Given —
\[A=\begin{bmatrix}2 & 3\\-4 & -6\end{bmatrix}\]
Find \(\operatorname{adj}(A)\)
For a \(2\times2\) matrix
\[ A= \begin{bmatrix} a & b\\ c & d \end{bmatrix}, \]
we have
\[ \operatorname{adj}(A)= \begin{bmatrix} d & -b\\ -c & a \end{bmatrix}. \]
Here,
\[a=2,\quad b=3,\quad c=-4,\quad d=-6\]
Therefore,
\[ \operatorname{adj}(A) = \begin{bmatrix} -6 & -3\\ 4 & 2 \end{bmatrix} \]
Calculate \(A(\operatorname{adj}A)\)
We have
\[ A(\operatorname{adj}A) = \begin{bmatrix} 2 & 3\\ -4 & -6 \end{bmatrix} \begin{bmatrix} -6 & -3\\ 4 & 2 \end{bmatrix} \]
Multiplying the matrices entry by entry,
\[ \begin{aligned} A(\operatorname{adj}A) &= \begin{bmatrix} (2)(-6)+(3)(4) & (2)(-3)+(3)(2)\\[8pt] (-4)(-6)+(-6)(4) & (-4)(-3)+(-6)(2) \end{bmatrix}\\[8pt] &= \begin{bmatrix} -12+12 & -6+6\\[8pt] 24-24 & 12-12 \end{bmatrix}\\[8pt] &= \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix} \end{aligned} \]
Thus,
\[ \boxed{ A(\operatorname{adj}A) = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}} \]
Calculate \((\operatorname{adj}A)A\)
Now reverse the order of multiplication:
\[ (\operatorname{adj}A)A = \begin{bmatrix} -6 & -3\\\ 4 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3\\\ -4 & -6 \end{bmatrix}. \]
Multiplying the matrices entry by entry,
\[ \begin{aligned} (\operatorname{adj}A)A &= \begin{bmatrix} (-6)(2)+(-3)(-4) & (-6)(3)+(-3)(-6)\\[8pt] (4)(2)+(2)(-4) & (4)(3)+(2)(-6) \end{bmatrix}\\[8pt] &= \begin{bmatrix} -12+12 & -18+18\\[8pt] 8-8 & 12-12 \end{bmatrix}\\[8pt] &= \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix} \end{aligned} \]
Hence,
\[ \boxed{\bbox[5pt]{ (\operatorname{adj}A)A = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}}} \]
Calculate \(|A|I\)
The determinant of \(A\) is
\[ |A| = \begin{vmatrix} 2 & 3\\ -4 & -6 \end{vmatrix} \]
For a \(2\times2\) determinant,
\[ \begin{vmatrix} a & b\\ c & d \end{vmatrix} =ad-bc \]
Therefore,
\[ \begin{aligned} |A| &=(2)(-6)-(3)(-4)\\ &=-12+12\\ &=0 \end{aligned} \]
Thus,
\[|A|=0\]
The identity matrix of order \(2\) is
\[ I= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \]
Hence,
\[ \begin{aligned} |A|I &= 0 \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\\ &= \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}. \end{aligned} \]
Compare the Results
\[ A(\operatorname{adj}A) = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}, \]

\[ (\operatorname{adj}A)A = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}, \]
and
\[ |A|I = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix} \]
Therefore,
\[ \boxed{ A(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|I = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}} \]
Verified

🎯 Exam Significance

Exam Significance

›

This question tests one of the most important identities of the Determinants chapter. Students should be able to calculate the adjoint, perform matrix multiplication accurately, evaluate the determinant, and compare the resulting matrices.

A frequent mistake is to write only \(A(\operatorname{adj}A)\) and assume that the reverse product has the same value without verification. Although the identity guarantees equality, writing both products explicitly is important in a verification question.

This problem also reinforces the distinction between a singular matrix and a non-singular matrix. Since \(|A|=0\), the given matrix is singular and its inverse does not exist.

Significance for Competitive Entrance Examinations

The identity

\[ A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I \]

is frequently used to simplify matrix expressions without carrying out lengthy matrix multiplication. It is especially useful in problems involving determinants, adjoints, inverse matrices, matrix equations, and singular matrices.

For example, if \(|A|=0\), the identity immediately gives

\[ A(\operatorname{adj}A)=0. \]

This can save significant time in objective and numerical-answer questions.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 7 points

For every square matrix \(A\), \(A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I\).
The identity remains valid even when \(A\) is singular.
For this matrix, \(|A|=0\), so \(A\) is singular.
Since \(|A|=0\), both \(A(\operatorname{adj}A)\) and \((\operatorname{adj}A)A\) are zero matrices.
The order of multiplication should be kept explicit while verifying the identity.
A singular matrix has no inverse because \(|A|=0\).
The adjoint identity is a powerful shortcut in competitive examination problems.

📘 Concept & Theory

Concept/Theory

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For every square matrix \(A\), the fundamental adjoint identity is

\[ A(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|I. \]

Here, \(|A|\) denotes the determinant of \(A\), and \(I\) denotes the identity matrix of the same order as \(A\).

To verify the identity for a \(3\times3\) matrix, we need to:

Find all nine cofactors of \(A\).
Form the cofactor matrix.
Transpose the cofactor matrix to obtain \(\operatorname{adj}(A)\).
Calculate \(A(\operatorname{adj}A)\).
Calculate \((\operatorname{adj}A)A\).
Calculate \(|A|I\).
Show that all three matrices are equal.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given matrix \(A\).
Calculate each minor and corresponding cofactor.
Construct the cofactor matrix \(C\).
Transpose \(C\) to obtain \(\operatorname{adj}(A)\).
Multiply \(A\) by \(\operatorname{adj}(A)\).
Multiply \(\operatorname{adj}(A)\) by \(A\).
Find \(|A|\) and calculate \(|A|I\).
Compare the three results.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 51 steps

Given —
\[ A= \begin{bmatrix} 1 & -1 & 2\\\ 3 & 0 & -2\\ 1 & 0 & 3 \end{bmatrix}\]
Find the Cofactors of \(A\)
The cofactor \(A_{ij}\) is given by
\[A_{ij}=(-1)^{i+j}M_{ij}\]
where \(M_{ij}\) is the minor corresponding to the element in the \(i\)-th row and \(j\)-th column.
The signs of the cofactors are
\[\begin{bmatrix} + & - & +\\ - & + & -\\ + & - & + \end{bmatrix}\]
First Row of Cofactors
For \(A_{11}\),
\[ \begin{aligned} A_{11} &= (-1)^{1+1} \begin{vmatrix} 0 & -2\\ 0 & 3 \end{vmatrix}\\ &= 1\left[(0)(3)-(-2)(0)\right]\\ &=0 \end{aligned} \]
For \(A_{12}\),
\[ \begin{aligned} A_{12} &= (-1)^{1+2} \begin{vmatrix} 3 & -2\\ 1 & 3 \end{vmatrix}\\ &= -1\left[(3)(3)-(-2)(1)\right]\\ &= -1(9+2)\\ &=-11 \end{aligned} \]
For \(A_{13}\),
\[ \begin{aligned} A_{13} &= (-1)^{1+3} \begin{vmatrix} 3 & 0\\ 1 & 0 \end{vmatrix}\\ &= 1\left[(3)(0)-(0)(1)\right]\\ &=0 \end{aligned} \]
Therefore, the first row of the cofactor matrix is
\[ \begin{bmatrix} A_{11} & A_{12} & A_{13} \end{bmatrix} = \begin{bmatrix} 0 & -11 & 0 \end{bmatrix}. \]
Second Row of Cofactors
For \(A_{21}\),
\[ \begin{aligned} A_{21} &= (-1)^{2+1} \begin{vmatrix} -1 & 2\\ 0 & 3 \end{vmatrix}\\ &= -1\left[(-1)(3)-(2)(0)\right]\\ &= -1(-3)\\ &=3 \end{aligned} \]
For \(A_{22}\),
\[ \begin{aligned} A_{22} &= (-1)^{2+2} \begin{vmatrix} 1 & 2\\ 1 & 3 \end{vmatrix}\\ &= 1\left[(1)(3)-(2)(1)\right]\\ &=3-2\\ &=1 \end{aligned} \]
For \(A_{23}\),
\[ \begin{aligned} A_{23} &= (-1)^{2+3} \begin{vmatrix} 1 & -1\\ 1 & 0 \end{vmatrix}\\ &= -1\left[(1)(0)-(-1)(1)\right]\\ &= -1(1)\\ &=-1 \end{aligned} \]
Important: The value of \(A_{23}\) is \(-1\), not \(+1\). The negative sign comes from the cofactor sign \((-1)^{2+3}=-1\).
Therefore, the second row of the cofactor matrix is
\[ \begin{bmatrix} A_{21} & A_{22} & A_{23} \end{bmatrix} = \begin{bmatrix} 3 & 1 & -1 \end{bmatrix}. \]
Third Row of Cofactors
For \(A_{31}\),
\[ \begin{aligned} A_{31} &= (-1)^{3+1} \begin{vmatrix} -1 & 2\\ 0 & -2 \end{vmatrix}\\ &= 1\left[(-1)(-2)-(2)(0)\right]\\ &=2. \end{aligned} \]
For \(A_{32}\),
\[ \begin{aligned} A_{32} &= (-1)^{3+2} \begin{vmatrix} 1 & 2\\ 3 & -2 \end{vmatrix}\\ &= -1\left[(1)(-2)-(2)(3)\right]\\ &= -1(-2-6)\\ &=8 \end{aligned} \]
For \(A_{33}\),
\[ \begin{aligned} A_{33} &= (-1)^{3+3} \begin{vmatrix} 1 & -1\\ 3 & 0 \end{vmatrix}\\ &= 1\left[(1)(0)-(-1)(3)\right]\\ &=3 \end{aligned} \]
Therefore, the third row of the cofactor matrix is
\[ \begin{bmatrix} A_{31} & A_{32} & A_{33} \end{bmatrix} = \begin{bmatrix} 2 & 8 & 3 \end{bmatrix} \]
Form the Cofactor Matrix
Collecting all the cofactors, we obtain
\[ C= \begin{bmatrix} A_{11} & A_{12} & A_{13}\\ A_{21} & A_{22} & A_{23}\\ A_{31} & A_{32} & A_{33} \end{bmatrix} = \begin{bmatrix} 0 & -11 & 0\\ 3 & 1 & -1\\ 2 & 8 & 3 \end{bmatrix} \]
Calculate \(\operatorname{adj}(A)\)
The adjoint is the transpose of the cofactor matrix:
\[\operatorname{adj}(A)=C^T\]
Therefore,
\[ \begin{aligned} \operatorname{adj}(A) &= \begin{bmatrix} 0 & 3 & 2\\ -11 & 1 & 8\\ 0 & -1 & 3 \end{bmatrix} \end{aligned} \]
Hence,
\[ \boxed{\bbox[5pt]{ \operatorname{adj}(A)= \begin{bmatrix} 0 & 3 & 2\\ -11 & 1 & 8\\ 0 & -1 & 3 \end{bmatrix}}} \]
Calculate \(A(\operatorname{adj}A)\)
We have
\[ A(\operatorname{adj}A) = \begin{bmatrix} 1 & -1 & 2\\ 3 & 0 & -2\\ 1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 0 & 3 & 2\\ -11 & 1 & 8\\ 0 & -1 & 3 \end{bmatrix} \]
Now calculate each entry of the product.
First row, first column:
\[(1)(0)+(-1)(-11)+(2)(0)=0+11+0=11\]
First row, second column:
\[(1)(3)+(-1)(1)+(2)(-1)=3-1-2=0\]
First row, third column:
\[(1)(2)+(-1)(8)+(2)(3)=2-8+6=0\]
Second row, first column:
\[(3)(0)+(0)(-11)+(-2)(0)=0\]
Second row, second column:
\[(3)(3)+(0)(1)+(-2)(-1)=9+0+2=11\]
Second row, third column:
\[(3)(2)+(0)(8)+(-2)(3)=6+0-6=0\]
Third row, first column:
\[(1)(0)+(0)(-11)+(3)(0)=0\]
Third row, second column:
\[(1)(3)+(0)(1)+(3)(-1)=3+0-3=0\]
Third row, third column:
\[(1)(2)+(0)(8)+(3)(3)=2+0+9=11\]
Therefore,
\[ \begin{aligned} A(\operatorname{adj}A) &= \begin{bmatrix} 11 & 0 & 0\\ 0 & 11 & 0\\ 0 & 0 & 11 \end{bmatrix}\\ &=11I \end{aligned} \]
Thus,
\[\boxed{A(\operatorname{adj}A)=11I}\]
Calculate \((\operatorname{adj}A)A\)
Now reverse the order of multiplication:
\[ (\operatorname{adj}A)A = \begin{bmatrix} 0 & 3 & 2\\ -11 & 1 & 8\\ 0 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2\\ 3 & 0 & -2\\ 1 & 0 & 3 \end{bmatrix} \]
First row, first column:
\[(0)(1)+(3)(3)+(2)(1)=0+9+2=11\]
First row, second column:
\[(0)(-1)+(3)(0)+(2)(0)=0\]
First row, third column:
\[(0)(2)+(3)(-2)+(2)(3)=0-6+6=0\]
Second row, first column:
\[(-11)(1)+(1)(3)+(8)(1)=-11+3+8=0\]
Second row, second column:
\[(-11)(-1)+(1)(0)+(8)(0)=11\]
Second row, third column:
\[(-11)(2)+(1)(-2)+(8)(3)=-22-2+24=0\]
Third row, first column:
\[(0)(1)+(-1)(3)+(3)(1)=0-3+3=0\]
Third row, second column:
\[(0)(-1)+(-1)(0)+(3)(0)=0\]
Third row, third column:
\[(0)(2)+(-1)(-2)+(3)(3)=0+2+9=11\]
Therefore,
\[ \begin{aligned} (\operatorname{adj}A)A &= \begin{bmatrix} 11 & 0 & 0\\ 0 & 11 & 0\\ 0 & 0 & 11 \end{bmatrix}\\ &=11I \end{aligned} \]
Thus,
\[\boxed{(\operatorname{adj}A)A=11I}\]
Calculate \(|A|I\)
The determinant of \(A\) is
\[ |A|= \begin{vmatrix} 1 & -1 & 2\\ 3 & 0 & -2\\ 1 & 0 & 3 \end{vmatrix} \]
Expanding along the first row,
\[ \begin{aligned} |A| &= 1 \begin{vmatrix} 0 & -2\\ 0 & 3 \end{vmatrix} -(-1) \begin{vmatrix} 3 & -2\\ 1 & 3 \end{vmatrix} +2 \begin{vmatrix} 3 & 0\\ 1 & 0 \end{vmatrix} \end{aligned} \]
Evaluating the three \(2\times2\) determinants,
\[ \begin{aligned} |A| &= 1[(0)(3)-(-2)(0)]\\ &\quad +1[(3)(3)-(-2)(1)]\\ &\quad +2[(3)(0)-(0)(1)]\\ &=0+(9+2)+2(0)\\ &=11 \end{aligned} \]
Hence,
\[|A|=11\]
The identity matrix of order \(3\) is
\[ I= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \]
Therefore,
\[ \begin{aligned} |A|I &= 11 \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\\ &= \begin{bmatrix} 11 & 0 & 0\\ 0 & 11 & 0\\ 0 & 0 & 11 \end{bmatrix} \end{aligned} \]
Compare the Three Results
We have obtained
\[ A(\operatorname{adj}A) = \begin{bmatrix} 11 & 0 & 0\\ 0 & 11 & 0\\ 0 & 0 & 11 \end{bmatrix} \]

\[ (\operatorname{adj}A)A = \begin{bmatrix} 11 & 0 & 0\\ 0 & 11 & 0\\ 0 & 0 & 11 \end{bmatrix}, \]
and
\[ |A|I = \begin{bmatrix} 11 & 0 & 0\\ 0 & 11 & 0\\ 0 & 0 & 11 \end{bmatrix} \]
Therefore,
\[ \boxed{\bbox[5pt]{ A(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|I = 11I }} \]
Hence, Verified

🎯 Exam Significance

Exam Significance

›

This question combines several important skills from the Determinants chapter: calculation of minors, cofactors, formation of the cofactor matrix, transposition to obtain the adjoint, matrix multiplication, and evaluation of a determinant.

The most important point is that the cofactor matrix must be transposed to obtain the adjoint. Another common error is incorrect handling of the cofactor sign. In this question, for example,

\[ A_{23}=-1, \]

because \((-1)^{2+3}=-1\).

Since

\[ |A|=11\neq0, \]

the matrix is non-singular and its inverse exists. The adjoint obtained here can therefore also be used to find the inverse:

\[ A^{-1} = \frac{1}{|A|}\operatorname{adj}(A) = \frac{1}{11} \begin{bmatrix} 0 & 3 & 2\\ -11 & 1 & 8\\ 0 & -1 & 3 \end{bmatrix} \]

Significance for Competitive Entrance Examinations

The identity

\[ A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I \]

is a powerful shortcut in competitive examinations. Once \(|A|\) is known, a potentially lengthy matrix product can often be replaced immediately by \(|A|I\).

For this question,

\[ |A|=11 \]

immediately implies

\[ A(\operatorname{adj}A)=(\operatorname{adj}A)A=11I. \]

This property is particularly useful in questions involving matrix identities, inverse matrices, determinants, adjoints, and matrix equations.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

The adjoint of a matrix is the transpose of its cofactor matrix.
For every square matrix, \(A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I\).
The cofactor sign pattern for a \(3\times3\) matrix is \(+\), \(−\), \(+\); \(−\), \(+\), \(−\); \(+\), \(−\), \(+\).
For this matrix, \(A_{23}=-1\), so the corresponding entry of \(\operatorname{adj}(A)\) is \(A_{32}=-1\).
The correct adjoint is \(\operatorname{adj}(A)=\begin{bmatrix}0&3&2\-11&1&8\0&-1&3\end{bmatrix}\).
The determinant is \(|A|=11\), so \(A\) is non-singular.
Both matrix products are equal to \(11I\), verifying the identity.
The result immediately gives \(A^{-1}=\frac{1}{11}\operatorname{adj}(A)\).

📘 Concept & Theory

Concept/Theory

›

The inverse of a square matrix \(A\) exists only when its determinant is non-zero. For a non-singular square matrix,

\[ A^{-1} = \frac{1}{|A|}\operatorname{adj}(A), \qquad |A|\neq0. \]

For a \(2\times2\) matrix

\[ A= \begin{bmatrix} a & b\\ c & d \end{bmatrix} \]

the adjoint is

\[ \operatorname{adj}(A)= \begin{bmatrix} d & -b\\ -c & a \end{bmatrix} \]

Thus, to find the inverse, we need to calculate two quantities:

The determinant \(|A|\).
The adjoint \(\operatorname{adj}(A)\).

If \(|A|\neq0\), substitute these values into

\[ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A). \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given matrix \(A\).
Calculate \(\operatorname{adj}(A)\).
Calculate the determinant \(|A|\).
Check that \(|A|\neq0\), ensuring that \(A^{-1}\) exists.
Apply the formula \(A^{-1}=\frac{1}{|A|}\operatorname{adj}(A)\).
Write the inverse matrix in simplified form.
Optionally verify the result using \(AA^{-1}=I\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 19 steps

Given —
\[ A= \begin{bmatrix} 2 & -2\\ 4 & 3 \end{bmatrix} \]
Find \(\operatorname{adj}(A)\)
For a \(2\times2\) matrix
\[ A= \begin{bmatrix} a & b\\ c & d \end{bmatrix} \]
we have
\[ \operatorname{adj}(A)= \begin{bmatrix} d & -b\\ -c & a \end{bmatrix} \]
Comparing with the given matrix,
\[a=2,\quad b=-2,\quad c=4,\quad d=3\]
Therefore,
\[ \begin{aligned} \operatorname{adj}(A) &= \begin{bmatrix} 3 & -(-2)\\ -4 & 2 \end{bmatrix}\\ &= \begin{bmatrix} 3 & 2\\ -4 & 2 \end{bmatrix} \end{aligned} \]
Hence,
\[ \boxed{\bbox[5pt]{ \operatorname{adj}(A)= \begin{bmatrix} 3 & 2\\ -4 & 2 \end{bmatrix}}} \]
Calculate the Determinant \(|A|\)
For a \(2\times2\) matrix,
\[ \begin{vmatrix} a & b\\ c & d \end{vmatrix} =ad-bc \]
Therefore,
\[ \begin{aligned} |A| &= \begin{vmatrix} 2 & -2\\ 4 & 3 \end{vmatrix}\\ &=(2)(3)-(-2)(4)\\ &=6+8\\ &=14 \end{aligned} \]
Thus,
\[|A|=14\neq0\]
Therefore, \(A\) is a non-singular matrix, and its inverse exists.
Apply the Inverse Formula
The inverse of a non-singular square matrix is given by
\[ A^{-1} = \frac{1}{|A|}\operatorname{adj}(A) \]
Substituting \(|A|=14\) and the adjoint obtained above,
\[ \begin{aligned} A^{-1} &= \frac{1}{14} \begin{bmatrix} 3 & 2\\ -4 & 2 \end{bmatrix} \end{aligned} \]
Hence,
\[ \boxed{\bbox[5pt]{ A^{-1} = \frac{1}{14} \begin{bmatrix} 3 & 2\\ -4 & 2 \end{bmatrix}}} \]
Equivalently, each entry may be divided by \(14\):
\[ \boxed{\bbox[5pt]{ A^{-1} = \begin{bmatrix} \frac{3}{14} & \frac{1}{7}\\ -\frac{2}{7} & \frac{1}{7} \end{bmatrix}}} \]
Verification of the Inverse
A matrix \(A^{-1}\) is the inverse of \(A\) if
\[AA^{-1}=A^{-1}A=I\]
We verify \(AA^{-1}=I\)
\[ \begin{aligned} AA^{-1} &= \begin{bmatrix} 2 & -2\\ 4 & 3 \end{bmatrix} \frac{1}{14} \begin{bmatrix} 3 & 2\\ -4 & 2 \end{bmatrix}\\ &= \frac{1}{14} \begin{bmatrix} (2)(3)+(-2)(-4) & (2)(2)+(-2)(2)\\ (4)(3)+(3)(-4) & (4)(2)+(3)(2) \end{bmatrix}\\ &= \frac{1}{14} \begin{bmatrix} 6+8 & 4-4\\ 12-12 & 8+6 \end{bmatrix}\\ &= \frac{1}{14} \begin{bmatrix} 14 & 0\\ 0 & 14 \end{bmatrix}\\ &= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\\ &=I \end{aligned} \]
Similarly, the adjoint identity gives
\[ (\operatorname{adj}A)A=|A|I=14I. \]
Dividing both sides by \(14\) gives
\[ A^{-1}A=I. \]
Hence, the calculated matrix is indeed the inverse of \(A\).

🎯 Exam Significance

Exam Significance

›

Finding the inverse of a \(2\times2\) matrix is a fundamental application of the concepts of determinants and adjoints. In board examinations, students are expected to know the formula

\[ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A), \qquad |A|\neq0. \]

The most important preliminary check is that the determinant must not be zero. If

\[ |A|=0, \]

the matrix is singular and its inverse does not exist.

In this question,

\[ |A|=14\neq0, \]

so the inverse exists.

Students should also take particular care with the negative entry \(b=-2\). When forming the adjoint, the upper-right entry becomes

\[ -b=-(-2)=2. \]

Such sign errors are common in examination calculations.

Significance for Competitive Entrance Examinations

For competitive entrance examinations, the \(2\times2\) inverse formula provides a very fast method. Once the determinant is found, the inverse can be written almost immediately.

The result is also useful in matrix equations. For example, if

\[ AX=B, \]

and \(A\) is non-singular, then

\[ X=A^{-1}B. \]

Therefore, fluency with adjoints and inverses can substantially reduce the time required for problems involving matrix equations, determinants, and linear systems.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

The inverse of a square matrix exists only when its determinant is non-zero.
The fundamental formula is \(A^{-1}=\frac{1}{|A|}\operatorname{adj}(A)\).
For \(A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\), \(\operatorname{adj}(A)=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\).
For the given matrix, \(\operatorname{adj}(A)=\begin{bmatrix}3&2\\-4&2\end{bmatrix}\).
The determinant is \(|A|=14\neq0\), so the matrix is non-singular.
The inverse is \(A^{-1}=\frac{1}{14}\begin{bmatrix}3&2\\-4&2\end{bmatrix}\).
The inverse can be verified by checking \(AA^{-1}=A^{-1}A=I\).
Careful handling of negative signs is essential when finding the adjoint.

📘 Concept & Theory

Concept/Theory

›

The inverse of a square matrix \(A\) exists only if its determinant is non-zero. For a non-singular square matrix,

\[ A^{-1} = \frac{1}{|A|}\operatorname{adj}(A), \qquad |A|\neq0. \]

For a \(2\times2\) matrix

\[ A= \begin{bmatrix} a & b\ c & d \end{bmatrix}, \]

the adjoint is

\[ \operatorname{adj}(A)= \begin{bmatrix} d & -b\\ -c & a \end{bmatrix} \]

Thus, finding an inverse by the adjoint method involves finding the determinant and adjoint, followed by substitution in the inverse formula.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given matrix \(A\).
Find \(\operatorname{adj}(A)\).
Calculate \(|A|\).
Verify that \(|A|\neq0\).
Use \(A^{-1}=\frac{1}{|A|}\operatorname{adj}(A)\).
Write the inverse in simplified form.
Verify the answer by multiplying \(A\) and \(A^{-1}\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 10 steps

Given — Matrix
\[ A= \begin{bmatrix} -1 & 5\\ -3 & 2 \end{bmatrix} \]
Find \(\operatorname{adj}(A)\)
For
\[ A= \begin{bmatrix} a & b\\ c & d \end{bmatrix}, \]
the adjoint is
\[ \operatorname{adj}(A)= \begin{bmatrix} d & -b\\ -c & a \end{bmatrix} \]
Here,
\[a=-1,\quad b=5,\quad c=-3,\quad d=2\]
Therefore,
\[ \begin{aligned} \operatorname{adj}(A) &= \begin{bmatrix} 2 & -5\\ -(-3) & -1 \end{bmatrix}\\ &= \begin{bmatrix} 2 & -5\\ 3 & -1 \end{bmatrix} \end{aligned} \]
Hence,
\[ \boxed{\bbox[5pt]{ \operatorname{adj}(A)= \begin{bmatrix} 2 & -5\\ 3 & -1 \end{bmatrix}}} \]
Calculate the Determinant \(|A|\)
For a \(2\times2\) matrix,
\[ \begin{vmatrix} a & b\\ c & d \end{vmatrix} =ad-bc. \]
Therefore,
\[ \begin{aligned} |A| &= \begin{vmatrix} -1 & 5\\ -3 & 2 \end{vmatrix}\\ &=(-1)(2)-(5)(-3)\\ &=-2+15\\ &=13 \end{aligned} \]
Thus, |A|=13\neq0
Hence, \(A\) is a non-singular matrix, and its inverse exists.
Apply the Inverse Formula
The inverse of a non-singular matrix is given by
\[ A^{-1} = \frac{1}{|A|}\operatorname{adj}(A). \]
Substituting \(|A|=13\) and the adjoint,
\[ \begin{aligned} A^{-1} &= \frac{1}{13} \begin{bmatrix} 2 & -5\\ 3 & -1 \end{bmatrix} \end{aligned} \]
Therefore,
\[ \boxed{\bbox[5pt]{ A^{-1} = \frac{1}{13} \begin{bmatrix} 2 & -5\\ 3 & -1 \end{bmatrix}}} \]
Equivalently,
\[ \boxed{\bbox[5pt]{ A^{-1} = \begin{bmatrix} \frac{2}{13} & -\frac{5}{13}\\ \frac{3}{13} & -\frac{1}{13} \end{bmatrix}}} \]

🎯 Exam Significance

Exam Significance

›

This question illustrates the standard adjoint method for finding the inverse of a \(2\times2\) matrix. It is important to show the determinant calculation because the existence of the inverse depends on the condition

\[ |A|\neq0. \]

In this problem,

\[ |A|=13\neq0, \]

so \(A^{-1}\) exists.

Students should pay particular attention to the signs while forming the adjoint. For example, since the lower-left element of \(A\) is \(-3\), the corresponding entry in the adjoint is

\[ -(-3)=3. \]

A complete board-examination solution should clearly show the determinant, adjoint, inverse formula, and final matrix.

Significance for Competitive Entrance Examinations

The \(2\times2\) inverse formula is particularly useful in time-bound examinations because it allows the inverse to be obtained quickly:

\[ \begin{bmatrix} a & b\\ c & d \end{bmatrix}^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b\\ -c & a \end{bmatrix}, \qquad ad-bc\neq0. \]

Such inverses frequently occur in matrix equations and systems of linear equations. Recognising the direct \(2\times2\) formula can save valuable time in objective and numerical-answer questions.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

The inverse of a matrix exists only when its determinant is non-zero.
For a \(2\times2\) matrix, \(\operatorname{adj}(A)=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\).
For the given matrix, \(\operatorname{adj}(A)=\begin{bmatrix}2&-5\\3&-1\end{bmatrix}\).
The determinant is \(|A|=13\neq0\), so \(A\) is non-singular.
The inverse is \(A^{-1}=\frac{1}{13}\begin{bmatrix}2&-5\\3&-1\end{bmatrix}\).
The inverse can be verified using \(AA^{-1}=A^{-1}A=I\).
Careful handling of negative signs is essential when calculating the adjoint.
The inverse formula is a useful shortcut for competitive examination problems involving \(2\times2\) matrices.

📘 Concept & Theory

Concept/Theory

›

For a non-singular square matrix \(A\), the inverse is given by

\[ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A), \qquad |A|\neq0. \]

To find the adjoint of a \(3\times3\) matrix, we first calculate all its cofactors. If \(A_{ij}\) denotes the cofactor corresponding to the element in the \(i\)-th row and \(j\)-th column, then

\[ A_{ij}=(-1)^{i+j}M_{ij}, \]

where \(M_{ij}\) is the corresponding minor.

The cofactor sign pattern is

\[ \begin{bmatrix} +&-&+\\ -&+&-\\ +&-&+ \end{bmatrix} \]

After obtaining the cofactor matrix, its transpose gives the adjoint:

\[ \operatorname{adj}(A)=(\text{cofactor matrix})^T. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given matrix \(A\).
Calculate all nine cofactors.
Form the cofactor matrix.
Transpose the cofactor matrix to obtain \(\operatorname{adj}(A)\).
Calculate \(|A|\).
Check that \(|A|\neq0\).
Use \(A^{-1}=\frac{1}{|A|}\operatorname{adj}(A)\).
Verify the result by checking \(AA^{-1}=I\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 31 steps

Given —
\[ A= \begin{bmatrix} 1&2&3\\ 0&2&4\\ 0&0&5 \end{bmatrix} \]
Find the Cofactors
We calculate each cofactor using
\[ A_{ij}=(-1)^{i+j}M_{ij}. \]
Cofactor \(A_{11}\)
Delete the first row and first column:
\[ M_{11}= \begin{vmatrix} 2&4\\ 0&5 \end{vmatrix} =(2)(5)-(4)(0)=10 \]
Therefore,
\[ A_{11}=(-1)^{1+1}M_{11} =(+1)(10)=10 \]
Cofactor \(A_{12}\)
\[ M_{12}= \begin{vmatrix} 0&4\\ 0&5 \end{vmatrix} =(0)(5)-(4)(0)=0 \]
Therefore,
\[A_{12}=(-1)^{1+2}(0)=-0=0\]
Cofactor \(A_{13}\)
\[ M_{13}= \begin{vmatrix} 0&2\\ 0&0 \end{vmatrix} =(0)(0)-(2)(0)=0 \]
Therefore,
\[A_{13}=(-1)^{1+3}(0)=0\]
Cofactor \(A_{21}\)
\[ M_{21}= \begin{vmatrix} 2&3\\ 0&5 \end{vmatrix} =(2)(5)-(3)(0)=10 \]
Since \(1+2=3\), the sign is negative:
\[ A_{21}=(-1)^{2+1}(10) =-10 \]
Cofactor \(A_{22}\)
\[ M_{22}= \begin{vmatrix} 1&3\\ 0&5 \end{vmatrix} =(1)(5)-(3)(0)=5 \]
Therefore,
\[A_{22}=(-1)^{2+2}(5)=5\]
Cofactor \(A_{23}\)
\[ M_{23}= \begin{vmatrix} 1&2\\ 0&0 \end{vmatrix} =(1)(0)-(2)(0)=0 \]
Therefore,
\[A_{23}=(-1)^{2+3}(0)=0\]
Cofactor \(A_{31}\)
\[ M_{31}= \begin{vmatrix} 2&3\\ 2&4 \end{vmatrix} =(2)(4)-(3)(2) =8-6 =2 \]
Therefore,
\[A_{31}=(-1)^{3+1}(2)=2\]
Cofactor \(A_{32}\)
\[ M_{32}= \begin{vmatrix} 1&3\\ 0&4 \end{vmatrix} =(1)(4)-(3)(0) =4 \]
Since \(3+2=5\), the sign is negative:
\[A_{32}=(-1)^{3+2}(4)=-4\]
Cofactor \(A_{33}\)
\[ M_{33}= \begin{vmatrix} 1&2\\ 0&2 \end{vmatrix} =(1)(2)-(2)(0)=2 \]
Therefore,
\[A_{33}=(-1)^{3+3}(2)=2\]
Form the Cofactor Matrix
Collecting all nine cofactors,
\[ \begin{aligned} A_{11}&=10,& A_{12}&=0,& A_{13}&=0,\\ A_{21}&=-10,& A_{22}&=5,& A_{23}&=0,\\ A_{31}&=2,& A_{32}&=-4,& A_{33}&=2 \end{aligned} \]
Hence, the cofactor matrix is
\[ C= \begin{bmatrix} 10&0&0\\ -10&5&0\\ 2&-4&2 \end{bmatrix} \]
Find the Adjoint
The adjoint is the transpose of the cofactor matrix:
\[\operatorname{adj}(A)=C^T\]
Therefore,
\[ \operatorname{adj}(A) = \begin{bmatrix} 10&-10&2\\ 0&5&-4\\ 0&0&2 \end{bmatrix} \]
Thus,
\[ \boxed{\bbox[5pt]{ \operatorname{adj}(A)= \begin{bmatrix} 10&-10&2\\ 0&5&-4\\ 0&0&2 \end{bmatrix}}} \]
Calculate the Determinant \(|A|\)
The given matrix is upper triangular. Therefore, its determinant is simply the product of its diagonal elements:
\[|A|=(1)(2)(5)=10\]
Alternatively, expanding along the first column,
\[ \begin{aligned} |A| &= 1 \begin{vmatrix} 2&4\\ 0&5 \end{vmatrix} -0 \begin{vmatrix} 2&4\\ 0&5 \end{vmatrix} +0 \begin{vmatrix} 2&4\\ 0&0 \end{vmatrix}\\ &=1(10)-0+0\\ &=10 \end{aligned} \]
Hence,
\[|A|=10\neq0\]
Therefore, \(A\) is non-singular and its inverse exists.
Find \(A^{-1}\)
Using
\[ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A), \]
we obtain
\[ \begin{aligned} A^{-1} &= \frac{1}{10} \begin{bmatrix} 10&-10&2\\ 0&5&-4\\ 0&0&2 \end{bmatrix} \end{aligned} \]
Dividing every entry by \(10\),
\[ \boxed{\bbox[5pt]{ A^{-1} = \begin{bmatrix} 1&-1&\frac15\\ 0&\frac12&-\frac25\\ 0&0&\frac15 \end{bmatrix}}} \]

🎯 Exam Significance

Exam Significance

›

This problem tests the complete adjoint method for finding the inverse of a \(3\times3\) matrix. A board-examination solution should clearly demonstrate the calculation of the cofactors, formation of the adjoint, determinant calculation, and application of the inverse formula.

An important point is that the cofactor matrix must be transposed to obtain the adjoint:

\[ \operatorname{adj}(A)=C^T. \]

The determinant can be calculated especially quickly here because \(A\) is upper triangular:

\[ |A|=(1)(2)(5)=10. \]

Since \(10\neq0\), the inverse exists.

Significance for Competitive Entrance Examinations

The upper-triangular structure of the matrix provides a useful shortcut for the determinant. In competitive examinations, recognising triangular matrices immediately can save considerable calculation time.

Also, the identity

\[ A\operatorname{adj}(A)=|A|I \]

provides a rapid verification technique. Here,

\[ A\operatorname{adj}(A)=10I. \]

Therefore,

\[ A^{-1}=\frac{1}{10}\operatorname{adj}(A). \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

The inverse of a square matrix exists only when its determinant is non-zero.
For a \(3\times3\) matrix, cofactors are calculated using \(A_{ij}=(-1)^{i+j}M_{ij}\).
The cofactor sign pattern is \(+\,-\,+;\,-\,+\,-;\,+\,-\,+\).
The adjoint is the transpose of the cofactor matrix.
For the given matrix,
\[ \operatorname{adj}(A)= \begin{bmatrix} 10&-10&2\\ 0&5&-4\\ 0&0&2 \end{bmatrix} \]
Since \(A\) is upper triangular, \(|A|\) is the product of its diagonal elements:
\[ |A|=(1)(2)(5)=10. \]
The inverse is
\[ \boxed{ A^{-1}= \begin{bmatrix} 1&-1&\frac15\\ 0&\frac12&-\frac25\\ 0&0&\frac15 \end{bmatrix}} \]
The result is verified by \(AA^{-1}=I\).
For competitive examinations, identifying triangular matrices can make determinant calculations much faster.

📘 Concept & Theory

Concept/Theory

›

For a non-singular square matrix \(A\), the inverse is given by

\[ A^{-1} = \frac{1}{|A|}\operatorname{adj}(A), \qquad |A|\neq0. \]

To find the adjoint of a \(3\times3\) matrix, we first calculate all its cofactors. The cofactor corresponding to the element in the \(i\)-th row and \(j\)-th column is

\[ A_{ij}=(-1)^{i+j}M_{ij}, \]

where \(M_{ij}\) is the minor obtained after deleting the \(i\)-th row and \(j\)-th column.

The cofactor sign pattern is

\[ \begin{bmatrix} +&-&+\\ -&+&-\\ +&-&+ \end{bmatrix} \]

Once all cofactors are obtained, they are arranged in the cofactor matrix \(C\). The adjoint is then the transpose of this matrix:

\[ \operatorname{adj}(A)=C^T. \]

Since the given matrix is lower triangular, its determinant can also be obtained directly as the product of its diagonal elements.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given matrix \(A\).
Calculate all nine cofactors.
Form the cofactor matrix.
Transpose it to obtain \(\operatorname{adj}(A)\).
Calculate \(|A|\).
Check that \(|A|\neq0\).
Apply \(A^{-1}=\frac{1}{|A|}\operatorname{adj}(A)\).
Verify the result using \(AA^{-1}=I\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 34 steps

Given — Matrix
\[ A= \begin{bmatrix} 1&0&0\\ 3&3&0\\ 5&2&-1 \end{bmatrix} \]
Calculate the Cofactors
We use
\[ A_{ij}=(-1)^{i+j}M_{ij}. \]
Cofactor \(A_{11}\)
Delete the first row and first column:
\[ M_{11} = \begin{vmatrix} 3&0\\ 2&-1 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{11} &=(-1)^{1+1} \begin{vmatrix} 3&0\\ 2&-1 \end{vmatrix}\\ &=1\left[(3)(-1)-(0)(2)\right]\\ &=-3 \end{aligned} \]
Hence,
\[A_{11}=-3\]
Cofactor \(A_{12}\)
Delete the first row and second column:
\[ M_{12} = \begin{vmatrix} 3&0\\ 5&-1 \end{vmatrix} \]
Thus,
\[ \begin{aligned} A_{12} &=(-1)^{1+2} \begin{vmatrix} 3&0\\ 5&-1 \end{vmatrix}\\ &=-\left[(3)(-1)-(0)(5)\right]\\ &=-(-3)\\ &=3 \end{aligned} \]
Hence,
\[A_{12}=3\]
Cofactor \(A_{13}\)
Delete the first row and third column:
\[ M_{13} = \begin{vmatrix} 3&3\\ 5&2 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{13} &=(-1)^{1+3} \begin{vmatrix} 3&3\\ 5&2 \end{vmatrix}\\ &=1\left[(3)(2)-(3)(5)\right]\\ &=6-15\\ &=-9 \end{aligned} \]
Hence,
\[A_{13}=-9\]
Cofactor \(A_{21}\)
Delete the second row and first column:
\[ M_{21} = \begin{vmatrix} 0&0\\ 2&-1 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{21} &=(-1)^{2+1} \begin{vmatrix} 0&0\\ 2&-1 \end{vmatrix}\\ &=-\left[(0)(-1)-(0)(2)\right]\\ &=0 \end{aligned} \]
Cofactor \(A_{22}\)
Delete the second row and second column:
\[ M_{22} = \begin{vmatrix} 1&0\\ 5&-1 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{22} &=(-1)^{2+2} \begin{vmatrix} 1&0\\ 5&-1 \end{vmatrix}\\ &=1\left[(1)(-1)-(0)(5)\right]\\ &=-1 \end{aligned} \]
Cofactor \(A_{23}\)
Delete the second row and third column:
\[ M_{23} = \begin{vmatrix} 1&0\\ 5&2 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{23} &=(-1)^{2+3} \begin{vmatrix} 1&0\\ 5&2 \end{vmatrix}\\ &=-\left[(1)(2)-(0)(5)\right]\\ &=-2 \end{aligned} \]
Cofactor \(A_{31}\)
Delete the third row and first column:
\[ M_{31} = \begin{vmatrix} 0&0\\ 3&0 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{31} &=(-1)^{3+1} \begin{vmatrix} 0&0\\ 3&0 \end{vmatrix}\\ &=0 \end{aligned} \]
Cofactor \(A_{32}\)
Delete the third row and second column:
\[ M_{32} = \begin{vmatrix} 1&0\\ 3&0 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{32} &=(-1)^{3+2} \begin{vmatrix} 1&0\\ 3&0 \end{vmatrix}\\ &=0 \end{aligned} \]
Cofactor \(A_{33}\)
Delete the third row and third column:
\[ M_{33} = \begin{vmatrix} 1&0\\ 3&3 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{33} &=(-1)^{3+3} \begin{vmatrix} 1&0\\ 3&3 \end{vmatrix}\\ &=1\left[(1)(3)-(0)(3)\right]\\ &=3 \end{aligned} \]
Hence,
\[A_{33}=3\]
Form the Cofactor Matrix
Collecting all nine cofactors,
\[ \begin{aligned} A_{11}&=-3,&A_{12}&=3,&A_{13}&=-9,\\ A_{21}&=0,&A_{22}&=-1,&A_{23}&=-2,\\ A_{31}&=0,&A_{32}&=0,&A_{33}&=3 \end{aligned} \]
Thus, the cofactor matrix is
\[ C= \begin{bmatrix} -3&3&-9\\ 0&-1&-2\\ 0&0&3 \end{bmatrix} \]
Find the Adjoint
The adjoint is the transpose of the cofactor matrix:
\[\operatorname{adj}(A)=C^T\]
Therefore,
\[ \operatorname{adj}(A) = \begin{bmatrix} -3&0&0\\ 3&-1&0\\ -9&-2&3 \end{bmatrix} \]
Hence,
\[ \boxed{\bbox[5pt]{ \operatorname{adj}(A)= \begin{bmatrix} -3&0&0\\ 3&-1&0\\ -9&-2&3 \end{bmatrix}}} \]
Calculate the Determinant \(|A|\)
The given matrix is lower triangular. Therefore, its determinant is the product of its diagonal elements:
\[ \begin{aligned} |A| &=(1)(3)(-1)\\ &=-3 \end{aligned} \]
Alternatively, expanding along the first row gives
\[ \begin{aligned} |A| &= 1 \begin{vmatrix} 3&0\\ 2&-1 \end{vmatrix} -0 \begin{vmatrix} 3&0\\ 5&-1 \end{vmatrix} +0 \begin{vmatrix} 3&3\\ 5&2 \end{vmatrix}\\ &=1(-3)-0+0\\ &=-3 \end{aligned} \]
Thus,
\[|A|=-3\neq0\]
Therefore, \(A\) is non-singular and its inverse exists.
Find \(A^{-1}\)
Using
\[ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A), \]
we get [ \begin{aligned} A^{-1} &= \frac{1}{-3} \begin{bmatrix} -3&0&0\\ 3&-1&0\\ -9&-2&3 \end{bmatrix}\\ &= -\frac13 \begin{bmatrix} -3&0&0\\ 3&-1&0\\ -9&-2&3 \end{bmatrix} \end{aligned} \]
Multiplying each entry by \(-\frac13\),
\[ \boxed{\bbox[5pt]{ A^{-1} = \begin{bmatrix} 1&0&0\\ -1&\frac13&0\\ 3&\frac23&-1 \end{bmatrix}}} \]

🎯 Exam Significance

Exam Significance

›

This question tests the complete adjoint method for finding the inverse of a \(3\times3\) matrix. The most important steps are the accurate calculation of all cofactors, correct transposition to obtain the adjoint, and the determinant check before applying the inverse formula.

In particular, students should carefully observe the alternating cofactor signs. For example, \(A_{12}\) has a negative cofactor sign, but its minor is \(-3\), so

\[ A_{12}=(-1)(-3)=3. \]

Missing either of these signs can change the entire inverse. It is also important to distinguish the cofactor matrix from the adjoint; the latter is the transpose of the former.

Significance for Competitive Entrance Examinations

The matrix is lower triangular, so its determinant can be obtained immediately from its diagonal entries:

\[ |A|=(1)(3)(-1)=-3. \]

Recognising triangular matrices is therefore a useful time-saving technique in competitive examinations.

The structure of the resulting inverse is also consistent with the original lower-triangular form. Such structural observations can be useful for checking an answer quickly and detecting computational errors.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

The inverse exists because \(|A|=-3\neq0\).
The cofactor formula is \(A_{ij}=(-1)^{i+j}M_{ij}\).
The cofactor matrix is
\[ C= \begin{bmatrix} -3&3&-9\\ 0&-1&-2\\ 0&0&3 \end{bmatrix} \]
The adjoint is the transpose of the cofactor matrix:
\[ \operatorname{adj}(A)= \begin{bmatrix} -3&0&0\\ 3&-1&0\\ -9&-2&3 \end{bmatrix} \]
Since \(A\) is lower triangular,
\[ |A|=(1)(3)(-1)=-3 \]
The inverse is
\[ \boxed{ A^{-1}= \begin{bmatrix} 1&0&0\\ -1&\frac13&0\\ 3&\frac23&-1 \end{bmatrix}} \]
The answer is verified by \(AA^{-1}=I\).
For competitive examinations, triangular structure provides a quick way to calculate determinants.

📘 Concept & Theory

Concept/Theory

›

For a non-singular square matrix \(A\), the inverse is given by

\[ A^{-1} = \frac{1}{|A|}\operatorname{adj}(A), \qquad |A|\neq0. \]

For a \(3\times3\) matrix, we first calculate all nine cofactors. The cofactor corresponding to the element in the \(i\)-th row and \(j\)-th column is

\[ A_{ij}=(-1)^{i+j}M_{ij}, \]

where \(M_{ij}\) is the corresponding minor.

The signs of the cofactors follow the pattern

\[ \begin{bmatrix} +&-&+\\ -&+&-\\ +&-& \end{bmatrix} \]

After calculating the cofactors, we form the cofactor matrix \(C\). The adjoint is obtained by transposing this matrix:

\[ \operatorname{adj}(A)=C^T. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given matrix \(A\).
Calculate all nine cofactors.
Form the cofactor matrix.
Transpose the cofactor matrix to obtain \(\operatorname{adj}(A)\).
Calculate the determinant \(|A|\).
Check that \(|A|\neq0\).
Apply \(A^{-1}=\frac{1}{|A|}\operatorname{adj}(A)\).
Simplify the resulting matrix.
Verify the inverse by checking \(AA^{-1}=I\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 40 steps

Given — Matrix
\[ A= \begin{bmatrix} 2&1&3\\ 4&-1&0\\ -7&2&1 \end{bmatrix} \]
Calculate the Cofactors
We use the formula
\[ A_{ij}=(-1)^{i+j}M_{ij}. \]
Cofactor \(A_{11}\)
Delete the first row and first column:
\[ M_{11} = \begin{vmatrix} -1&0\\ 2&1 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{11} &=(-1)^{1+1} \begin{vmatrix} -1&0\\ 2&1 \end{vmatrix}\\ &=1\left[(-1)(1)-(0)(2)\right]\\ &=-1. \end{aligned} \]
Hence,
\[A_{11}=-1\]
Cofactor \(A_{12}\)
Delete the first row and second column:
\[ M_{12} = \begin{vmatrix} 4&0\\ -7&1 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{12} &=(-1)^{1+2} \begin{vmatrix} 4&0\\ -7&1 \end{vmatrix}\\ &=-\left[(4)(1)-(0)(-7)\right]\\ &=-4 \end{aligned} \]
Hence,
\[A_{12}=-4\]
Cofactor \(A_{13}\)
Delete the first row and third column:
\[ M_{13} = \begin{vmatrix} 4&-1\\ -7&2 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{13} &=(-1)^{1+3} \begin{vmatrix} 4&-1\\ -7&2 \end{vmatrix}\\ &=(4)(2)-(-1)(-7)\\ &=8-7\\ &=1 \end{aligned} \]
Hence,
\[A_{13}=1\]
Cofactor \(A_{21}\)
Delete the second row and first column:
\[ M_{21} = \begin{vmatrix} 1&3\\ 2&1 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{21} &=(-1)^{2+1} \begin{vmatrix} 1&3\\ 2&1 \end{vmatrix}\\ &=-\left[(1)(1)-(3)(2)\right]\\ &=-\left(1-6\right)\\ &=5 \end{aligned} \]
Hence,
\[A_{21}=5\]
Cofactor \(A_{22}\)
Delete the second row and second column:
\[ M_{22} = \begin{vmatrix} 2&3\\ -7&1 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{22} &=(-1)^{2+2} \begin{vmatrix} 2&3\\ -7&1 \end{vmatrix}\\ &=(2)(1)-(3)(-7)\\ &=2+21\\ &=23 \end{aligned} \]
Hence,
\[A_{22}=23\]
Cofactor \(A_{23}\)
Delete the second row and third column:
\[ M_{23} = \begin{vmatrix} 2&1\\ -7&2 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{23} &=(-1)^{2+3} \begin{vmatrix} 2&1\\ -7&2 \end{vmatrix}\\ &=-\left[(2)(2)-(1)(-7)\right]\\ &=-\left(4+7\right)\\ &=-11 \end{aligned} \]
Hence,
\[A_{23}=-11\]
Cofactor \(A_{31}\)
Delete the third row and first column:
\[ M_{31} = \begin{vmatrix} 4&3\\ 2&1 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{31} &=(-1)^{3+1} \begin{vmatrix} 4&3\\ 2&1 \end{vmatrix}\\ &=(4)(1)-(3)(2)\\ &=4-6\\ &=-2 \end{aligned} \]
Hence,
\[A_{31}=-2\]
Cofactor \(A_{32}\)
Delete the third row and second column:
\[ M_{32} = \begin{vmatrix} 4&3\\ 2&1 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{32} &=(-1)^{3+2} \begin{vmatrix} 4&3\\ 2&1 \end{vmatrix}\\ &=-\left[(4)(1)-(3)(2)\right]\\ &=-\left(4-6\right)\\ &=2 \end{aligned} \]
Hence,
\[A_{32}=2\]
Cofactor \(A_{33}\)
Delete the third row and third column:
\[ M_{33} = \begin{vmatrix} 4&3\\ 2&1 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{33} &=(-1)^{3+3} \begin{vmatrix} 4&3\\ 2&1 \end{vmatrix}\\ &=(4)(1)-(3)(2)\\ &=4-6\\ &=-2 \end{aligned} \]
Hence,
\[A_{33}=-2\]
Form the Cofactor Matrix
Collecting all nine cofactors, we have
\[\begin{aligned} A_{11}&=-1,&A_{12}&=-4,&A_{13}&=1,\\ A_{21}&=5,&A_{22}&=23,&A_{23}&=-11,\\ A_{31}&=3,&A_{32}&=12,&A_{33}&=-6 \end{aligned} \]
Therefore, the cofactor matrix is
\[ C= \begin{bmatrix} -1&-4&1\\ 5&23&-11\\ 3&12&-6 \end{bmatrix} \]
Find the Adjoint
The adjoint is the transpose of the cofactor matrix:
\[\operatorname{adj}(A)=C^T\]
Thus,
\[ \operatorname{adj}(A) = \begin{bmatrix} -1&5&3\\ -4&23&12\\ 1&-11&-6 \end{bmatrix} \]
Hence,
\[ \boxed{ \operatorname{adj}(A)= \begin{bmatrix} -1&5&3\\ -4&23&12\\ 1&-11&-6 \end{bmatrix}} \]
Calculate the Determinant \(|A|\)
We expand the determinant along the first row:
\[ |A| = 2 \begin{vmatrix} -1&0\\ 2&1 \end{vmatrix} - 1 \begin{vmatrix} 4&0\\ -7&1 \end{vmatrix} + 3 \begin{vmatrix} 4&-1\\ -7&2 \end{vmatrix} \]
Calculate each minor separately:
\[ \begin{vmatrix} -1&0\\ 2&1 \end{vmatrix} =(-1)(1)-(0)(2) =-1, \]

\[ \begin{vmatrix} 4&0\\ -7&1 \end{vmatrix} =(4)(1)-(0)(-7) =4, \]
and
\[ \begin{vmatrix} 4&-1\\ -7&2 \end{vmatrix} =(4)(2)-(-1)(-7) =8-7 =1 \]
Therefore,
\[ \begin{aligned} |A| &=2(-1)-1(4)+3(1)\\ &=-2-4+3\\ &=-3 \end{aligned} \]
Thus,
\[|A|=-3\neq0\]
Hence, \(A\) is a non-singular matrix, and its inverse exists.
Find \(A^{-1}\)
Using the inverse formula,
\[ A^{-1} = \frac{1}{|A|}\operatorname{adj}(A). \]
Substituting \(|A|=-3\) and the adjoint,
\[ \begin{aligned} A^{-1} &= \frac{1}{-3} \begin{bmatrix} -1&5&3\\ -4&23&12\\ 1&-11&-6 \end{bmatrix}\\[8pt] &= -\frac13 \begin{bmatrix} -1&5&3\\ -4&23&12\\ 1&-11&-6 \end{bmatrix} \end{aligned} \]
Multiplying each element by \(-\frac13\),
\[ \boxed{\bbox[5pt]{ A^{-1} = \begin{bmatrix} \frac13&-\frac53&-1\\ \frac43&-\frac{23}{3}&-4\\ -\frac13&\frac{11}{3}&2 \end{bmatrix}}} \]

🎯 Exam Significance

Exam Significance

›

This problem tests the complete procedure for finding the inverse of a \(3\times3\) matrix using the adjoint method. The essential stages are the calculation of cofactors, formation of the adjoint, evaluation of the determinant, and application of the inverse formula.

Students should be especially careful with the alternating signs of cofactors. For example, \(A_{12}\) has a negative sign:

\[ A_{12} = -\begin{vmatrix} 4&0\\ -7&1 \end{vmatrix} =-4. \]

Similarly,

\[ A_{21} = -\begin{vmatrix} 1&3\\ 2&1 \end{vmatrix} =5. \]

These sign changes are a common source of errors. It is also important not to confuse the cofactor matrix with the adjoint: the adjoint is the transpose of the cofactor matrix.

Significance for Competitive Entrance Examinations

In competitive examinations, the cofactor and determinant calculations can often be organised systematically using the sign pattern

\[ \begin{bmatrix} +&-&+\\ -&+&-\\ +&-&+ \end{bmatrix}. \]

Once the determinant is known to be non-zero, the inverse can be obtained directly from

\[ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A). \]

The identity

\[ A\operatorname{adj}(A)=|A|I \]

also gives a powerful way of checking calculations. For this matrix,

\[ A\operatorname{adj}(A)=-3I. \]

Therefore,

\[ A^{-1}=-\frac13\operatorname{adj}(A). \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

The inverse exists because \(|A|=-3\neq0\).
Every cofactor must include the correct alternating sign.
The cofactor matrix is
\[ C= \begin{bmatrix} -1&-4&1\\ 5&23&-11\\ 3&12&-6 \end{bmatrix} \]
The adjoint is the transpose of the cofactor matrix:
\[ \operatorname{adj}(A)= \begin{bmatrix} -1&5&3\\ -4&23&12\\ 1&-11&-6 \end{bmatrix} \]
The determinant is
\[ |A|=-3. \]
The inverse is
\[ \boxed{ A^{-1} = \begin{bmatrix} \frac13&-\frac53&-1\\ \frac43&-\frac{23}{3}&-4\\ -\frac13&\frac{11}{3}&2 \end{bmatrix}} \]
The answer is verified by \(AA^{-1}=I\).
The identity \(A\operatorname{adj}(A)=|A|I\) provides an efficient verification method.

📘 Concept & Theory

Concept/Theory

›

For a non-singular square matrix \(A\), the inverse is given by

\[ A^{-1} = \frac{1}{|A|}\operatorname{adj}(A), \qquad |A|\neq0. \]

For a \(3\times3\) matrix, the adjoint is obtained by first calculating all nine cofactors, arranging them into the cofactor matrix, and then taking its transpose.

The cofactor corresponding to the element in the \(i\)-th row and \(j\)-th column is

\[ A_{ij}=(-1)^{i+j}M_{ij}, \]

where \(M_{ij}\) is the corresponding minor.

The cofactor signs follow the pattern

\[ \begin{bmatrix} +&-&+\\ -&+&-\\ +&-&+ \end{bmatrix} \]

After obtaining the cofactor matrix \(C\),

\[ \operatorname{adj}(A)=C^T. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given matrix \(A\).
Calculate all nine cofactors.
Form the cofactor matrix.
Transpose the cofactor matrix to obtain \(\operatorname{adj}(A)\).
Calculate the determinant \(|A|\).
Verify that \(|A|\neq0\).
Apply \(A^{-1}=\frac{1}{|A|}\operatorname{adj}(A)\).
Simplify the inverse.
Verify the result by checking \(AA^{-1}=I\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 30 steps

Given — Matrix
\[ A= \begin{bmatrix} 1&-1&2\\ 0&2&-3\\ 3&-2&4 \end{bmatrix} \]
Calculate the Cofactors
We calculate each cofactor using
\[ A_{ij}=(-1)^{i+j}M_{ij}. \]
Cofactor \(A_{11}\)
Delete the first row and first column:
\[ M_{11} = \begin{vmatrix} 2&-3\\ -2&4 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{11} &=(-1)^{1+1} \begin{vmatrix} 2&-3\\ -2&4 \end{vmatrix}\\ &=(2)(4)-(-3)(-2)\\ &=8-6\\ &=2 \end{aligned} \]
Cofactor \(A_{12}\)
Delete the first row and second column:
\[ M_{12} = \begin{vmatrix} 0&-3\\ 3&4 \end{vmatrix} \]
Since the cofactor sign is negative,
\[ \begin{aligned} A_{12} &=(-1)^{1+2} \begin{vmatrix} 0&-3\\ 3&4 \end{vmatrix}\\ &=-\left[(0)(4)-(-3)(3)\right]\\ &=-\left(0+9\right)\\ &=-9 \end{aligned} \]
Cofactor \(A_{13}\)
Delete the first row and second column:
\[ M_{13} = \begin{vmatrix} 0&2\\ 3&-2 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{13} &=(-1)^{1+3} \begin{vmatrix} 0&2\\ 3&-2 \end{vmatrix}\\ &=(0)(-2)-(2)(3)\\ &=-6 \end{aligned} \]
Cofactor \(A_{21}\)
Delete the second row and first column:
\[ M_{21} = \begin{vmatrix} -1&2\\ -2&4 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{21} &=(-1)^{2+1} \begin{vmatrix} -1&2\\ -2&4 \end{vmatrix}\\ &=-\left[(-1)(4)-(2)(-2)\right]\\ &=-(-4+4)\\ &=0 \end{aligned} \]
Cofactor \(A_{22}\)
Delete the second row and second column:
\[ M_{22} = \begin{vmatrix} 1&2\\ 3&4 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{22} &=(-1)^{2+2} \begin{vmatrix} 1&2\\ 3&4 \end{vmatrix}\\ &=(1)(4)-(2)(3)\\ &=4-6\\ &=-2 \end{aligned} \]
Cofactor \(A_{23}\)
Delete the second row and third column:
\[ M_{23} = \begin{vmatrix} 1&-1\\ 3&-2 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{23} &=(-1)^{2+3} \begin{vmatrix} 1&-1\\ 3&-2 \end{vmatrix}\\ &=-\left[(1)(-2)-(-1)(3)\right]\\ &=-(-2+3)\\ &=-1 \end{aligned} \]
Cofactor \(A_{31}\)
Delete the third row and first column:
\[ M_{31} = \begin{vmatrix} -1&2\\ 2&-3 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{31} &=(-1)^{3+1} \begin{vmatrix} -1&2\\ 2&-3 \end{vmatrix}\\ &=(-1)(-3)-(2)(2)\\ &=3-4\\ &=-1 \end{aligned} \]
Cofactor \(A_{32}\)
Delete the third row and second column:
\[ M_{32} = \begin{vmatrix} 1&2\\ 0&-3 \end{vmatrix} \]
Since the cofactor sign is negative,
\[ \begin{aligned} A_{32} &=(-1)^{3+2} \begin{vmatrix} 1&2\\ 0&-3 \end{vmatrix}\\ &=-\left[(1)(-3)-(2)(0)\right]\\ &=-(-3)\\ &=3 \end{aligned} \]
Cofactor \(A_{33}\)
Delete the third row and second column:
\[ M_{33} = \begin{vmatrix} 1&-1\\ 0&2 \end{vmatrix} \]
Therefore,
\[ \begin{aligned} A_{33} &=(-1)^{3+3} \begin{vmatrix} 1&-1\\ 0&2 \end{vmatrix}\\ &=(1)(2)-(-1)(0)\\ &=2 \end{aligned} \]
Form the Cofactor Matrix
Collecting all nine cofactors,
\[ \begin{aligned} A_{11}&=2,&A_{12}&=-9,&A_{13}&=-6,\\ A_{21}&=0,&A_{22}&=-2,&A_{23}&=-1,\\ A_{31}&=-1,&A_{32}&=3,&A_{33}&=2. \end{aligned} \]
Thus, the cofactor matrix is
\[ C= \begin{bmatrix} 2&-9&-6\\ 0&-2&-1\\ -1&3&2 \end{bmatrix} \]
Find the Adjoint
The adjoint is the transpose of the cofactor matrix:
\[\operatorname{adj}(A)=C^T\]
Therefore,
\[ \operatorname{adj}(A) = \begin{bmatrix} 2&0&-1\\ -9&-2&3\\ -6&-1&2 \end{bmatrix} \]
Hence,
\[ \boxed{\bbox[5pt]{ \operatorname{adj}(A)= \begin{bmatrix} 2&0&-1\\ -9&-2&3\\ -6&-1&2 \end{bmatrix}}} \]
Calculate the Determinant \(|A|\)
Expand the determinant along the first row:
\[ \begin{aligned} |A| &= 1 \begin{vmatrix} 2&-3\\ -2&4 \end{vmatrix} -(-1) \begin{vmatrix} 0&-3\\ 3&4 \end{vmatrix} +2 \begin{vmatrix} 0&2\\ 3&-2 \end{vmatrix} \end{aligned} \]
Now evaluate each minor:
\[ \begin{vmatrix} 2&-3\\ -2&4 \end{vmatrix} =(2)(4)-(-3)(-2) =8-6 =2, \]

\[ \begin{vmatrix} 0&-3\\ 3&4 \end{vmatrix} =(0)(4)-(-3)(3) =9, \]
and
\[ \begin{vmatrix} 0&2\\ 3&-2 \end{vmatrix} =(0)(-2)-(2)(3) =-6 \]
Therefore,
\[ \begin{aligned} |A| &=1(2)-(-1)(9)+2(-6)\\ &=2+9-12\\ &=-1 \end{aligned} \]
Since
\[ |A|=-1\neq0, \]
the matrix \(A\) is non-singular and its inverse exists.
Find \(A^{-1}\)
Using
\[ A^{-1} = \frac{1}{|A|}\operatorname{adj}(A), \]
We obtain
\[ \begin{aligned} A^{-1} &= \frac{1}{-1} \begin{bmatrix} 2&0&-1\\ -9&-2&3\\ -6&-1&2 \end{bmatrix}\\ &= - \begin{bmatrix} 2&0&-1\\ -9&-2&3\\ -6&-1&2 \end{bmatrix} \end{aligned} \]
Therefore,
\[ \boxed{\bbox[5pt]{ A^{-1} = \begin{bmatrix} -2&0&1\ 9&2&-3\ 6&1&-2 \end{bmatrix}}} \]

🎯 Exam Significance

Exam Significance

›

This question tests the complete adjoint method for finding the inverse of a \(3\times3\) matrix. Students should present the cofactors systematically, form the cofactor matrix, transpose it to obtain the adjoint, and then calculate the determinant before applying the inverse formula.

A particularly important lesson from this problem is the need to maintain the correct signs while expanding a determinant. The correct expansion along the first row is

\[ |A| = 1(2)-(-1)(9)+2(-6) = 2+9-12 = -1. \]

Thus, the determinant is \(-1\), not \(-7\). Because the determinant is non-zero, the inverse exists.

Significance for Competitive Entrance Examinations

In competitive examinations, careful sign management is essential because one incorrect sign can change the determinant, the adjoint, and consequently the entire inverse.

The identity

\[ A\operatorname{adj}(A)=|A|I \]

provides an efficient way to verify the cofactor calculations. In this problem,

\[ A\operatorname{adj}(A)=-I. \]

Since \(|A|=-1\), it follows immediately that

\[ A^{-1}=-\operatorname{adj}(A). \]

\] This makes the final calculation particularly simple.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

The inverse formula is
\[ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A), \qquad |A|\neq0. \]
The cofactor sign pattern is
\[ \begin{bmatrix} +&-&+\\ -&+&-\\ +&-&+ \end{bmatrix}. \]
The cofactor matrix is
\[ C= \begin{bmatrix} 2&-9&-6\\ 0&-2&-1\\ -1&3&2 \end{bmatrix}. \]
The adjoint is
\[ \operatorname{adj}(A)= \begin{bmatrix} 2&0&-1\\ -9&-2&3\\ -6&-1&2 \end{bmatrix}. \]
The correct determinant is
\[ |A|=-1. \]
The supplied determinant value \(-7\) is incorrect; the error comes from the sign in the second term of the determinant expansion.
The inverse is
\[ \boxed{ A^{-1} = \begin{bmatrix} -2&0&1\\ 9&2&-3\\ 6&1&-2 \end{bmatrix}}. \]
The result is verified by
\[ AA^{-1}=I. \]
Since \(|A|=-1\), the inverse is simply the negative of the adjoint.

📘 Concept & Theory

Concept/Theory

›

For a square matrix \(A\), the inverse exists if and only if

\[ |A|\neq0. \]

When the determinant is non-zero, the inverse is given by

\[ A^{-1} = \frac{1}{|A|}\operatorname{adj}(A). \]

To find \(\operatorname{adj}(A)\), we first calculate the nine cofactors \(A_{ij}\), form the cofactor matrix, and then transpose it.

The cofactor is defined by

\[ A_{ij}=(-1)^{i+j}M_{ij}, \]

where \(M_{ij}\) is the minor corresponding to the element in the \(i\)-th row and \(j\)-th column.

The cofactor signs are arranged as

\[ \begin{bmatrix} +&-&+\\ -&+&-\\ +&-&+ \end{bmatrix} \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Calculate the determinant \(|A|\).
Calculate all nine cofactors.
Form the cofactor matrix.
Transpose it to obtain \(\operatorname{adj}(A)\).
Use \(A^{-1}=\frac{1}{|A|}\operatorname{adj}(A)\).
Simplify the result.
Verify that the resulting matrix is indeed the inverse.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 21 steps

Calculate the Determinant
We have
\[ A= \begin{bmatrix} 1&0&0\\ 0&\cos\alpha&\sin\alpha\\ 0&\sin\alpha&-\cos\alpha \end{bmatrix} \]
Expanding along the first row,
\[ \begin{aligned} |A| &= 1 \begin{vmatrix} \cos\alpha&\sin\alpha\\ \sin\alpha&-\cos\alpha \end{vmatrix} -0+0\\ &= 1\left[ (\cos\alpha)(-\cos\alpha) - (\sin\alpha)(\sin\alpha) \right]\\ &= -\cos^2\alpha-\sin^2\alpha \end{aligned} \]
Using the identity
\[\sin^2\alpha+\cos^2\alpha=1\]
we get
\[\boxed{|A|=-1}\]
Since
\[ |A|=-1\neq0, \]
the matrix \(A\) is non-singular and its inverse exists.
Calculate the Cofactors
Cofactor \(A_{11}\)
\[ \begin{aligned} A_{11} &= (-1)^{1+1} \begin{vmatrix} \cos\alpha&\sin\alpha\\ \sin\alpha&-\cos\alpha \end{vmatrix}\\ &= (\cos\alpha)(-\cos\alpha) - (\sin\alpha)(\sin\alpha)\\ &= -\cos^2\alpha-\sin^2\alpha\\ &=-1 \end{aligned} \]
Cofactor \(A_{12}\)
\[ \begin{aligned} A_{12} &= (-1)^{1+2} \begin{vmatrix} 0&\sin\alpha\\ 0&-\cos\alpha \end{vmatrix}\\ &= -\left[ (0)(-\cos\alpha) - (\sin\alpha)(0) \right]\\ &=0 \end{aligned} \]
Cofactor \(A_{13}\)
\[ \begin{aligned} A_{13} &= (-1)^{1+3} \begin{vmatrix} 0&\cos\alpha\\ 0&\sin\alpha \end{vmatrix}\\ &= (0)(\sin\alpha) - (\cos\alpha)(0)\\ &=0 \end{aligned} \]
Cofactor \(A_{21}\)
\[ \begin{aligned} A_{21} &= (-1)^{2+1} \begin{vmatrix} 0&0\\ \sin\alpha&-\cos\alpha \end{vmatrix}\\ &= -\left[ (0)(-\cos\alpha)-(0)(\sin\alpha) \right]\\ &=0 \end{aligned} \]
Cofactor \(A_{22}\)
\[ \begin{aligned} A_{22} &= (-1)^{2+2} \begin{vmatrix} 1&0\\ 0&-\cos\alpha \end{vmatrix}\\ &= (1)(-\cos\alpha)-(0)(0)\\ &=-\cos\alpha \end{aligned} \]
Cofactor \(A_{23}\)
\[ \begin{aligned} A_{23} &= (-1)^{2+3} \begin{vmatrix} 1&0\\ 0&\sin\alpha \end{vmatrix}\\ &= -\left[(1)(\sin\alpha)-(0)(0)\right]\\ &=-\sin\alpha \end{aligned} \]
Cofactor \(A_{31}\)
\[ \begin{aligned} A_{31} &= (-1)^{3+1} \begin{vmatrix} 0&0\\ \cos\alpha&\sin\alpha \end{vmatrix}\\ &= (0)(\sin\alpha)-(0)(\cos\alpha)\\ &=0 \end{aligned} \]
Cofactor \(A_{32}\)
\[ \begin{aligned} A_{32} &= (-1)^{3+2} \begin{vmatrix} 1&0\\ 0&\sin\alpha \end{vmatrix}\\ &= -\left[(1)(\sin\alpha)-(0)(0)\right]\\ &=-\sin\alpha \end{aligned} \]
Cofactor \(A_{33}\)
\[ \begin{aligned} A_{33} &= (-1)^{3+3} \begin{vmatrix} 1&0\\ 0&\cos\alpha \end{vmatrix}\\ &= (1)(\cos\alpha)-(0)(0)\\ &=\cos\alpha \end{aligned} \]
Form the Cofactor Matrix
The nine cofactors are therefore [ \begin{aligned} A_{11}&=-1,& A_{12}&=0,& A_{13}&=0,\\ A_{21}&=0,& A_{22}&=-\cos\alpha,& A_{23}&=-\sin\alpha,\\ A_{31}&=0,& A_{32}&=-\sin\alpha,& A_{33}&=\cos\alpha. \end{aligned} \]
Hence, the cofactor matrix is
\[ C= \begin{bmatrix} -1&0&0\\ 0&-\cos\alpha&-\sin\alpha\\ 0&-\sin\alpha&\cos\alpha \end{bmatrix} \]
Find the Adjoint
By definition,
\[ \operatorname{adj}(A)=C^T. \]
In this case, the cofactor matrix is symmetric because
\[C^T=C\]
Therefore,
\[ \boxed{ \operatorname{adj}(A)= \begin{bmatrix} -1&0&0\\ 0&-\cos\alpha&-\sin\alpha\\ 0&-\sin\alpha&\cos\alpha \end{bmatrix}} \]
Find the Inverse
We use
\[ A^{-1} = \frac{1}{|A|}\operatorname{adj}(A). \]
Since
\[ |A|=-1, \]
we have
\[ \begin{aligned} A^{-1} &= -\operatorname{adj}(A)\\ &= - \begin{bmatrix} -1&0&0\\ 0&-\cos\alpha&-\sin\alpha\\ 0&-\sin\alpha&\cos\alpha \end{bmatrix} \end{aligned} \]
Multiplying every entry by \(-1\),
\[ \boxed{\bbox[5pt]{ A^{-1}= \begin{bmatrix} 1&0&0\ 0&\cos\alpha&\sin\alpha\ 0&\sin\alpha&-\cos\alpha \end{bmatrix}}} \]
Thus, we obtain the particularly elegant result
\[\boxed{A^{-1}=A}\]

🎯 Exam Significance

Exam Significance

›

This problem is an excellent example of how the trigonometric identity

\[ \sin^2\alpha+\cos^2\alpha=1 \]

simplifies determinant and inverse calculations.

For full marks, students should clearly show the determinant, all required cofactors, the cofactor matrix, the adjoint, and the inverse formula. The final statement

\[ \boxed{A^{-1}=A} \]

should be highlighted.

Significance for Competitive Entrance Examinations

The matrix has an important structural property:

\[ A^2=I. \]

Such a matrix is called involutory. Whenever a non-singular matrix satisfies

\[ A^2=I, \]

multiplying both sides by \(A^{-1}\) gives

\[ A=A^{-1}. \]

Therefore, in a competitive examination, one can recognize the structure of the matrix and obtain the inverse rapidly without calculating all nine cofactors.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

The given matrix is
\[ A= \begin{bmatrix} 1&0&0\\ 0&\cos\alpha&\sin\alpha\\ 0&\sin\alpha&-\cos\alpha \end{bmatrix}. \]
Its determinant is
\[ \boxed{|A|=-1}. \]
Since \(|A|\neq0\), the inverse exists.
The cofactor matrix is
\[ C= \begin{bmatrix} -1&0&0\\ 0&-\cos\alpha&-\sin\alpha\\ 0&-\sin\alpha&\cos\alpha \end{bmatrix}. \]
The cofactor matrix is symmetric, so
\[ \operatorname{adj}(A)=C^T=C. \]
The inverse is
\[ \boxed{ A^{-1}= \begin{bmatrix} 1&0&0\\ 0&\cos\alpha&\sin\alpha\\ 0&\sin\alpha&-\cos\alpha \end{bmatrix}}. \]
Thus,
\[ \boxed{A^{-1}=A}. \]
The matrix satisfies
\[ \boxed{A^2=I}, \]
so it is an involutory matrix.
The result follows from the fundamental trigonometric identity
\[ \boxed{\sin^2\alpha+\cos^2\alpha=1}. \]

📘 Concept & Theory

Concept/Theory

›

For two non-singular square matrices \(A\) and \(B\), the inverse of their product is given by

\[ \boxed{(AB)^{-1}=B^{-1}A^{-1}}. \]

The most important point is that the order of the matrices is reversed. In general,

\[ (AB)^{-1}\neq A^{-1}B^{-1}. \]

For a \(2\times2\) matrix

\[ M= \begin{bmatrix} a&b\\ c&d \end{bmatrix}, \]

its inverse, provided \(ad-bc\neq0\), is

\[ M^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d&-b\\ -c&a \end{bmatrix} \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Calculate \(AB\).
Find \((AB)^{-1}\).
Find \(A^{-1}\).
Find \(B^{-1}\).
Calculate \(B^{-1}A^{-1}\), keeping the order exactly as stated.
Compare the two matrices.
Conclude that \((AB)^{-1}=B^{-1}A^{-1}\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 25 steps

Calculate \(AB\)
We have
\[ A= \begin{bmatrix} 3&7\\ 2&5 \end{bmatrix}, \qquad B= \begin{bmatrix} 6&8\\ 7&9 \end{bmatrix} \]
Therefore,
\[ AB= \begin{bmatrix} 3&7\\ 2&5 \end{bmatrix} \begin{bmatrix} 6&8\\ 7&9 \end{bmatrix} \]
Multiplying row by column,
\[ \begin{aligned} (1,1)&=(3)(6)+(7)(7)=18+49=67,\\ (1,2)&=(3)(8)+(7)(9)=24+63=87,\\ (2,1)&=(2)(6)+(5)(7)=12+35=47,\\ (2,2)&=(2)(8)+(5)(9)=16+45=61 \end{aligned} \]
Hence,
\[ \boxed{\bbox[5pt]{ AB= \begin{bmatrix} 67&87\\ 47&61 \end{bmatrix}}} \]
Find \((AB)^{-1}\)
First calculate the determinant of \(AB\):
\[ \begin{aligned} |AB| &= \begin{vmatrix} 67&87\\ 47&61 \end{vmatrix}\\ &=(67)(61)-(87)(47)\\ &=4087-4089\\ &=-2 \end{aligned} \]
Since
\[ |AB|=-2\neq0, \]
the inverse of \(AB\) exists.
The adjoint of \(AB\) is
\[ \operatorname{adj}(AB) = \begin{bmatrix} 61&-87\\ -47&67 \end{bmatrix}. \]
Therefore,
\[ \begin{aligned} (AB)^{-1} &= \frac{1}{|AB|} \operatorname{adj}(AB)\\ &= -\frac12 \begin{bmatrix} 61&-87\\ -47&67 \end{bmatrix} \end{aligned} \]
Thus,
\[ \boxed{\bbox[5pt]{ (AB)^{-1} = \begin{bmatrix} -\frac{61}{2}&\frac{87}{2}\\ \frac{47}{2}&-\frac{67}{2} \end{bmatrix}}} \]
Find \(A^{-1}\)
For
\[ A= \begin{bmatrix} 3&7\\ 2&5 \end{bmatrix}, \]
its determinant is
\[ \begin{aligned} |A| &= (3)(5)-(7)(2)\\ &=15-14\\ &=1 \end{aligned} \]
Hence,
\[ \operatorname{adj}(A) = \begin{bmatrix} 5&-7\\ -2&3 \end{bmatrix} \]
Therefore,
\[ \begin{aligned} A^{-1} &= \frac{1}{|A|} \operatorname{adj}(A)\\ &= \begin{bmatrix} 5&-7\\ -2&3 \end{bmatrix} \end{aligned} \]
Thus,
\[ \boxed{ A^{-1}= \begin{bmatrix} 5&-7\\ -2&3 \end{bmatrix}} \]
Find \(B^{-1}\)
For
\[ B= \begin{bmatrix} 6&8\\ 7&9 \end{bmatrix} \]
the determinant is
\[ \begin{aligned} |B| &= (6)(9)-(8)(7)\\ &=54-56\\ &=-2 \end{aligned} \]
Hence,
\[ \operatorname{adj}(B) = \begin{bmatrix} 9&-8\\ -7&6 \end{bmatrix} \]
Therefore,
\[ \begin{aligned} B^{-1} &= \frac{1}{|B|} \operatorname{adj}(B)\\ &= -\frac12 \begin{bmatrix} 9&-8\\ -7&6 \end{bmatrix}\\ &= \begin{bmatrix} -\frac92&4\\ \frac72&-3 \end{bmatrix} \end{aligned} \]
Thus,
\[ \boxed{\bbox[5pt]{ B^{-1}= \begin{bmatrix} -\frac92&4\\ \frac72&-3 \end{bmatrix}}} \]
Calculate \(B^{-1}A^{-1}\)
The required identity is
\[ (AB)^{-1}=B^{-1}A^{-1}. \]
Therefore, we must multiply \(B^{-1}\) by \(A^{-1}\) in this order.
\[ B^{-1}A^{-1} = \begin{bmatrix} -\frac92&4\\ \frac72&-3 \end{bmatrix} \begin{bmatrix} 5&-7\\ -2&3 \end{bmatrix} \]
Calculate each entry:
\[ \begin{aligned} (1,1) &= \left(-\frac92\right)(5)+(4)(-2)\\ &=-\frac{45}{2}-8\\ &=-\frac{45}{2}-\frac{16}{2}\\ &=-\frac{61}{2}, \end{aligned} \]
\[ \begin{aligned} (1,2) &= \left(-\frac92\right)(-7)+(4)(3)\\ &=\frac{63}{2}+12\\ &=\frac{63}{2}+\frac{24}{2}\\ &=\frac{87}{2} \end{aligned} \]

\[ \begin{aligned} (2,1) &= \left(\frac72\right)(5)+(-3)(-2)\\ &=\frac{35}{2}+6\\ &=\frac{35}{2}+\frac{12}{2}\\ &=\frac{47}{2}, \end{aligned} \]

\[ \begin{aligned} (2,2) &= \left(\frac72\right)(-7)+(-3)(3)\\ &=-\frac{49}{2}-9\\ &=-\frac{49}{2}-\frac{18}{2}\\ &=-\frac{67}{2} \end{aligned} \]
Therefore,
\[ \boxed{\bbox[5pt]{ B^{-1}A^{-1} = \begin{bmatrix} -\frac{61}{2}&\frac{87}{2}\\ \frac{47}{2}&-\frac{67}{2} \end{bmatrix}}} \]
Compare Both Sides
From Step 2,
\[ (AB)^{-1} = \begin{bmatrix} -\frac{61}{2}&\frac{87}{2}\\ \frac{47}{2}&-\frac{67}{2} \end{bmatrix} \]
From Step 5,
\[ B^{-1}A^{-1} = \begin{bmatrix} -\frac{61}{2}&\frac{87}{2}\\ \frac{47}{2}&-\frac{67}{2} \end{bmatrix} \]
Hence, both sides are equal:
\[ \boxed{\bbox[5pt]{ (AB)^{-1}=B^{-1}A^{-1}}} \]
Hence, Verified

🎯 Exam Significance

Exam Significance

›

This problem directly tests the important property of inverse matrices:

\[ \boxed{(AB)^{-1}=B^{-1}A^{-1}}. \]

Students should remember that the order is reversed. The statement

\[ (AB)^{-1}=A^{-1}B^{-1} \]

is generally incorrect.

For a complete board-examination solution, calculate both sides separately and show that the resulting matrices are identical.

Significance for Competitive Entrance Examinations

The reversal of order is a fundamental property of matrix algebra and is frequently tested in objective questions.

A fast conceptual proof is

\[ (AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1} = I. \]

Thus, without explicitly calculating the matrices, we can immediately identify \(B^{-1}A^{-1}\) as the inverse of \(AB\).

The example also demonstrates why matrix multiplication cannot be rearranged arbitrarily. The order of factors is essential.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 10 points

\[ A= \begin{bmatrix} 3&7\\ 2&5 \end{bmatrix}, \qquad |A|=1. \]
\[ A^{-1}= \begin{bmatrix} 5&-7\\ -2&3 \end{bmatrix}. \]
\[ B= \begin{bmatrix} 6&8\\ 7&9 \end{bmatrix}, \qquad |B|=-2. \]
\[ B^{-1} = \begin{bmatrix} -\frac92&4\\ \frac72&-3 \end{bmatrix}. \]
\[ AB= \begin{bmatrix} 67&87\\ 47&61 \end{bmatrix}, \qquad |AB|=-2. \]
\[ (AB)^{-1} = \begin{bmatrix} -\frac{61}{2}&\frac{87}{2}\\ \frac{47}{2}&-\frac{67}{2} \end{bmatrix}. \]
Direct multiplication gives
\[ B^{-1}A^{-1} = \begin{bmatrix} -\frac{61}{2}&\frac{87}{2}\\ \frac{47}{2}&-\frac{67}{2} \end{bmatrix}. \]
Therefore,
\[ \boxed{(AB)^{-1}=B^{-1}A^{-1}}. \]
The order of the inverse factors is reversed:
\[ \boxed{(AB)^{-1}=B^{-1}A^{-1}}, \]
not \(A^{-1}B^{-1}\).
The supplied solution's final multiplication used \(A^{-1}B^{-1}\), so that step must be corrected to \(B^{-1}A^{-1}\).

📘 Concept & Theory

Concept/Theory

›

A matrix can satisfy a polynomial equation in \(A\), just as a number can satisfy an algebraic equation. Here, the required identity is

\[ A^2-5A+7I=0. \]

Once this identity is established, we can rearrange it to isolate a term containing \(A^{-1}\). This provides an efficient alternative to directly using the adjoint formula.

In particular,

\[ A^2-5A+7I=0 \]

gives

\[ A^2-5A=-7I. \]

Since \(A\) is non-singular, multiplication by \(A^{-1}\) allows us to obtain the inverse directly.

🗺️ Solution Roadmap

Step-by-step Plan

›

Calculate \(A^2\).
Calculate \(5A\).
Calculate \(7I\).
Evaluate \(A^2-5A+7I\) and verify that it is the zero matrix.
Use the established identity to derive \(A^{-1}\).
Verify the resulting inverse by multiplication.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 23 steps

Calculate \(A^2\)
We have
\[ A= \begin{bmatrix} 3&1\\ -1&2 \end{bmatrix} \]
Therefore,
\[ A^2=A\cdot A = \begin{bmatrix} 3&1\\ -1&2 \end{bmatrix} \begin{bmatrix} 3&1\\ -1&2 \end{bmatrix} \]
Multiplying row by column,
\[ \begin{aligned} (1,1)&=(3)(3)+(1)(-1)=9-1=8,\\ (1,2)&=(3)(1)+(1)(2)=3+2=5,\\ (2,1)&=(-1)(3)+(2)(-1)=-3-2=-5,\\ (2,2)&=(-1)(1)+(2)(2)=-1+4=3 \end{aligned} \]
Hence,
\[ \boxed{ A^2= \begin{bmatrix} 8&5\\ -5&3 \end{bmatrix}} \]
Calculate \(5A\)
\[ 5A = 5 \begin{bmatrix} 3&1\\ -1&2 \end{bmatrix} \]
Multiplying every entry by \(5\),
\[ \begin{aligned} 5A &= \begin{bmatrix} 5(3)&5(1)\\ 5(-1)&5(2) \end{bmatrix}\\ &= \boxed{ \begin{bmatrix} 15&5\\ -5&10 \end{bmatrix}} \end{aligned} \]
Calculate \(7I\)
For a \(2\times2\) matrix, the identity matrix is
\[ I= \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix} \]
Therefore,
\[ 7I = 7 \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix} = \boxed{ \begin{bmatrix} 7&0\\ 0&7 \end{bmatrix}} \]
Verify \(A^2-5A+7I=0\)
\[ \begin{aligned} A^2-5A+7I &= \begin{bmatrix} 8&5\\ -5&3 \end{bmatrix} - \begin{bmatrix} 15&5\\ -5&10 \end{bmatrix} + \begin{bmatrix} 7&0\\ 0&7 \end{bmatrix} \end{aligned} \]
Performing the operations entry by entry,
\[ \begin{aligned} A^2-5A+7I &= \begin{bmatrix} 8-15+7&5-5+0\\ -5-(-5)+0&3-10+7 \end{bmatrix}\\ &= \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix} \end{aligned} \]
Since the resulting matrix is the zero matrix,
\[\boxed{A^2-5A+7I=0}\]
Hence, Proved
Derive \(A^{-1}\) from the Identity
We have
\[A^2-5A+7I=0.\]
Rearranging,
\[A^2-5A=-7I\]
Alternatively,
\[A^2+7I=5A\]
Multiply both sides by \(A^{-1}\) on the right:
\[(A^2+7I)A^{-1}=5AA^{-1}\]
Using \(AA^{-1}=I\),
\[A^2A^{-1}+7IA^{-1}=5I\]
Since
\[A^2A^{-1}=A(AA^{-1})=AI=A,\]
and
\[IA^{-1}=A^{-1},\]
we obtain
\[A+7A^{-1}=5I\]
Therefore,
\[7A^{-1}=5I-A\]
Calculate \(5I-A\)
We have
\[ 5I= \begin{bmatrix} 5&0\\ 0&5 \end{bmatrix} \]
and
\[ A= \begin{bmatrix} 3&1\\ -1&2 \end{bmatrix} \]
Hence,
\[ \begin{aligned} 5I-A &= \begin{bmatrix} 5&0\\ 0&5 \end{bmatrix} - \begin{bmatrix} 3&1\\ -1&2 \end{bmatrix}\\ &= \begin{bmatrix} 5-3&0-1\\ 0-(-1)&5-2 \end{bmatrix}\\ &= \begin{bmatrix} 2&-1\\ 1&3 \end{bmatrix} \end{aligned} \]
Therefore,
\[ 7A^{-1} = \begin{bmatrix} 2&-1\\ 1&3 \end{bmatrix} \]
Dividing both sides by \(7\),
\[ \boxed{\bbox[5pt]{ A^{-1} = \frac17 \begin{bmatrix} 2&-1\\ 1&3 \end{bmatrix}}} \]

🎯 Exam Significance

Exam Significance

›

This question illustrates an important technique: a polynomial identity involving a matrix can be used to find its inverse without directly calculating the adjoint.

The key sequence to remember is

\[ A^2-5A+7I=0 \]

followed by

\[ A^2+7I=5A, \]

and multiplication by \(A^{-1}\):

\[ A+7A^{-1}=5I. \]

Therefore,

\[ \boxed{A^{-1}=\frac17(5I-A)}. \]

Showing the intermediate matrices \(A^2\), \(5A\), and \(7I\) makes the verification complete and helps secure method marks.

Significance for Competitive Entrance Examinations

For objective examinations, the polynomial identity provides a much faster route than calculating nine cofactors.

From

\[ A^2-5A+7I=0, \]

directly multiply by \(A^{-1}\) to obtain

\[ A-5I+7A^{-1}=0. \]

Hence, immediately,

\[ \boxed{A^{-1}=\frac{1}{7}(5I-A)}. \]

This is an efficient application of matrix algebra and is particularly useful when a problem asks for the inverse of a matrix satisfying a given polynomial equation.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

The given matrix is
\[ A= \begin{bmatrix} 3&1\\ -1&2 \end{bmatrix}. \]
Its square is
\[ A^2= \begin{bmatrix} 8&5\\ -5&3 \end{bmatrix}. \]
We have
\[ 5A= \begin{bmatrix} 15&5\\ -5&10 \end{bmatrix}, \qquad 7I= \begin{bmatrix} 7&0\\ 0&7 \end{bmatrix}. \]
Therefore,
\[ \boxed{A^2-5A+7I=0}. \]
Rearranging the identity gives
\[ A+7A^{-1}=5I. \]
Hence,
\[ \boxed{A^{-1}=\frac17(5I-A)}. \]
Since
\[ 5I-A= \begin{bmatrix} 2&-1\\ 1&3 \end{bmatrix}, \]
we obtain
\[ \boxed{ A^{-1} = \frac17 \begin{bmatrix} 2&-1\\ 1&3 \end{bmatrix}}. \]
Direct verification confirms
\[ AA^{-1}=I. \]
As an independent check,
\[ |A|=7\neq0, \]
confirming that \(A\) is non-singular and its inverse exists.

📘 Concept & Theory

Concept/Theory

›

Here \(a\) and \(b\) are scalar constants, while \(A\) is a matrix. The equation

\[ A^2+aA+bI=0 \]

means that the matrix obtained by adding \(A^2\), \(aA\), and \(bI\) must be the zero matrix.

The zero matrix of order \(2\) is

\[ 0= \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix} \]

Since corresponding entries of two equal matrices are equal, we can compare the entries of

\[ A^2+aA+bI \]

with the corresponding entries of the zero matrix to determine \(a\) and \(b\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Calculate \(A^2\).
Calculate \(aA\).
Calculate \(bI\).
Substitute these into \(A^2+aA+bI=0\).
Compare corresponding matrix entries.
Solve the resulting equations for \(a\) and \(b\).
Substitute the values back to verify the matrix equation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 16 steps

Given —
\[ A= \begin{bmatrix} 3&2\\ 1&1 \end{bmatrix} \]
Calculate \(A^2\)
\[ A^2=A\cdot A = \begin{bmatrix} 3&2\\ 1&1 \end{bmatrix} \begin{bmatrix} 3&2\\ 1&1 \end{bmatrix} \]
Multiplying row by column,
\[ \begin{aligned} (1,1)&=(3)(3)+(2)(1)=9+2=11,\\ (1,2)&=(3)(2)+(2)(1)=6+2=8,\\ (2,1)&=(1)(3)+(1)(1)=3+1=4,\\ (2,2)&=(1)(2)+(1)(1)=2+1=3 \end{aligned} \]
Hence,
\[ \boxed{ A^2= \begin{bmatrix} 11&8\\ 4&3 \end{bmatrix}} \]
Calculate \(aA\)
Multiplying every entry of \(A\) by the scalar \(a\),
\[ \begin{aligned} aA &= a \begin{bmatrix} 3&2\\ 1&1 \end{bmatrix}\\ &= \boxed{ \begin{bmatrix} 3a&2a\\ a&a \end{bmatrix}} \end{aligned} \]
Calculate \(bI\)
The identity matrix of order \(2\) is
\[ I= \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix} \]
Therefore,
\[ bI = b \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix} = \boxed{ \begin{bmatrix} b&0\\ 0&b \end{bmatrix}} \]
Substitute in \(A^2+aA+bI=0\)
We require
\[ A^2+aA+bI = \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix} \]
Substituting the matrices obtained above,
\[ \begin{aligned} & \begin{bmatrix} 11&8\\ 4&3 \end{bmatrix} + \begin{bmatrix} 3a&2a\\ a&a \end{bmatrix} + \begin{bmatrix} b&0\\ 0&b \end{bmatrix}\\ &= \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix} \end{aligned} \]
Adding corresponding entries,
\[ \boxed{\bbox[5pt]{ \begin{bmatrix} 11+3a+b&8+2a\\ 4+a&3+a+b \end{bmatrix} = \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix}}} \]
Compare Corresponding Entries
From the \((2,1)\)-entry,
\[4+a=0\]
Therefore,
\[\boxed{a=-4}\]
Now use the \((2,2)\)-entry:
\[3+a+b=0\]
Substituting \(a=-4\),
\[ \begin{aligned} 3-4+b&=0,\\ b-1&=0,\\ \boxed{b=1} \end{aligned} \]
Thus, the required values are
\[\boxed{a=-4,\qquad b=1}\]

🎯 Exam Significance

Exam Significance

›

This problem tests the ability to perform matrix multiplication, scalar multiplication, and matrix addition, followed by comparison of corresponding entries.

The most direct board-examination approach is to calculate \(A^2\), form \(A^2+aA+bI\), and equate it to the zero matrix:

\[ \begin{bmatrix} 11+3a+b&8+2a\\ 4+a&3+a+b \end{bmatrix} = \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix} \]

From the entries we obtain

\[ 4+a=0 \]

and

\[ 3+a+b=0, \]

giving

\[ \boxed{a=-4,\quad b=1}. \]

Significance for Competitive Entrance Examinations

For competitive examinations, the Cayley–Hamilton theorem gives a very efficient solution. For a \(2\times2\) matrix,

\[ A^2-(\operatorname{tr}A)A+|A|I=0. \]

Here,

\[ \operatorname{tr}(A)=4,\qquad |A|=1. \]

Therefore,

\[ A^2-4A+I=0. \]

Thus, without explicitly multiplying \(A\) by itself, we can immediately identify

\[ \boxed{a=-4,\qquad b=1}. \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

The given matrix is
\[ A= \begin{bmatrix} 3&2\\ 1&1 \end{bmatrix} \]
Its square is
\[ A^2= \begin{bmatrix} 11&8\\ 4&3 \end{bmatrix} \]
The required equation becomes
\[ \begin{bmatrix} 11+3a+b&8+2a\\ 4+a&3+a+b \end{bmatrix} = \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix} \]
From \(4+a=0\),
\[ \boxed{a=-4}. \]
From \(3+a+b=0\),
\[ \boxed{b=1}. \]
Therefore,
\[ \boxed{A^2-4A+I=0}. \]
The result can also be obtained from the Cayley–Hamilton theorem because
\[ \operatorname{tr}(A)=4,\qquad |A|=1. \]
Final answer:
\[ \boxed{a=-4,\qquad b=1}. \]

📘 Concept & Theory

Concept/Theory

›

A matrix may satisfy a polynomial equation in \(A\). If such an equation contains a non-zero constant multiple of \(I\), it can often be rearranged to obtain an expression for \(A^{-1}\).

Here we need to establish

\[ A^3-6A^2+5A+11I=0. \]

Once this identity is proved, multiplying it by \(A^{-1}\) gives

\[ A^2-6A+5I+11A^{-1}=0, \]

from which \(A^{-1}\) can be obtained directly.

🗺️ Solution Roadmap

Step-by-step Plan

›

Calculate \(A^2\).
Calculate \(A^3=A^2A\).
Evaluate \(A^3-6A^2+5A+11I\) and verify that it is the zero matrix.
Use the established identity to derive \(A^{-1}\).
Calculate the resulting matrix explicitly.
Verify the inverse by checking \(AA^{-1}=I\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 29 steps

Given —
\[ A= \begin{bmatrix} 1&1&1\\ 1&2&-3\\ 2&-1&3 \end{bmatrix} \]
Calculate \(A^2\)
\[ A^2=A\cdot A = \begin{bmatrix} 1&1&1\\ 1&2&-3\\ 2&-1&3 \end{bmatrix} \begin{bmatrix} 1&1&1\\ 1&2&-3\\ 2&-1&3 \end{bmatrix} \]
Calculate each entry carefully:
\[ \begin{aligned} (1,1)&=(1)(1)+(1)(1)+(1)(2)=1+1+2=4,\\ (1,2)&=(1)(1)+(1)(2)+(1)(-1)=1+2-1=2,\\ (1,3)&=(1)(1)+(1)(-3)+(1)(3)=1-3+3=1 \end{aligned} \]
For the second row:
\[ \begin{aligned} (2,1)&=(1)(1)+(2)(1)+(-3)(2)=1+2-6=-3,\\ (2,2)&=(1)(1)+(2)(2)+(-3)(-1)=1+4+3=8,\\ (2,3)&=(1)(1)+(2)(-3)+(-3)(3)=1-6-9=-14 \end{aligned} \]
For the third row:
\[ \begin{aligned} (3,1)&=(2)(1)+(-1)(1)+(3)(2)=2-1+6=7,\\ (3,2)&=(2)(1)+(-1)(2)+(3)(-1)=2-2-3=-3,\\ (3,3)&=(2)(1)+(-1)(-3)+(3)(3)=2+3+9=14 \end{aligned} \]
Hence,
\[ \boxed{\bbox[5pt]{ A^2= \begin{bmatrix} 4&2&1\\ -3&8&-14\\ 7&-3&14 \end{bmatrix}}} \]
Calculate \(A^3\)
Since
\[ A^3=A^2A, \]
we have
\[ A^3= \begin{bmatrix} 4&2&1\\\ -3&8&-14\\ 7&-3&14 \end{bmatrix} \begin{bmatrix} 1&1&1\\ 1&2&-3\\ 2&-1&3 \end{bmatrix} \]
Calculate the first row:
\[ \begin{aligned} (1,1)&=(4)(1)+(2)(1)+(1)(2)=4+2+2=8,\\ (1,2)&=(4)(1)+(2)(2)+(1)(-1)=4+4-1=7,\\ (1,3)&=(4)(1)+(2)(-3)+(1)(3)=4-6+3=1 \end{aligned} \]
Calculate the second row:
\[ \begin{aligned} (2,1)&=(-3)(1)+(8)(1)+(-14)(2)=-3+8-28=-23,\\ (2,2)&=(-3)(1)+(8)(2)+(-14)(-1)=-3+16+14=27,\\ (2,3)&=(-3)(1)+(8)(-3)+(-14)(3)=-3-24-42=-69 \end{aligned} \]
Calculate the third row:
\[ \begin{aligned} (3,1)&=(7)(1)+(-3)(1)+(14)(2)=7-3+28=32,\\ (3,2)&=(7)(1)+(-3)(2)+(14)(-1)=7-6-14=-13,\\ (3,3)&=(7)(1)+(-3)(-3)+(14)(3)=7+9+42=58 \end{aligned} \]
Thus,
\[ \boxed{ A^3= \begin{bmatrix} 8&7&1\\ -23&27&-69\\ 32&-13&58 \end{bmatrix}} \]
Calculate \(6A^2\), \(5A\), and \(11I\)<
First,
\[ 6A^2 = 6 \begin{bmatrix} 4&2&1\\ -3&8&-14\\ 7&-3&14 \end{bmatrix} = \begin{bmatrix} 24&12&6\\ -18&48&-84\\ 42&-18&84 \end{bmatrix} \]
Next,
\[ 5A = 5 \begin{bmatrix} 1&1&1\\ 1&2&-3\\ 2&-1&3 \end{bmatrix} = \begin{bmatrix} 5&5&5\\ 5&10&-15\\ 10&-5&15 \end{bmatrix} \]
The identity matrix of order \(3\) is
\[ I= \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} \]
Therefore,
\[ 11I= \begin{bmatrix} 11&0&0\\ 0&11&0\\ 0&0&11 \end{bmatrix} \]
Verify \(A^3-6A^2+5A+11I=0\)
Substitute the matrices obtained above:
\[ \begin{aligned} A^3-6A^2+5A+11I &= \begin{bmatrix} 8&7&1\\ -23&27&-69\\ 32&-13&58 \end{bmatrix} - \begin{bmatrix} 24&12&6\\ -18&48&-84\\ 42&-18&84 \end{bmatrix}\\ &\quad+ \begin{bmatrix} 5&5&5\\ 5&10&-15\\ 10&-5&15 \end{bmatrix} + \begin{bmatrix} 11&0&0\\ 0&11&0\\ 0&0&11 \end{bmatrix} \end{aligned} \]
Now calculate every entry:
\[ \begin{aligned} (1,1)&=8-24+5+11=0,\\ (1,2)&=7-12+5+0=0,\\ (1,3)&=1-6+5+0=0,\\ (2,1)&=-23-(-18)+5+0=-23+18+5=0,\\ (2,2)&=27-48+10+11=0,\\ (2,3)&=-69-(-84)-15+0=-69+84-15=0,\\ (3,1)&=32-42+10+0=0,\\ (3,2)&=-13-(-18)-5+0=-13+18-5=0,\\ (3,3)&=58-84+15+11=0 \end{aligned} \]
Therefore,
\[ A^3-6A^2+5A+11I = \begin{bmatrix} 0&0&0\\ 0&0&0\\ 0&0&0 \end{bmatrix} \]
Hence, Proved
Derive \(A^{-1}\)
Starting with
\[A^3-6A^2+5A+11I=0,\]
multiply both sides by \(A^{-1}\):
\[(A^3-6A^2+5A+11I)A^{-1}=0\]
Distributing \(A^{-1}\),
\[A^3A^{-1}-6A^2A^{-1}+5AA^{-1}+11IA^{-1}=0\]
Using
\[ A^3A^{-1}=A^2,\quad A^2A^{-1}=A,\quad AA^{-1}=I,\quad IA^{-1}=A^{-1}, \]
we get
\[A^2-6A+5I+11A^{-1}=0\]
Therefore,
\[11A^{-1}=-A^2+6A-5I\]
Hence,
\[ \boxed{\bbox[8pt]{ A^{-1} = \frac{1}{11} \left(-A^2+6A-5I\right)}} \]
Calculate \(-A^2+6A-5I\)
We already have
\[ A^2= \begin{bmatrix} 4&2&1\\ -3&8&-14\\ 7&-3&14 \end{bmatrix} \]
Thus,
\[ -A^2= \begin{bmatrix} -4&-2&-1\\ 3&-8&14\\ -7&3&-14 \end{bmatrix} \]
Also,
\[ 6A= \begin{bmatrix} 6&6&6\\ 6&12&-18\\ 12&-6&18 \end{bmatrix}, \]
and
\[ -5I= \begin{bmatrix} -5&0&0\\ 0&-5&0\\ 0&0&-5 \end{bmatrix} \]
Therefore,
\[ \begin{aligned} -A^2+6A-5I &= \begin{bmatrix} -4&-2&-1\\ 3&-8&14\\ -7&3&-14 \end{bmatrix} + \begin{bmatrix} 6&6&6\\ 6&12&-18\\ 12&-6&18 \end{bmatrix} + \begin{bmatrix} -5&0&0\\ 0&-5&0\\ 0&0&-5 \end{bmatrix}\\ &= \begin{bmatrix} -4+6-5&-2+6&-1+6\\ 3+6&-8+12-5&14-18\\ -7+12&3-6&-14+18-5 \end{bmatrix}\\ &= \begin{bmatrix} -3&4&5\\ 9&-1&-4\\ 5&-3&-1 \end{bmatrix} \end{aligned} \]
Consequently,
\[ \boxed{\bbox[5pt]{ A^{-1} = \frac{1}{11} \begin{bmatrix} -3&4&5\\ 9&-1&-4\\ 5&-3&-1 \end{bmatrix}}} \]

🎯 Exam Significance

Exam Significance

›

This question combines matrix multiplication with the use of a polynomial identity to determine the inverse. The essential derivation is:

\[ A^3-6A^2+5A+11I=0. \]

Multiplying by \(A^{-1}\),

\[ A^2-6A+5I+11A^{-1}=0. \]

Therefore,

\[ \boxed{ A^{-1}=\frac{1}{11}\left(-A^2+6A-5I\right)}. \]

For a board examination, clearly showing the calculation of \(A^2\), \(A^3\), the zero matrix verification, and the final substitution makes the solution complete and methodologically sound.

Significance for Competitive Entrance Examinations

The polynomial identity provides a useful shortcut. Once

\[ A^3-6A^2+5A+11I=0 \]

is known, there is no need to calculate all nine cofactors to obtain the inverse.

Multiplying by \(A^{-1}\) immediately gives

\[ \boxed{ A^{-1}=\frac{1}{11}(-A^2+6A-5I)}. \]

This is especially useful in objective-type questions where a matrix polynomial is provided or can be identified from the characteristic equation.

2-mark standard board question.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

The given matrix is
\[ A= \begin{bmatrix} 1&1&1\\ 1&2&-3\\ 2&-1&3 \end{bmatrix}. \]
Its square is
\[ A^2= \begin{bmatrix} 4&2&1\\ -3&8&-14\\ 7&-3&14 \end{bmatrix}. \]
Its cube is
\[ A^3= \begin{bmatrix} 8&7&1\\ -23&27&-69\\ 32&-13&58 \end{bmatrix}. \]
Direct substitution verifies
\[ \boxed{A^3-6A^2+5A+11I=0}. \]
Multiplication by \(A^{-1}\) gives
\[ A^2-6A+5I+11A^{-1}=0. \]
Hence,
\[ \boxed{ A^{-1} = \frac{1}{11} \left(-A^2+6A-5I\right)}. \]
Therefore,
\[ \boxed{ A^{-1} = \frac{1}{11} \begin{bmatrix} -3&4&5\\ 9&-1&-4\\ 5&-3&-1 \end{bmatrix}}. \]
The determinant check gives
\[ |A|=-11\neq0, \]
confirming that \(A^{-1}\) exists.
Direct multiplication confirms
\[ AA^{-1}=I. \]

📘 Concept & Theory

Concept/Theory

›

A matrix can satisfy a polynomial equation in \(A\). Such an identity can be used to express \(A^{-1}\) as a polynomial in \(A\).

Here, after verifying

\[ A^3-6A^2+9A-4I=0, \]

we multiply the equation by \(A^{-1}\). This reduces the highest power from \(A^3\) to \(A^2\) and produces an expression containing \(A^{-1}\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Calculate \(A^2\).
Calculate \(A^3=A^2A\).
Evaluate \(A^3-6A^2+9A-4I\) and verify that it is the zero matrix.
Multiply the verified identity by \(A^{-1}\).
Obtain an expression for \(A^{-1}\) in terms of \(A^2\), \(A\), and \(I\).
Substitute the calculated matrices to obtain the explicit inverse.
Verify the result by checking \(AA^{-1}=I\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 28 steps

Calculate \(A^2\)
We have
\[ A= \begin{bmatrix} 2&-1&1\\ -1&2&-1\\ 1&-1&2 \end{bmatrix} \]
Therefore,
\[ A^2= \begin{bmatrix} 2&-1&1\\ -1&2&-1\\ 1&-1&2 \end{bmatrix} \begin{bmatrix} 2&-1&1\\ -1&2&-1\\ 1&-1&2 \end{bmatrix} \]
Calculate the entries row by row.
First row:
\[ \begin{aligned} (1,1)&=(2)(2)+(-1)(-1)+(1)(1)=4+1+1=6,\\ (1,2)&=(2)(-1)+(-1)(2)+(1)(-1)=-2-2-1=-5,\\ (1,3)&=(2)(1)+(-1)(-1)+(1)(2)=2+1+2=5 \end{aligned} \]
Second row:
\[ \begin{aligned} (2,1)&=(-1)(2)+(2)(-1)+(-1)(1)=-2-2-1=-5,\\ (2,2)&=(-1)(-1)+(2)(2)+(-1)(-1)=1+4+1=6,\\ (2,3)&=(-1)(1)+(2)(-1)+(-1)(2)=-1-2-2=-5 \end{aligned} \]
Third row:
\[ \begin{aligned} (3,1)&=(1)(2)+(-1)(-1)+(2)(1)=2+1+2=5,\\ (3,2)&=(1)(-1)+(-1)(2)+(2)(-1)=-1-2-2=-5,\\ (3,3)&=(1)(1)+(-1)(-1)+(2)(2)=1+1+4=6 \end{aligned} \]
Hence,
\[ \boxed{ A^2= \begin{bmatrix} 6&-5&5\\ -5&6&-5\\ 5&-5&6 \end{bmatrix}} \]
Calculate \(A^3\)
Since,
\[A^3=A^2A,\]
we have
\[ A^3= \begin{bmatrix} 6&-5&5\\ -5&6&-5\\ 5&-5&6 \end{bmatrix} \begin{bmatrix} 2&-1&1\\ -1&2&-1\\ 1&-1&2 \end{bmatrix} \]
First row:
\[ \begin{aligned} (1,1)&=(6)(2)+(-5)(-1)+(5)(1)=12+5+5=22,\\ (1,2)&=(6)(-1)+(-5)(2)+(5)(-1)=-6-10-5=-21,\\ (1,3)&=(6)(1)+(-5)(-1)+(5)(2)=6+5+10=21 \end{aligned} \]
Second row:
\[ \begin{aligned} (2,1)&=(-5)(2)+(6)(-1)+(-5)(1)=-10-6-5=-21,\\ (2,2)&=(-5)(-1)+(6)(2)+(-5)(-1)=5+12+5=22,\\ (2,3)&=(-5)(1)+(6)(-1)+(-5)(2)=-5-6-10=-21 \end{aligned} \]
Third row:
\[ \begin{aligned} (3,1)&=(5)(2)+(-5)(-1)+(6)(1)=10+5+6=21,\\ (3,2)&=(5)(-1)+(-5)(2)+(6)(-1)=-5-10-6=-21,\\ (3,3)&=(5)(1)+(-5)(-1)+(6)(2)=5+5+12=22 \end{aligned} \]
Thus,
\[ \boxed{ A^3= \begin{bmatrix} 22&-21&21\\ -21&22&-21\\ 21&-21&22 \end{bmatrix}} \]
Calculate \(6A^2\), \(9A\), and \(4I\)
From the value of \(A^2\),
\[ 6A^2 = 6 \begin{bmatrix} 6&-5&5\\ -5&6&-5\\ 5&-5&6 \end{bmatrix} = \begin{bmatrix} 36&-30&30\\ -30&36&-30\\ 30&-30&36 \end{bmatrix} \]
Also,
\[ 9A = 9 \begin{bmatrix} 2&-1&1\\ -1&2&-1\\ 1&-1&2 \end{bmatrix} = \begin{bmatrix} 18&-9&9\\ -9&18&-9\\ 9&-9&18 \end{bmatrix} \]
The identity matrix of order \(3\) is
\[ I= \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} \]
Hence,
\[ 4I= \begin{bmatrix} 4&0&0\\ 0&4&0\\ 0&0&4 \end{bmatrix} \]
Hence, Proved
Derive the Formula for \(A^{-1}\)
We have
\[A^3-6A^2+9A-4I=0\]
Multiply both sides by \(A^{-1}\)
\[(A^3-6A^2+9A-4I)A^{-1}=0\]
Therefore,
\[A^3A^{-1}-6A^2A^{-1}+9AA^{-1}-4IA^{-1}=0\]
Using
\[ A^3A^{-1}=A^2,\qquad A^2A^{-1}=A,\qquad AA^{-1}=I,\qquad IA^{-1}=A^{-1}, \]
we obtain
\[A^2-6A+9I-4A^{-1}=0\]
Hence,
\[4A^{-1}=A^2-6A+9I\]
Therefore,
\[\boxed{A^{-1}=\frac14\left(A^2-6A+9I\right)}\]
Substitute the Matrices
We already have
\[ A^2= \begin{bmatrix} 6&-5&5\\ -5&6&-5\\ 5&-5&6 \end{bmatrix} \]
Also,
\[ -6A = \begin{bmatrix} -12&6&-6\\ 6&-12&6\\ -6&6&-12 \end{bmatrix}, \]
and
\[ 9I= \begin{bmatrix} 9&0&0\\ 0&9&0\\ 0&0&9 \end{bmatrix} \]
Hence,
\[ \begin{aligned} A^2-6A+9I &= \begin{bmatrix} 6&-5&5\\ -5&6&-5\\ 5&-5&6 \end{bmatrix} + \begin{bmatrix} -12&6&-6\\ 6&-12&6\\ -6&6&-12 \end{bmatrix} + \begin{bmatrix} 9&0&0\\ 0&9&0\\ 0&0&9 \end{bmatrix}\\ &= \begin{bmatrix} 6-12+9&-5+6&5-6\\ -5+6&6-12+9&-5+6\\ 5-6&-5+6&6-12+9 \end{bmatrix}\\ &= \begin{bmatrix} 3&1&-1\\ 1&3&1\\ -1&1&3 \end{bmatrix} \end{aligned} \]
Therefore,
\[ \boxed{\bbox[5pt]{ A^{-1} = \frac14 \begin{bmatrix} 3&1&-1\ 1&3&1\ -1&1&3 \end{bmatrix}}} \]
or equivalently,
\[ \boxed{ A^{-1} = \begin{bmatrix} \frac34&\frac14&-\frac14\\ \frac14&\frac34&\frac14\\ -\frac14&\frac14&\frac34 \end{bmatrix}} \]

🎯 Exam Significance

Exam Significance

›

This question tests the use of matrix algebra to verify a polynomial identity and then derive the inverse without directly calculating the adjoint.

The key step after verification is

\[ A^3-6A^2+9A-4I=0. \]

Multiplying by \(A^{-1}\) gives

\[ A^2-6A+9I-4A^{-1}=0. \]

Hence,

\[ \boxed{ A^{-1}=\frac14(A^2-6A+9I)}. \]

In a board examination, all matrix multiplications should be shown clearly, particularly when verifying the given polynomial identity.

Significance for Competitive Entrance Examinations

Once the polynomial identity is known, the inverse follows immediately:

\[ \boxed{ A^{-1}=\frac14(A^2-6A+9I)}. \]

This is considerably faster than calculating nine cofactors and then transposing the cofactor matrix. Such polynomial identities are particularly useful in matrix-based objective questions and problems involving higher powers of \(A\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

Given
\[ A= \begin{bmatrix} 2&-1&1\\ -1&2&-1\\ 1&-1&2 \end{bmatrix}. \]
The square of \(A\) is
\[ A^2= \begin{bmatrix} 6&-5&5\\ -5&6&-5\\ 5&-5&6 \end{bmatrix}. \]
The cube of \(A\) is
\[ A^3= \begin{bmatrix} 22&-21&21\\ -21&22&-21\\ 21&-21&22 \end{bmatrix}. \]
Direct substitution verifies
\[ \boxed{A^3-6A^2+9A-4I=0}. \]
Multiplying by \(A^{-1}\) gives
\[ A^2-6A+9I-4A^{-1}=0. \]
Therefore,
\[ \boxed{ A^{-1}=\frac14(A^2-6A+9I)}. \]
Hence,
\[ \boxed{ A^{-1} = \frac14 \begin{bmatrix} 3&1&-1\\ 1&3&1\\ -1&1&3 \end{bmatrix}} \]
or
\[ \boxed{ A^{-1} = \begin{bmatrix} \frac34&\frac14&-\frac14\\ \frac14&\frac34&\frac14\\ -\frac14&\frac14&\frac34 \end{bmatrix}}. \]
The determinant check gives
\[ |A|=4\neq0, \]
confirming that \(A^{-1}\) exists.
Direct multiplication confirms
\[ AA^{-1}=I. \]

📘 Concept & Theory

Concept/Theory

›

For any square matrix \(A\), the fundamental adjoint identity is

\[ \boxed{A\,\operatorname{adj}(A)=\operatorname{adj}(A)\,A=|A|I}. \]

Since \(A\) is nonsingular,

\[ |A|\neq0 \]

and \(A^{-1}\) exists.

The inverse of a matrix is given by

\[ \boxed{ A^{-1}=\frac{\operatorname{adj}(A)}{|A|} }. \]

Equivalently,

\[ \boxed{\operatorname{adj}(A)=|A|A^{-1}}. \]

✏️ Solution

Complete Solution

›

Step-by-step Solution · 7 steps

Since \(A\) is nonsingular,
\[|A|\neq0.\]
From the fundamental property of the adjoint,
\[ A\,\operatorname{adj}(A)=|A|I. \]
Multiplying both sides on the left by \(A^{-1}\), we get
\[ A^{-1}A\,\operatorname{adj}(A) = A^{-1}|A|I \]
Since
\[ A^{-1}A=I \]
and
\[ A^{-1}I=A^{-1}, \]
therefore
\[\operatorname{adj}(A)=|A|A^{-1}\]
Hence the expression for \(\operatorname{adj}(A)\) is
\[\boxed{\operatorname{adj}(A)=|A|A^{-1}}\]
Important Observation
The fact that \(A\) is of order \(3\times3\) does not mean that \(\operatorname{adj}(A)\) is \(3|A|\), \(|A|^2\), or \(|A|^3\). The order of the matrix is irrelevant to this particular identity.

The correct general relation for every nonsingular square matrix is

\[ \boxed{\operatorname{adj}(A)=|A|A^{-1}}. \]

💡 Answer

Final Answer

›

The correct answer is:

\[\boxed{\text{(A)}\quad |A|A^{-1}}\]

🎯 Exam Significance

Exam Significance

›

Remember the fundamental formula:

\[ \boxed{A^{-1}=\frac{1}{|A|}\operatorname{adj}(A)} \qquad(|A|\neq0). \]

Rearranging it immediately gives

\[ \boxed{\operatorname{adj}(A)=|A|A^{-1}}. \]

This identity is frequently used in short-answer and multiple-choice questions.

Significance for Competitive Entrance Examinations

For an MCQ, the fastest route is to recall

\[ \boxed{A\,\operatorname{adj}(A)=|A|I}. \]

For nonsingular \(A\), multiply by \(A^{-1}\):

\[ \boxed{\operatorname{adj}(A)=|A|A^{-1}}. \]

No calculation of cofactors is required.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

For a nonsingular square matrix,
\[ |A|\neq0. \]
The fundamental adjoint identity is
\[ A\,\operatorname{adj}(A)=|A|I. \]
Therefore,
\[ \boxed{\operatorname{adj}(A)=|A|A^{-1}}. \]
The order \(3\times3\) does not change this formula.
Hence, among the listed choices, the intended correct answer is
\[ \boxed{\text{(A)}} \]
only if option (A) is intended to mean \(|A|A^{-1}\). If it literally reads \(A|A|\), none of the four options is correct.

📘 Concept & Theory

Concept/Theory

›

For any invertible square matrix \(A\),

\[ AA^{-1}=I. \]

Taking determinants on both sides gives

\[ \det(AA^{-1})=\det(I). \]

Using the property

\[ \det(AB)=\det(A)\det(B), \]

we obtain

\[ \det(A)\det(A^{-1})=\det(I). \]

Since the determinant of the identity matrix is \(1\),

\[ \det(I)=1. \]

Therefore,

\[ \det(A)\det(A^{-1})=1. \]

Since \(A\) is invertible,

\[ \det(A)\neq0. \]

Hence division by \(\det(A)\) is valid.

🗺️ Solution Roadmap

Step-by-step Plan

›

Use the identity \(AA^{-1}=I\).
Take determinants on both sides.
Apply \(\det(AB)=\det(A)\det(B)\).
Use \(\det(I)=1\).
Solve for \(\det(A^{-1})\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 6 steps

Since \(A\) is invertible,
\[AA^{-1}=I\]
Taking determinant on both sides,
\[\det(AA^{-1})=\det(I)\]
By the multiplicative property of determinants,
\[\det(A)\det(A^{-1})=\det(I)\]
For the identity matrix of order \(2\times2\),
\[ I= \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix}, \]
and therefore
\[ \det(I) = \begin{vmatrix} 1&0\\ 0&1 \end{vmatrix} =1(1)-0(0)=1 \]
Thus,
\[\det(A)\det(A^{-1})=1\]
Dividing both sides by \(\det(A)\), which is non-zero because \(A\) is invertible,
\[\boxed{\bbox[5pt]{\det(A^{-1})=\frac{1}{\det(A)}}}.\]

💡 Answer

Final Answer

›

The correct option is

\[\boxed{\text{(B)}\quad \dfrac{1}{\det(A)}}\]

🎯 Exam Significance

Exam Significance

›

This is a direct application of the determinant property

\[ \det(AB)=\det(A)\det(B). \]

For an invertible matrix, the identity

\[ AA^{-1}=I \]

immediately leads to

\[ \boxed{\det(A^{-1})=\frac{1}{\det(A)}}. \]

This result is useful in questions involving determinants of inverse matrices, matrix inverses, and verification of matrix identities.

Significance for Competitive Entrance Examinations

For an MCQ, the quickest method is:

\[ AA^{-1}=I \]

\[ \Rightarrow\det(A)\det(A^{-1})=1 \]

\[ \boxed{\therefore\det(A^{-1})=\frac{1}{\det(A)}}. \]

No explicit calculation of \(A^{-1}\) is necessary.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

\(A\) is invertible if and only if
\[ \det(A)\neq0. \]
The inverse satisfies
\[ AA^{-1}=I. \]
The determinant of a product satisfies
\[ \det(AB)=\det(A)\det(B). \]
Therefore,
\[ \boxed{\det(A^{-1})=\frac{1}{\det(A)}}. \]
The result is valid for every invertible square matrix, not only for matrices of order \(2\times2\).

Determinants

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

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Determinants — Learning Resources

Frequently Asked Questions

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Determinants

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Chapter Complete!

Recent posts

Determinants — Learning Resources

Frequently Asked Questions

Q 01 What is the main topic of these NCERT Class 12 Mathematics MCQs?

Q 02 What is the adjoint of a matrix?

Q 03 What is the fundamental property of the adjoint of a matrix?

Q 04 When does the inverse of a matrix exist?

Q 05 What is the formula for the inverse of a matrix using its adjoint?

Q 06 What is the determinant of the inverse of a matrix?

Q 07 Does the order of a matrix affect the formula for det(A?¹)?

Q 08 Why is det(A) non-zero for an invertible matrix?

Q 09 Are these Determinants MCQs useful for CBSE Board examinations?

Q 10 How can these Determinants MCQs help in competitive examinations?

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