NCERT Class 12 Determinants Miscellaneous Exercise Solutions

Q1

NUMERIC3 marks

Prove that the determinant

\[ \begin{vmatrix} x & \sin\theta & \cos\theta\\ -\sin\theta & -x & 1\\ \cos\theta & 1 & x \end{vmatrix} \] is independent of \(\theta\)

📘 Concept & Theory

Concept/Theory

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A determinant is said to be independent of a variable if, after simplification, its value does not contain that variable. Here, the determinant contains \(\sin\theta\) and \(\cos\theta\), so the objective is to simplify it and show that all terms involving \(\theta\) either cancel or combine using a trigonometric identity.

The most suitable method is expansion along the first row. For a determinant

\[ \begin{vmatrix} a & b & c\\ d & e & f\\ g & h & i \end{vmatrix}, \]

expansion along the first row gives

\[ a \begin{vmatrix} e & f\\ h & i \end{vmatrix} - b \begin{vmatrix} d & f\\ g & i \end{vmatrix} + c \begin{vmatrix} d & e\\ g & h \end{vmatrix} \]

We will also use the fundamental trigonometric identity

\[ \sin^2\theta+\cos^2\theta=1. \]

Thus, if the final expression contains only \(x\) and no \(\theta\), the determinant will be proved independent of \(\theta\).

🗺️ Solution Roadmap

Step-by-step Plan

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Expand the determinant along the first row.
Evaluate each of the three \(2\times2\) minors carefully.
Multiply each minor by its corresponding cofactor term, observing the signs \(+,-,+\).
Combine the resulting terms.
Cancel the mixed terms containing \(\sin\theta\cos\theta\).
Use \(\sin^2\theta+\cos^2\theta=1\).
Show that the final value contains no \(\theta\), establishing the required result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 18 steps

Let
\[ D= \begin{vmatrix} x & \sin\theta & \cos\theta\\ -\sin\theta & -x & 1\\ \cos\theta & 1 & x \end{vmatrix} \]
We expand \(D\) along the first row. The signs of the cofactors in the first row are
\[ +,\quad -,\quad +. \]
Therefore,
\[ D=x\begin{vmatrix}-x & 1\\1 & x\end{vmatrix}- \sin\theta\begin{vmatrix}-\sin\theta & 1\\\cos\theta & x\end{vmatrix}+ \cos\theta\begin{vmatrix}-\sin\theta & -x\\\cos\theta & 1\end{vmatrix}\]
First minor
\[\begin{aligned}\begin{vmatrix}-x & 1\\1 & x\end{vmatrix}&=(-x)(x)-(1)(1)\\&=-x^2-1\end{aligned}\]
Hence, the first term is
\[x(-x^2-1)\]
Second minor
\[\begin{aligned}\begin{vmatrix}-\sin\theta & 1\\\cos\theta & x\end{vmatrix}&=(-\sin\theta)(x)-(1)(\cos\theta)\\&=-x\sin\theta-\cos\theta\end{aligned}\]
Therefore, the second term is
\[-\sin\theta(-x\sin\theta-\cos\theta)\]
Third minor
\[\begin{aligned}\begin{vmatrix}-\sin\theta & -x\\\cos\theta & 1\end{vmatrix} &=(-\sin\theta)(1)-(-x)(\cos\theta)\\ &=-\sin\theta+x\cos\theta \end{aligned} \]
Therefore, the third term is
\[\cos\theta(-\sin\theta+x\cos\theta)\]
Substituting all three evaluated minors, we obtain
\[ D=x(-x^2-1) -\sin\theta(-x\sin\theta-\cos\theta) +\cos\theta(-\sin\theta+x\cos\theta). \]
Now distribute the factors term by term:
\[x(-x^2-1)=-x^3-x\]

\[ -\sin\theta(-x\sin\theta-\cos\theta) = x\sin^2\theta+\sin\theta\cos\theta, \]

\[ \cos\theta(-\sin\theta+x\cos\theta) = -\sin\theta\cos\theta+x\cos^2\theta. \]
Hence,
\[ D=-x^3-x+x\sin^2\theta+\sin\theta\cos\theta -\sin\theta\cos\theta+x\cos^2\theta. \]
The two mixed terms cancel each other:
\[\sin\theta\cos\theta-\sin\theta\cos\theta=0\]
Therefore,
\[D=-x^3-x+x\sin^2\theta+x\cos^2\theta.\]
Taking \(x\) common from the last two terms
\[D=-x^3-x+x(\sin^2\theta+\cos^2\theta)\]
Using the identity

\[ \sin^2\theta+\cos^2\theta=1, \]
we get
\[D=-x^3-x+x(1)\]
Thus,
\[D=-x^3-x+x\]
Therefore,
\[\boxed{D=-x^3}\]
Since the final value
\[ -x^3 \]
contains no \(\theta\), the value of the determinant is independent of \(\theta\).

🎯 Exam Significance

Exam Significance

›

This problem tests several important Class 12 determinant skills simultaneously: expansion along a row, correct application of cofactor signs, evaluation of \(2\times2\) determinants, algebraic simplification, cancellation of terms, and application of the identity

\[ \sin^2\theta+\cos^2\theta=1. \]

For CBSE Board examinations, writing these intermediate steps is important because the question asks for a proof, and the method carries substantial value even before the final conclusion is reached.

A particularly important examination point is that the expression should not merely be simplified to a numerical-looking result. The final statement must explicitly establish that the resulting determinant contains no \(\theta\). Therefore,

\[ D=-x^3 \]

directly proves the required independence.

Significance for Competitive Entrance Examinations

For competitive examinations, this question develops the ability to recognize cancellation patterns quickly. The terms

\[ \sin\theta\cos\theta \quad\text{and}\quad -\sin\theta\cos\theta \]

cancel, while the remaining trigonometric terms combine as

\[ x\sin^2\theta+x\cos^2\theta=x. \]

Recognizing this structure can substantially reduce calculation time.

The problem also reinforces an important determinant strategy: when a determinant contains trigonometric expressions, do not attempt to assign particular values to the angle. Instead, simplify symbolically and look for standard identities such as

\[ \sin^2\theta+\cos^2\theta=1. \]

This approach is useful in JEE and other entrance examinations where speed and algebraic pattern recognition are important.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 9 points

For a \(3\times3\) determinant, expansion along the first row uses the cofactor signs \(+,-,+\).
Evaluate every \(2\times2\) minor carefully before performing the final simplification.
When a determinant contains \(\sin\theta\) and \(\cos\theta\), look for cancellation and the identity \(\sin^2\theta+\cos^2\theta=1\).
To prove independence of a variable, simplify completely and verify that the variable disappears from the final expression.
The mixed terms \(\sin\theta\cos\theta\) cancel exactly in this determinant.
The trigonometric terms reduce to \(x(\sin^2\theta+\cos^2\theta)=x\).
The determinant has the remarkably simple value \(\boxed{-x^3}\), regardless of the value of \(\theta\).
For Board examinations, show the expansion, minors, simplification, identity, and final conclusion explicitly.
For competitive examinations, learn to identify cancellation patterns and standard identities rapidly.

📘 Concept & Theory

Concept/Theory

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To evaluate a determinant efficiently, we should choose the row or column that contains the most convenient entries. In this determinant, the third column is

\[ \begin{pmatrix} -\sin\alpha\\ 0\\ \cos\alpha \end{pmatrix}, \]

which contains a zero. Therefore, expansion along the third column is the most efficient approach because the zero makes the middle cofactor vanish.

For expansion along the third column of a \(3\times3\) determinant, the cofactor signs are

\[ +,\quad -,\quad +. \]

Thus, if the third-column entries are \(c_{13},c_{23},c_{33}\), the expansion is

\[ c_{13}C_{13}+c_{23}C_{23}+c_{33}C_{33}. \]

Since \(c_{23}=0\), only the first and third terms need to be calculated.

The calculation also uses the fundamental trigonometric identity

\[ \sin^2\theta+\cos^2\theta=1. \]

We will apply this identity first to \(\beta\) and then to \(\alpha\).

🗺️ Solution Roadmap

Step-by-step Plan

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Observe that the third column contains a zero.
Expand the determinant along the third column.
Evaluate the two resulting \(2\times2\) determinants separately.
Use \(\sin^2\beta+\cos^2\beta=1\) to simplify both minors.
Carefully account for the negative sign associated with the entry \(-\sin\alpha\).
Obtain an expression involving \(\sin^2\alpha+\cos^2\alpha\).
Use \(\sin^2\alpha+\cos^2\alpha=1\) to obtain the final value.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 16 steps

Let
\[ D= \begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta & -\sin\alpha\\ -\sin\beta & \cos\beta & 0\\ \sin\alpha\cos\beta & \sin\alpha\sin\beta & \cos\alpha \end{vmatrix} \]
Since the third column contains the zero entry \(0\), we expand \(D\) along the third column.
\[ D= (-\sin\alpha) \begin{vmatrix} -\sin\beta & \cos\beta\\ \sin\alpha\cos\beta & \sin\alpha\sin\beta \end{vmatrix} + 0 \begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta\\ \sin\alpha\cos\beta & \sin\alpha\sin\beta \end{vmatrix} + \cos\alpha \begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta\\ -\sin\beta & \cos\beta \end{vmatrix} \]
The middle term is zero because it is multiplied by \(0\). Hence,
\[ D= (-\sin\alpha) \begin{vmatrix} -\sin\beta & \cos\beta\\ \sin\alpha\cos\beta & \sin\alpha\sin\beta \end{vmatrix} + \cos\alpha \begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta\\ -\sin\beta & \cos\beta \end{vmatrix} \]
Evaluating the First Minor
\[ \begin{vmatrix} a & b\\ c & d \end{vmatrix} =ad-bc, \]
we get
\[ \begin{aligned} \begin{vmatrix} -\sin\beta & \cos\beta\\ \sin\alpha\cos\beta & \sin\alpha\sin\beta \end{vmatrix} &= (-\sin\beta)(\sin\alpha\sin\beta) - (\cos\beta)(\sin\alpha\cos\beta)\\ &= -\sin\alpha\sin^2\beta - \sin\alpha\cos^2\beta \end{aligned} \]
Taking \(-\sin\alpha\) common,
\[ = -\sin\alpha \left(\sin^2\beta+\cos^2\beta\right). \]
Using

\[ \sin^2\beta+\cos^2\beta=1, \]
the first minor becomes
\[-\sin\alpha\]
Therefore, the contribution of the first entry of the third column is
\[ (-\sin\alpha)(-\sin\alpha) = \sin^2\alpha. \]
Evaluating the Second Non-zero Minor
Now consider
\[ \begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta\\ -\sin\beta & \cos\beta \end{vmatrix}. \]
Using the \(2\times2\) determinant formula,
\[ \begin{aligned} \begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta\\ -\sin\beta & \cos\beta \end{vmatrix} &= (\cos\alpha\cos\beta)(\cos\beta) - (\cos\alpha\sin\beta)(-\sin\beta)\\ &= \cos\alpha\cos^2\beta + \cos\alpha\sin^2\beta \end{aligned} \]
Taking \(\cos\alpha\) common,
\[ = \cos\alpha \left(\cos^2\beta+\sin^2\beta\right). \]
Using

\[ \cos^2\beta+\sin^2\beta=1, \]
the minor becomes
\[\cos\alpha\]
Therefore, the contribution of the third entry of the third column is
\[ (\cos\alpha)(\cos\alpha) = \cos^2\alpha. \]
Combining the Two Contributions. Thus,
\[D=\sin^2\alpha+\cos^2\alpha\]
Again, using the fundamental identity

\[ \sin^2\alpha+\cos^2\alpha=1, \]
we obtain
\[D=1\]

🎯 Exam Significance

Exam Significance

›

This problem is an excellent application of cofactor expansion. A key examination skill is choosing the row or column that minimizes computation. Since the third column contains a zero, expansion along this column reduces the \(3\times3\) determinant to only two \(2\times2\) determinants.

The problem also tests accurate handling of signs. In particular, the entry in the first position of the third column is \(-\sin\alpha\), and the corresponding cofactor has a positive sign. Missing either sign can lead to an incorrect result. Writing the \(2\times2\) determinant calculations explicitly makes the solution logically complete and helps secure method marks.

Significance for Competitive Entrance Examinations

For competitive examinations, the main lesson is strategic expansion. Instead of expanding along the first or second column, immediately notice the zero in the third column. This reduces the number of calculations and makes the determinant almost mechanical to evaluate.

There is also a useful structural observation: each \(2\times2\) minor contains either

\[ \sin^2\beta+\cos^2\beta \]

or

\[ \cos^2\beta+\sin^2\beta. \]

Both expressions equal \(1\). Recognizing this pattern quickly is particularly valuable in JEE and other entrance examinations, where determinant questions may be designed to reward pattern recognition rather than lengthy expansion.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 10 points

Choose the row or column containing the most zeros or simplest entries for cofactor expansion.
The third column is the natural choice here because it contains the zero entry.
For expansion along the third column, the cofactor signs are \(+,-,+\).
A \(2\times2\) determinant is evaluated using \(ad-bc\).
Use \(\sin^2\beta+\cos^2\beta=1\) to simplify the two minors.
Keep track of the negative sign in the entry \(-\sin\alpha\).
After simplification, the determinant becomes \(\sin^2\alpha+\cos^2\alpha\).
Finally, \(\sin^2\alpha+\cos^2\alpha=1\).
The value of the determinant is therefore \(\boxed{1}\).
For entrance examinations, identifying the best expansion route can save substantial calculation time.

📘 Concept & Theory

Concept/Theory

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The question directly tests the important inverse-matrix property

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

provided \(A\) and \(B\) are invertible.

The matrix \(A^{-1}\) is already given. Therefore, there is no need to determine \(A\) or calculate \(A^{-1}\) again. We only need to calculate \(B^{-1}\) and then multiply it by the given matrix \(A^{-1}\).

To calculate the inverse of \(B\), we use

\[ B^{-1}=\frac{1}{|B|}\operatorname{adj}(B), \]

where \(\operatorname{adj}(B)\) is the transpose of the cofactor matrix of \(B\).

Thus, the solution involves three main stages:

Find \(|B|\) and verify that \(B\) is invertible.
Find the cofactors of \(B\), form the cofactor matrix, and transpose it to obtain \(\operatorname{adj}(B)\).
Use \((AB)^{-1}=B^{-1}A^{-1}\) and perform the matrix multiplication.

🗺️ Solution Roadmap

Step-by-step Plan

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Write the inverse-product property.
Calculate the determinant \(|B|\).
Since \(|B|\neq0\), conclude that \(B^{-1}\) exists.
Calculate all nine cofactors of \(B\).
Form the cofactor matrix.
Transpose the cofactor matrix to obtain \(\operatorname{adj}(B)\).
Calculate \(B^{-1}\).
Multiply \(B^{-1}\) by the given \(A^{-1}\).
State the resulting matrix as \((AB)^{-1}\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 33 steps

We know that
\[\boxed{(AB)^{-1}=B^{-1}A^{-1}}\]
Given — We are already given
\[ A^{-1}= \begin{bmatrix} 3 & -1 & 1\\ -15 & 6 & -5\\ 5 & -2 & 2 \end{bmatrix} \]
Therefore, our first task is to find \(B^{-1}\).
Find the Determinant of \(B\)
Given —
\[B=\begin{bmatrix}1 & 2 & -2\\-1 & 3 & 0\\0 & -2 & 1\end{bmatrix}\]
Expanding \(|B|\) along the first row,
\[\begin{aligned}|B|&=1\begin{vmatrix}3 & 0\\-2 & 1\end{vmatrix}- 2\begin{vmatrix}-1 & 0\\0 & 1\end{vmatrix}+ (-2)\begin{vmatrix}-1 & 3\\0 & -2\end{vmatrix}\end{aligned}\]
Evaluating the \(2\times2\) determinants,
\[ \begin{aligned} |B| &= 1(3\times1-0\times(-2)) - 2((-1)\times1-0\times0)\\ &\quad +(-2)((-1)\times(-2)-3\times0)\\ &=1(3)-2(-1)+(-2)(2)\\ &=3+2-4\\ &=1 \end{aligned} \]
Since
\[ |B|=1\neq0, \]
the matrix \(B\) is non-singular and hence \(B^{-1}\) exists.
Find the Cofactors of \(B\)
Let \(C_{ij}\) denote the cofactor corresponding to the element in the \(i\)-th row and \(j\)-th column. We use
\[ C_{ij}=(-1)^{i+j}M_{ij}, \]
where \(M_{ij}\) is the corresponding minor.
The cofactor signs for a \(3\times3\) matrix are
\[ \begin{bmatrix} + & - & +\\ - & + & -\\ + & - & + \end{bmatrix} \]
First Row Cofactors
For \(C_{11}\),
\[\begin{aligned}C_{11}&=(+)\begin{vmatrix}3 & 0\\-2 & 1\end{vmatrix}\\&=3(1)-0(-2)\\&=3\end{aligned}\]
For \(C_{12}\),
\[\begin{aligned}C_{12}&=(-)\begin{vmatrix}-1 & 0\\0 & 1\end{vmatrix}\\&=-\left((-1)(1)-0(0)\right)\\&=-(-1)\\&=1\end{aligned}\]
For \(C_{13}\),
\[\begin{aligned}C_{13}&=(+)\begin{vmatrix}-1 & 3\\0 & -2\end{vmatrix}\\&=(-1)(-2)-3(0)\\&=2\end{aligned}\]
Thus, the first row of the cofactor matrix is
\[\begin{bmatrix}3 & 1 & 2\end{bmatrix}\]
Second Row Cofactors
For \(C_{21}\),
\[\begin{aligned}C_{21}&=(-)\begin{vmatrix}2 & -2\\-2 & 1\end{vmatrix}\\&=-\left(2(1)-(-2)(-2)\right)\\&=-(2-4)\\&=2\end{aligned}\]
For \(C_{22}\),
\[\begin{aligned}C_{22}&=(+)\begin{vmatrix}1 & -2\\0 & 1\end{vmatrix}\\&=1(1)-(-2)(0)\\&=1\end{aligned}\]
For \(C_{23}\),
\[\begin{aligned}C_{23}&=(-)\begin{vmatrix}1 & 2\\0 & -2\end{vmatrix}\\&=-\left(1(-2)-2(0)\right)\\&=-(-2)\\&=2\end{aligned}\]
Thus, the second row of the cofactor matrix is
\[\begin{bmatrix}2 & 1 & 2\end{bmatrix}\]
Third Row Cofactors
For \(C_{31}\),
\[\begin{aligned}C_{31}&=(+)\begin{vmatrix}2 & -2\\3 & 0\end{vmatrix}\\&=2(0)-(-2)(3)\\&=6\end{aligned}\]
For \(C_{32}\),
\[\begin{aligned}C_{32}&=(-)\begin{vmatrix}1 & -2\\-1 & 0\end{vmatrix}\\&=-\left(1(0)-(-2)(-1)\right)\\&=-(-2)\\&=2\end{aligned}\]
For \(C_{33}\),
\[\begin{aligned}C_{33}&=(+)\begin{vmatrix}1 & 2\\-1 & 3\end{vmatrix}\\&=1(3)-2(-1)\\&=3+2\\&=5\end{aligned}\]
Thus, the third row of the cofactor matrix is
\[\begin{bmatrix}6 & 2 & 5\end{bmatrix}\]
Form the Cofactor Matrix
Therefore, the cofactor matrix of \(B\) is
\[C=\begin{bmatrix}3 & 1 & 2\\2 & 1 & 2\\6 & 2 & 5\end{bmatrix}\]
Find the Adjoint of \(B\)
The adjoint of a matrix is the transpose of its cofactor matrix:
\[ \operatorname{adj}(B)=C^{T}. \]
Hence,
\[\operatorname{adj}(B)=\begin{bmatrix}3 & 2 & 6\\1 & 1 & 2\\2 & 2 & 5\end{bmatrix}\]
Find \(B^{-1}\)
use
\[ B^{-1}=\frac{1}{|B|}\operatorname{adj}(B). \]
Since \(|B|=1\)
\[\begin{aligned}B^{-1}&=\frac{1}{1}\begin{bmatrix}3 & 2 & 6\\1 & 1 & 2\\2 & 2 & 5\end{bmatrix}\\ &=\begin{bmatrix}3 & 2 & 6\\1 & 1 & 2\\2 & 2 & 5\end{bmatrix}\end{aligned}\]
Calculate \((AB)^{-1}\)
Using
\[ (AB)^{-1}=B^{-1}A^{-1}, \]
\[\begin{aligned}(AB)^{-1}&=\begin{bmatrix}3 & 2 & 6\\1 & 1 & 2\\2 & 2 & 5\end{bmatrix}\begin{bmatrix}3 & -1 & 1\\-15 & 6 & -5\\5 & -2 & 2\end{bmatrix}.\end{aligned}\]
We now calculate each entry of the resulting matrix.
First row, first column:
\[(3)(3)+(2)(-15)+(6)(5)=9-30+30=9\]
First row, second column:
\[(3)(-1)+(2)(6)+(6)(-2)=-3+12-12=-3\]
First row, third column:
\[(3)(1)+(2)(-5)+(6)(2)=3-10+12=5\]
Second row, first column:
\[(1)(3)+(1)(-15)+(2)(5)=3-15+10=-2\]
Second row, second column:
\[(1)(-1)+(1)(6)+(2)(-2)=-1+6-4=1\]
Second row, third column:
\[(1)(1)+(1)(-5)+(2)(2)=1-5+4=0\]
Third row, first column:
\[(2)(3)+(2)(-15)+(5)(5)=6-30+25=1\]
Third row, second column:
\[(2)(-1)+(2)(6)+(5)(-2)=-2+12-10=0\]
Third row, third column:
\[(2)(1)+(2)(-5)+(5)(2)=2-10+10=2\]
Therefore,
\[ \boxed{\bbox[5pt]{ (AB)^{-1} = \begin{bmatrix} 9 & -3 & 5\\ -2 & 1 & 0\\ 1 & 0 & 2 \end{bmatrix}}} \]

💡 Answer

Final Answer

›

\[ \boxed{\bbox[5pt]{ (AB)^{-1} = \begin{bmatrix} 9 & -3 & 5\\ -2 & 1 & 0\\ 1 & 0 & 2 \end{bmatrix} }} \]

🎯 Exam Significance

Exam Significance

›

This problem combines several high-value concepts from the chapter Determinants and the related matrix-inverse methods: determinant evaluation, cofactors, adjoint, inverse of a matrix, and the inverse of a product.

For Board examinations, the most important point is to distinguish clearly between the cofactor matrix and the adjoint matrix. The adjoint is obtained by taking the transpose of the cofactor matrix:

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

Writing the cofactor calculations explicitly also demonstrates the method and helps avoid losing marks through sign errors.

Another important examination point is the order of multiplication:

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

not \(A^{-1}B^{-1}\). Matrix multiplication is generally not commutative, so reversing the order changes the answer.

Significance for Competitive Entrance Examinations

This question illustrates an important time-saving strategy. Since \(A^{-1}\) is already supplied, there is no reason to calculate \(A\) first. Directly use

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

Also, because \(|B|=1\), the inverse of \(B\) is simply its adjoint:

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

Recognizing this immediately eliminates an unnecessary scalar division.

For JEE and other competitive entrance examinations, remembering the reversed order in

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

and identifying determinants equal to \(1\) can significantly reduce calculation time.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 9 points

The inverse of a product is given by
\[ (AB)^{-1}=B^{-1}A^{-1}. \]
The order of the matrices is reversed when taking the inverse of a product.
The inverse of \(B\) exists because
\[ |B|=1\neq0. \]
The inverse formula is
\[ B^{-1}=\frac{1}{|B|}\operatorname{adj}(B). \]
The cofactor matrix and adjoint matrix are different:
\[ \operatorname{adj}(B)=\left(\text{cofactor matrix of }B\right)^T. \]
Here,
\[ \operatorname{adj}(B)= \begin{bmatrix} 3 & 2 & 6\\ 1 & 1 & 2\\ 2 & 2 & 5 \end{bmatrix}. \]
Since \(|B|=1\),
\[ B^{-1}=\operatorname{adj}(B). \]
The final result is
\[ \boxed{ (AB)^{-1}= \begin{bmatrix} 9 & -3 & 5\\ -2 & 1 & 0\\ 1 & 0 & 2 \end{bmatrix} }. \]
For competitive examinations, use the given \(A^{-1}\) directly rather than reconstructing \(A\).

Q4

NUMERIC3 marks

Let

\[ A= \begin{bmatrix} 1 & 2 & 1\\ 2 & 3 & 1\\ 1 & 1 & 5 \end{bmatrix} \]

Verify that

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

📘 Concept & Theory

Concept/Theory

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The adjoint of a square matrix is obtained by taking the transpose of its cofactor matrix. An important identity connecting a matrix with its adjoint is

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

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

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

Another fundamental property is

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

In this question, we first calculate the cofactors of \(A\), form \(\operatorname{adj}A\), and calculate \(|A|\). These results allow us to verify the required inverse relationships directly.

🗺️ Solution Roadmap

Step-by-step Plan

›

Calculate all nine cofactors of \(A\).
Form the cofactor matrix.
Transpose it to obtain \(\operatorname{adj}A\).
Calculate \(|A|\).
Use \((\operatorname{adj}A)A=|A|I\).
Use this result to evaluate \(\left[(\operatorname{adj}A)A\right]^{-1}\).
Calculate \(A^{-1}\) and then \(\operatorname{adj}(A^{-1})\) to examine part (i).
Use the inverse-of-an-inverse property to verify part (ii).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 29 steps

Given —
\[A=\begin{bmatrix}1 & 2 & 1\\2 & 3 & 1\\1 & 1 & 5\end{bmatrix}\]
Find the Cofactors of \(A\)
The cofactor \(C_{ij}\) is given by
\[ C_{ij}=(-1)^{i+j}M_{ij} \]
The cofactor signs are
\[ \begin{bmatrix} + & - & +\\ - & + & -\\ + & - & + \end{bmatrix}\]
First Row Cofactors
For \(C_{11}\),
\[\begin{aligned}C_{11}&=(+)\begin{vmatrix}3 & 1\\1 & 5\end{vmatrix}\\&=3(5)-1(1)\\&=15-1\\&=14\end{aligned}\]
For \(C_{12}\),
\[\begin{aligned}C_{12}&=(-)\begin{vmatrix}2 & 1\\1 & 5\end{vmatrix}\\&=-\left[2(5)-1(1)\right]\\&=-(10-1)\\&=-9\end{aligned}\]
For \(C_{13}\),
\[\begin{aligned}C_{13}&=(+)\begin{vmatrix}2 & 3\\1 & 1\end{vmatrix}\\&=2(1)-3(1)\\&=2-3\\&=-1\end{aligned}\]
Second Row Cofactors
For \(C_{21}\),
\[\begin{aligned}C_{21}&=(-)\begin{vmatrix}2 & 1\\1 & 5\end{vmatrix}\\&=-\left[2(5)-1(1)\right]\\&=-(10-1)\\&=-9\end{aligned}\]
For \(C_{22}\),
\[\begin{aligned}C_{22}&=(+)\begin{vmatrix}1 & 1\\1 & 5\end{vmatrix}\\&=1(5)-1(1)\\&=5-1\\&=4\end{aligned}\]
For \(C_{23}\),
\[\begin{aligned}C_{23}&=(-)\begin{vmatrix}1 & 2\\1 & 1\end{vmatrix}\\&=-\left[1(1)-2(1)\right]\\&=-(1-2)\\&=1\end{aligned}\]
Third Row Cofactors
For \(C_{31}\),
\[\begin{aligned}C_{31}&=(+)\begin{vmatrix}2 & 1\\3 & 1\end{vmatrix}\\&=2(1)-1(3)\\&=2-3\\&=-1\end{aligned}\]
For \(C_{32}\),
\[\begin{aligned}C_{32}&=(-)\begin{vmatrix}1 & 1\\2 & 1\end{vmatrix}\\&=-\left[1(1)-1(2)\right]\\&=-(1-2)\\&=1\end{aligned}\]
For \(C_{33}\),
\[\begin{aligned}C_{33}&=(+)\begin{vmatrix}1 & 2\\2 & 3\end{vmatrix}\\&=1(3)-2(2)\\&=3-4\\&=-1\end{aligned}\]
Form the Cofactor Matrix
Therefore, the cofactor matrix is
\[C=\begin{bmatrix}14 & -9 & -1\\-9 & 4 & 1\\-1 & 1 & -1\end{bmatrix}\]
Since \(C\) is symmetric, its transpose is the same matrix. Hence,
\[ \operatorname{adj}A=C^T=C. \]
Therefore,
\[\boxed{\bbox[5pt]{\operatorname{adj}A=\begin{bmatrix}14 & -9 & -1\\-9 & 4 & 1\\-1 & 1 & -1\end{bmatrix}}}\]
Find \(|A|\)
Expanding \(|A|\) along the first row,
\[\begin{aligned}|A|&=1\begin{vmatrix}3 & 1\\1 & 5\end{vmatrix}- 2\begin{vmatrix}2 & 1\\1 & 5\end{vmatrix}+ 1\begin{vmatrix}2 & 3\\1 & 1\end{vmatrix}\\&=1(15-1)-2(10-1)+(2-3)\\&=14-18-1\\&=-5\end{aligned}\]
Thus,
\[\boxed{|A|=-5\neq0}\]
Therefore, \(A\) is non-singular and \(A^{-1}\) exists.
Part (i)
Using the fundamental identity
\[ (\operatorname{adj}A)A=|A|I, \]
and \(|A|=-5\),
we obtain
\[(\operatorname{adj}A)A=-5I\]
Hence,
\[\begin{aligned}\left[(\operatorname{adj}A)A\right]^{-1}&=(-5I)^{-1}\\&=-\frac{1}{5}I\end{aligned}\]
Therefore,
\[ \boxed{ \left[(\operatorname{adj}A)A\right]^{-1} = -\frac{1}{5}I } \]
Now, from
\[ A^{-1}=\frac{1}{|A|}\operatorname{adj}A, \]
we get
\[ A^{-1} = -\frac{1}{5} \begin{bmatrix} 14 & -9 & -1\\ -9 & 4 & 1\\ -1 & 1 & -1 \end{bmatrix} \]
For any non-singular \(3\times3\) matrix \(M\),
\[ \operatorname{adj}(M)=|M|M^{-1} \]
Taking \(M=A^{-1}\), we have
\[ \operatorname{adj}(A^{-1}) = |A^{-1}|(A^{-1})^{-1} \]
Since
\[|A^{-1}|=\frac{1}{|A|}=-\frac{1}{5}\]
and
\[(A^{-1})^{-1}=A\]
it follows that
\[ \operatorname{adj}(A^{-1}) = -\frac{1}{5}A \]
Thus,
\[\operatorname{adj}(A^{-1})=-\frac{1}{5}\begin{bmatrix}1 & 2 & 1\\2 & 3 & 1\\1 & 1 & 5\end{bmatrix}\]
Consequently,
\[ \left[(\operatorname{adj}A)A\right]^{-1} = -\frac15 I \]
whereas
\[ \operatorname{adj}(A^{-1}) = -\frac15 A. \]
These two matrices are not equal for the given matrix \(A\).
Hence, the identity in part (i), as written, does not hold for the given matrix. The correct identity obtained from the standard adjoint properties is
\[ \boxed{\bbox[5pt]{ \left[(\operatorname{adj}A)A\right]^{-1} = \frac{1}{|A|}I }} \]
and, separately,
\[ \boxed{\bbox[5pt]{ \operatorname{adj}(A^{-1}) = \frac{1}{|A|}A }} \]
Part (ii)
We know that the inverse of \(A\) is the unique matrix satisfying
\[ AA^{-1}=A^{-1}A=I. \]
Therefore, \(A\) is itself the inverse of \(A^{-1}\). Hence,
\[\boxed{(A^{-1})^{-1}=A}\]
Thus, part (ii) is verified.

🎯 Exam Significance

Exam Significance

›

This question brings together several fundamental concepts: cofactors, adjoint, determinant, inverse of a matrix, and inverse-of-an-inverse. For Board examinations, it is important to maintain the distinction between the cofactor matrix and the adjoint matrix.

The central identity

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

is especially important. Once \(|A|\) has been calculated, it can often simplify an apparently complicated matrix expression immediately.

Significance for Competitive Entrance Examinations

Competitive examinations frequently test whether a student can recognize matrix identities without performing unnecessary multiplication. Here,

\[ (\operatorname{adj}A)A=|A|I=-5I \]

can be obtained immediately from the standard theorem.

Another important result is

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

for a non-singular matrix \(A\). Such identities are considerably faster to use than calculating every cofactor from scratch.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

\[ A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I. \]
\[ A^{-1}=\frac{1}{|A|}\operatorname{adj}A. \]
For the given matrix,
\[ |A|=-5. \]
The adjoint is
\[ \operatorname{adj}A= \begin{bmatrix} 14 & -9 & -1\\ -9 & 4 & 1\\ -1 & 1 & -1 \end{bmatrix} \]
\[ (\operatorname{adj}A)A=-5I. \]
\[ \left[(\operatorname{adj}A)A\right]^{-1} =-\frac15I. \]
\[ \operatorname{adj}(A^{-1})=-\frac15A. \]
The identity stated in part (i) is not valid for the given matrix as written.
For every non-singular matrix,
\[ (A^{-1})^{-1}=A. \]

📘 Concept & Theory

Concept/Theory

›

To evaluate a determinant of order \(3\), we may expand it along any row or column. When expanding along the first row, the corresponding cofactor signs are

\[+,\quad -,\quad +\]

Thus, for

\[\begin{vmatrix}a & b & c\\d & e & f\\g & h & i\end{vmatrix},\]

expansion along the first row gives

\[a\begin{vmatrix}e & f\\h & i\end{vmatrix}- b\begin{vmatrix}d & f\\g & i\end{vmatrix}+ c\begin{vmatrix}d & e\\g & h\end{vmatrix} \]

For a \(2\times2\) determinant,

\[\begin{vmatrix}a & b\\c & d\end{vmatrix}=ad-bc\]

In this problem, careful expansion followed by collection of like terms leads to complete cancellation of the mixed terms \(x^2y\) and \(xy^2\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Expand the determinant along the first row.
Evaluate each of the three \(2\times2\) minors.
Multiply each minor by its corresponding first-row element and cofactor sign.
Expand and simplify each contribution separately.
Combine all three contributions.
Collect like terms and obtain the final expression.
Optionally factor the result using the sum-of-cubes identity.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 26 steps

Let
\[D=\begin{vmatrix}x & y & x+y\\y & x+y & x\\x+y & x & y\end{vmatrix}\]
Expanding \(D\) along the first row, using the cofactor signs \(+,-,+\),
\[\begin{aligned}D&=x\begin{vmatrix}x+y & x\\x & y\end{vmatrix}- y\begin{vmatrix}y & x\\x+y & y\end{vmatrix}+ (x+y)\begin{vmatrix}y & x+y\\x+y & x\end{vmatrix}\end{aligned}\]
First Contribution
Consider
\[x\begin{vmatrix}x+y & x\\x & y\end{vmatrix}\]
Evaluating the \(2\times2\) determinant,
\[\begin{aligned}\begin{vmatrix}x+y & x\\x & y\end{vmatrix}&=(x+y)y-x(x)\\&=xy+y^2-x^2\end{aligned}\]
Therefore,
\[\begin{aligned}x(xy+y^2-x^2)&=x^2y+xy^2-x^3\end{aligned}\]
Thus, the first contribution is
\[\boxed{x^2y+xy^2-x^3}\]
Second Contribution
Consider
\[-y\begin{vmatrix}y & x\\x+y & y\end{vmatrix}\]
Evaluating the \(2\times2\) determinant,
\[\begin{aligned}\begin{vmatrix}y & x\\x+y & y\end{vmatrix}&=y(y)-x(x+y)\\&=y^2-x^2-xy\end{aligned}\]
Hence,
\[\begin{aligned}-y(y^2-x^2-xy)&=-y^3+x^2y+xy^2\end{aligned}\]
Thus, the second contribution is
\[\boxed{-y^3+x^2y+xy^2}\]
Third Contribution
Consider
\[(x+y)\begin{vmatrix}y & x+y\\x+y & x\end{vmatrix}\]
Evaluating the \(2\times2\) determinant,
\[\begin{aligned}\begin{vmatrix}y & x+y\\x+y & x\end{vmatrix}&=xy-(x+y)(x+y)\\&=xy-(x+y)^2\end{aligned}\]
Now,
\[(x+y)^2=x^2+2xy+y^2\]
Therefore,
\[\begin{aligned}xy-(x+y)^2&=xy-(x^2+2xy+y^2)\\&=xy-x^2-2xy-y^2\\&=-x^2-xy-y^2\end{aligned}\]
Hence, the third contribution becomes
\[\begin{aligned}(x+y)(-x^2-xy-y^2)&=-\left(x+y\right)\left(x^2+xy+y^2\right)\end{aligned}\]
Expanding the product,
\[\begin{aligned}&(x+y)(x^2+xy+y^2)\\ &=x(x^2+xy+y^2)+y(x^2+xy+y^2)\\ &=x^3+x^2y+xy^2+x^2y+xy^2+y^3\\ &=x^3+2x^2y+2xy^2+y^3\end{aligned}\]
Therefore,
\[\begin{aligned}(x+y)(-x^2-xy-y^2)&=-x^3-2x^2y-2xy^2-y^3\end{aligned}\]
Thus, the third contribution is
\[\boxed{-x^3-2x^2y-2xy^2-y^3}\]
Combine the Three Contributions
Therefore,
\[\begin{aligned}D&=\left(x^2y+xy^2-x^3\right)+\left(-y^3+x^2y+xy^2\right)\\&\quad+\left(-x^3-2x^2y-2xy^2-y^3\right)\end{aligned}\]
Now collect the \(x^3\) terms:
\[-x^3-x^3=-2x^3\]
Collect the \(y^3\) terms:
\[-y^3-y^3=-2y^3\]
Collect the \(x^2y\) terms:
\[x^2y+x^2y-2x^2y=0\]
Collect the \(xy^2\) terms:
\[xy^2+xy^2-2xy^2=0\]
Hence,
\[\begin{aligned}D&=-2x^3-2y^3\\&=-2(x^3+y^3)\end{aligned}\]
Using the identity
\[x^3+y^3=(x+y)(x^2-xy+y^2)\]
we may also write
\[D=-2(x+y)(x^2-xy+y^2)\]

🎯 Exam Significance

Exam Significance

›

This problem tests three important skills: expansion of a determinant, accurate handling of cofactor signs, and algebraic simplification. The first-row expansion must use the sign pattern

\[ +,-,+. \]

A particularly important step is the simplification of the third minor:

\[ xy-(x+y)^2=-x^2-xy-y^2. \]

Writing this step explicitly reduces the possibility of sign errors.

The final cancellation

\[ x^2y+x^2y-2x^2y=0 \]

and

\[ xy^2+xy^2-2xy^2=0 \]

provides a useful internal check on the solution.

Significance for Competitive Entrance Examinations

For competitive entrance examinations, recognizing cancellation patterns is valuable because it prevents unnecessary expansion. Although direct first-row expansion is systematic and reliable, the cyclic structure of the determinant indicates that the mixed terms are likely to cancel.

The final result can also be immediately factorised using

\[ x^3+y^3=(x+y)(x^2-xy+y^2). \]

This form can be particularly useful in questions asking when the determinant is zero.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

For expansion along the first row, use the cofactor signs \(+,-,+\).
Evaluate every \(2\times2\) minor separately before simplifying the complete determinant.
Remember that
\[ (x+y)^2=x^2+2xy+y^2. \]
The mixed terms \(x^2y\) and \(xy^2\) cancel completely.
The determinant simplifies to
\[ -2x^3-2y^3. \]
Therefore,
\[ \boxed{D=-2(x^3+y^3)}. \]
Using the sum-of-cubes identity,
\[ \boxed{D=-2(x+y)(x^2-xy+y^2)}. \]
When solving under examination conditions, collecting like terms separately is an effective way to avoid algebraic errors.

📘 Concept & Theory

Concept/Theory

›

A determinant of order \(3\) can be evaluated by expanding along any row or column. When expanding along the first row, the cofactor signs are

\[+,\quad -,\quad +\]

Therefore,

\[\begin{vmatrix}a & b & c\\d & e & f\\g & h & i\end{vmatrix}= a\begin{vmatrix}e & f\\h & i\end{vmatrix}- b\begin{vmatrix}d & f\\g & i\end{vmatrix}+ c\begin{vmatrix}d & e\\g & h\end{vmatrix}\]

For a \(2\times2\) determinant, we use

\[\begin{vmatrix}a & b\\c & d\end{vmatrix}=ad-bc\]

In this problem, direct expansion along the first row is convenient because the first entry is \(1\), and the resulting minors simplify quickly.

🗺️ Solution Roadmap

Step-by-step Plan

›

Expand the determinant along the first row.
Evaluate each of the three \(2\times2\) minors.
Apply the correct cofactor signs \(+,-,+\).
Simplify each resulting algebraic expression.
Expand \((x+y)^2\) and collect like terms.
Obtain the final value of the determinant.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 15 steps

Let
\[D=\begin{vmatrix}1 & x & y\\1 & x+y & y\\1 & x & x+y\end{vmatrix}\]
Expanding \(D\) along the first row, whose cofactor signs are \(+,-,+\),
\[\begin{aligned}D&=1\begin{vmatrix}x+y & y\\x & x+y\end{vmatrix}- x\begin{vmatrix}1 & y\\1 & x+y\end{vmatrix}+ y\begin{vmatrix}1 & x+y\\1 & x\end{vmatrix}\end{aligned}\]
First Minor
Consider the first \(2\times2\) determinant:
\[\begin{aligned}\begin{vmatrix}x+y & y\\x & x+y\end{vmatrix}&=(x+y)(x+y)-xy\\&=(x+y)^2-xy\end{aligned}\]
Therefore, the first contribution is
\[\boxed{(x+y)^2-xy}\]
Second Minor
Consider the second \(2\times2\) determinant:
\[\begin{aligned}\begin{vmatrix}1 & y\\1 & x+y\end{vmatrix}&=1(x+y)-y(1)\\&=x+y-y\\&=x\end{aligned}\]
Since this term appears with the cofactor sign \(-\), its contribution is
\[\begin{aligned}-x(x)&=-x^2\end{aligned}\]
Thus, the second contribution is
\[\boxed{-x^2}\]
Third Minor
Consider the third \(2\times2\) determinant:
\[\begin{aligned}\begin{vmatrix}1 & x+y\\1 & x\end{vmatrix}&=1(x)-(x+y)(1)\\&=x-x-y\\&=-y\end{aligned}\]
Since this term appears with the cofactor sign \(+\), its contribution is
\[\begin{aligned}y(-y)&=-y^2\end{aligned}\]
Thus, the third contribution is
\[\boxed{-y^2}\]
Combine the Three Contributions
Therefore,
\[\begin{aligned}D&=(x+y)^2-xy-x^2-y^2\end{aligned}\]
Now expand the square:
\[(x+y)^2=x^2+2xy+y^2\]
Substituting this into the expression for \(D\),
\[\begin{aligned}D&=x^2+2xy+y^2-xy-x^2-y^2\end{aligned}\]
Collecting like terms,
\[\begin{aligned}D&=(x^2-x^2)+(y^2-y^2)+(2xy-xy)\\&=0+0+xy\\&=xy\end{aligned}\]

💡 Answer

Final Answer

›

\[\boxed{\bbox[5pt]{\begin{vmatrix}1 & x & y\\1 & x+y & y\\1 & x & x+y\end{vmatrix}=xy}}\]

🎯 Exam Significance

Exam Significance

›

This question is a straightforward application of determinant expansion. It is particularly useful for practicing the cofactor sign pattern

\[ +,-,+. \]

A sign error in the second or third term changes the final answer, so writing the complete expansion before simplifying is good examination practice.

The calculation also demonstrates the importance of keeping the algebra organized. The expression

\[ (x+y)^2-xy-x^2-y^2 \]

becomes simple only after expanding the square and collecting like terms.

For a Board examination, the solution should not jump directly from the determinant to \(xy\). Showing the minors and the intermediate simplification establishes the complete logical sequence and makes the answer easy to award marks for.

Significance for Competitive Entrance Examinations

For competitive examinations, the determinant can be evaluated rapidly by noticing that the first column consists entirely of \(1\)'s. One may also perform row operations before expansion. However, direct first-row expansion is already efficient here because the resulting \(2\times2\) determinants are simple.

The cancellation pattern

\[ x^2-x^2=0,\qquad y^2-y^2=0 \]

leaves only

\[ 2xy-xy=xy. \]

Recognizing such cancellation is useful when solving determinant questions under time constraints.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

For first-row expansion, the cofactor signs are \(+,-,+\).
For a \(2\times2\) determinant,
\[ \begin{vmatrix} a & b\\ c & d \end{vmatrix} =ad-bc. \]
The first minor gives
\[ (x+y)^2-xy. \]
The second contribution is
\[ -x^2. \]
The third contribution is
\[ -y^2. \]
Hence,
\[ D=(x+y)^2-xy-x^2-y^2. \]
Using
\[ (x+y)^2=x^2+2xy+y^2, \]
all \(x^2\) and \(y^2\) terms cancel.
The final value is
\[ \boxed{D=xy}. \]

📘 Concept & Theory

Concept/Theory

›

A system of three linear equations can be written in matrix form as

\[ AX=B. \]

If

\[ |A|\neq0, \]

then \(A\) is non-singular and the system has a unique solution. The solution is obtained using

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

The given equations are not directly linear in \(x,y,z\), but they become a linear system if we introduce

\[ u=\frac1x,\qquad v=\frac1y,\qquad w=\frac1z. \]

We can then solve for \(u,v,w\), and finally take reciprocals to obtain \(x,y,z\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Introduce the variables
\[u=\frac1x,\quad v=\frac1y,\quad w=\frac1z\]
Write the resulting linear system in matrix form \(AX=B\).
Calculate \(|A|\) to check whether \(A^{-1}\) exists.
Find all cofactors of \(A\).
Form the adjoint by transposing the cofactor matrix.
Calculate \(A^{-1}\).
Use \(X=A^{-1}B\) to determine \(u,v,w\).
Take reciprocals to obtain \(x,y,z\).
Verify the solution in the original equations.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 32 steps

Convert the System into a Linear System
Put
\[u=\frac1x,\quad v=\frac1y,\quad w=\frac1z\]
The given equations become
\[\begin{aligned}2u+3v+10w&=4\\4u-6v+5w&=1\\6u+9v-20w&=2\end{aligned}\]
In matrix form,
\[\begin{bmatrix}2&3&10\\4&-6&5\\6&9&-20\end{bmatrix}\begin{bmatrix}u\\v\\w\end{bmatrix}=\begin{bmatrix}4\\1\\2\end{bmatrix}\]
Let
\[ A= \begin{bmatrix} 2&3&10\\ 4&-6&5\\ 6&9&-20 \end{bmatrix}, \qquad X=\begin{bmatrix}u\\v\\w\end{bmatrix}, \qquad B=\begin{bmatrix}4\\1\\2\end{bmatrix}\]
Thus,
\[AX=B\]
Since \(|A|\neq0\),
\[X=A^{-1}B\]
Calculate the Determinant of \(A\)
We have
\[\begin{aligned}|A|&=\begin{vmatrix}2&3&10\\4&-6&5\\6&9&-20\end{vmatrix}\end{aligned}\]
Expanding along the first row,
\[\begin{aligned}|A|&=2\begin{vmatrix}-6&5\\9&-20\end{vmatrix}- 3\begin{vmatrix}4&5\\6&-20\end{vmatrix}+ 10\begin{vmatrix}4&-6\\6&9\end{vmatrix}\\ &=2\left[(-6)(-20)-(5)(9)\right] -3\left[(4)(-20)-(5)(6)\right]\\ &\quad +10\left[(4)(9)-(-6)(6)\right]\\ &=2(120-45)-3(-80-30)+10(36+36)\\ &=2(75)-3(-110)+10(72)\\ &=150+330+720\\ &=1200 \end{aligned} \]
Therefore,
\[\boxed{|A|=1200\neq0}\]
Hence \(A\) is non-singular, \(A^{-1}\) exists, and the system has a unique solution.
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.
First Row Cofactors
\[\begin{aligned}A_{11}&=(+1)\begin{vmatrix}-6&5\\9&-20\end{vmatrix}\\&=(-6)(-20)-(5)(9)\\&=120-45\\&=75\end{aligned}\]
\[\begin{aligned}A_{12}&=(-1)\begin{vmatrix}4&5\\6&-20\end{vmatrix}\\&=-\left[(4)(-20)-(5)(6)\right]\\&=-(-80-30)\\&=110\end{aligned}\]
\[\begin{aligned}A_{13}&=(+1)\begin{vmatrix}4&-6\\6&9\end{vmatrix}\\&=(4)(9)-(-6)(6)\\&=36+36\\&=72\end{aligned}\]
Second Row Cofactors
\[\begin{aligned}A_{21}&=(-1)\begin{vmatrix}3&10\\9&-20\end{vmatrix}\\&=-\left[(3)(-20)-(10)(9)\right]\\&=-(-60-90)\\&=150\end{aligned}\]
\[\begin{aligned}A_{22}&=(+1)\begin{vmatrix}2&10\\6&-20\end{vmatrix}\\&=(2)(-20)-(10)(6)\\&=-40-60\\&=-100\end{aligned}\]
\[\begin{aligned}A_{23}&=(-1)\begin{vmatrix}2&3\\6&9\end{vmatrix}\\&=-\left[(2)(9)-(3)(6)\right]\\&=-(18-18)\\&=0\end{aligned}\]
Third Row Cofactors
\[\begin{aligned}A_{31}&=(+1)\begin{vmatrix}3&10\\-6&5\end{vmatrix}\\&=(3)(5)-(10)(-6)\\&=15+60\\&=75\end{aligned}\]
Note carefully that the sign of \(A_{31}\) is positive because
\[ (-1)^{3+1}=(-1)^4=+1. \]
Hence,
\[ \boxed{A_{31}=75}. \]
\[\begin{aligned}A_{32}&=(-1)\begin{vmatrix}2&10\\4&5\end{vmatrix}\\&=-\left[(2)(5)-(10)(4)\right]\\&=-(10-40)\\&=30\end{aligned}\]
\[\begin{aligned}A_{33}&=(+1)\begin{vmatrix}2&3\\4&-6\end{vmatrix}\\&=(2)(-6)-(3)(4)\\&=-12-12\\&=-24\end{aligned}\]
Form the Cofactor Matrix
Therefore, the cofactor matrix is
\[C=\begin{bmatrix}75&110&72\\150&-100&0\\75&30&-24\end{bmatrix}\]
Find the Adjoint of \(A\)
The adjoint of a matrix is the transpose of its cofactor matrix:
\[ \operatorname{adj}(A)=C^T. \]
Hence,
\[\begin{aligned}\operatorname{adj}(A)&=\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}\end{aligned}\]
Calculate \(A^{-1}\)
We know that
\[ A^{-1} = \frac{1}{|A|} \operatorname{adj}(A)\]
Since \(|A|=1200\),
\[\begin{aligned}A^{-1}&=\frac1{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}\end{aligned}\]
Thus,
\[\boxed{\bbox[5pt]{A^{-1}=\frac1{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}}}\]
Calculate \(X=A^{-1}B\)
We have
\[\begin{aligned}X&=A^{-1}B\\&=\frac1{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}\begin{bmatrix}4\\1\\2\end{bmatrix}\end{aligned}\]
Multiplying the matrices,
\[\begin{aligned}X &=\frac1{1200}\begin{bmatrix}75(4)+150(1)+75(2)\\110(4)-100(1)+30(2)\\72(4)+0(1)-24(2)\end{bmatrix}\\ &=\frac1{1200}\begin{bmatrix}300+150+150\\440-100+60\\288+0-48\end{bmatrix}\\ &=\frac1{1200}\begin{bmatrix}600\\400\\240\end{bmatrix}\end{aligned}\]
Therefore,
\[\begin{aligned}X&=\begin{bmatrix}\frac{600}{1200}\\\frac{400}{1200}\\\frac{240}{1200}\end{bmatrix}\\ &=\begin{bmatrix}\frac12\\\frac13\\\frac15\end{bmatrix}\end{aligned}\]
Since
\[X=\begin{bmatrix}u\\v\\w\end{bmatrix}=\begin{bmatrix}\frac1x\\\frac1y\\\frac1z\end{bmatrix},\]
we obtain
\[\begin{bmatrix}\frac1x\\\frac1y\\\frac1z\end{bmatrix}=\begin{bmatrix}\frac12\\\frac13\\\frac15\end{bmatrix}\]
Taking reciprocals,
\[\boxed{x=2,\quad y=3,\quad z=5}\]

🎯 Exam Significance

Exam Significance

›

This problem combines two important Class 12 concepts: solving a system of linear equations using matrices and finding the inverse of a matrix using cofactors and adjoints. It is therefore an excellent question for practicing the complete matrix-inverse method.

The most important point is to recognize that the equations become linear after substituting

\[ u=\frac1x,\qquad v=\frac1y,\qquad w=\frac1z. \]

A clear answer should explicitly state this substitution before writing the matrix equation.

When calculating the inverse, students must distinguish carefully between the cofactor matrix and its transpose. The adjoint is

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

not the cofactor matrix itself.

Another critical examination point is the sign of \(A_{31}\). Since

\[ (-1)^{3+1}=+1, \]

we have

\[ A_{31}=75, \]

not \(-75\). This single sign error changes the final value of \(x\).

Significance for Competitive Entrance Examinations

This problem is useful for developing accuracy in matrix inversion and systems of equations. In a competitive examination, calculating every cofactor may be time-consuming, so recognizing alternative methods such as direct elimination can be advantageous.

Nevertheless, when the question specifically tests the inverse-matrix method, the sequence

\[ AX=B \quad\Longrightarrow\quad X=A^{-1}B \]

is the central strategy.

Competitive examinations also frequently test whether a matrix is singular or non-singular. The calculation

\[ |A|=1200\neq0 \]

immediately establishes that \(A^{-1}\) exists and that the associated linear system has a unique solution.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 10 points

Convert reciprocal variables into linear variables by putting
\[ u=\frac1x,\quad v=\frac1y,\quad w=\frac1z. \]
Write the system as
\[ AX=B. \]
If
\[ |A|\neq0, \]
then \(A^{-1}\) exists and the system has a unique solution.
The inverse of \(A\) is
\[ A^{-1}=\frac1{|A|}\operatorname{adj}(A). \]
The adjoint is the transpose of the cofactor matrix:
\[ \operatorname{adj}(A)=C^T. \]
For the present matrix,
\[ |A|=1200. \]
The critical cofactor is
\[ A_{31}=75, \]
because its cofactor sign is positive.
The reciprocal variables are
\[ \frac1x=\frac12,\qquad \frac1y=\frac13,\qquad \frac1z=\frac15. \]
Therefore,
\[ \boxed{x=2,\ y=3,\ z=5}. \]
Always verify the final values in the original equations when solving reciprocal systems.

📘 Concept & Theory

Concept/Theory

›

A square matrix \(A\) has an inverse if and only if its determinant is non-zero:

\[ |A|\neq0. \]

When the inverse exists, it can be calculated using

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

where \(\operatorname{adj}(A)\) is the transpose of the cofactor matrix of \(A\).

The given matrix is a diagonal matrix because all its non-diagonal entries are zero. Its determinant is simply the product of its diagonal entries:

\[ |A|=xyz. \]

Since \(x,y,z\) are non-zero,

\[ xyz\neq0. \]

Therefore, \(A^{-1}\) exists.

🗺️ Solution Roadmap

Step-by-step Plan

›

Find the determinant \(|A|\) and establish that it is non-zero.
Find all nine cofactors of \(A\).
Form the cofactor matrix.
Transpose the cofactor matrix to obtain \(\operatorname{adj}(A)\).
Use
\[ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A). \]
Simplify the resulting diagonal entries.
Optionally verify the result by multiplying \(A\) and \(A^{-1}\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 22 steps

Calculate the Determinant of \(A\)
We have
\[A=\begin{bmatrix}x&0&0\\0&y&0\\0&0&z\end{bmatrix}\]
Therefore,
\[\begin{aligned}|A|&=\begin{vmatrix}x&0&0\\0&y&0\\0&0&z\end{vmatrix}\end{aligned}\]
Expanding along the first row,
\[\begin{aligned}|A|&=x\begin{vmatrix}y&0\\0&z\end{vmatrix}- 0\begin{vmatrix}0&0\\0&z\end{vmatrix}+ 0\begin{vmatrix}0&y\\0&0\end{vmatrix}\\&=x(yz-0)-0+0\\&=xyz\end{aligned}\]
Since \(x,y,z\neq0\),
\[\boxed{|A|=xyz\neq0}\]
Hence, \(A\) is non-singular and its inverse exists.
Find the Cofactors of \(A\)
The cofactor of the element \(a_{ij}\) is
\[ A_{ij}=(-1)^{i+j}M_{ij} \]
The cofactor signs for a \(3\times3\) matrix are
\[\begin{bmatrix}+&-&+\\-&+&-\\+&-&+\end{bmatrix}\]
Cofactors of the First Row
\[\begin{aligned}A_{11}&=(+1)\begin{vmatrix}y&0\\0&z\end{vmatrix}\\&=yz-0\\&=yz\end{aligned}\]
\[\begin{aligned}A_{12}&=(-1)\begin{vmatrix}0&0\\0&z\end{vmatrix}\\&=-(0-0)\\&=0\end{aligned}\]
\[\begin{aligned}A_{13}&=(+1)\begin{vmatrix}0&y\\0&0\end{vmatrix}\\&=0-0\\&=0\end{aligned}\]
Cofactors of the Second Row
\[\begin{aligned}A_{21}&=(-1)\begin{vmatrix}0&0\\0&z\end{vmatrix}\\&=0\end{aligned}\]
\[\begin{aligned}A_{22}&=(+1)\begin{vmatrix}x&0\\0&z\end{vmatrix}\\&=xz-0\\&=xz\end{aligned}\]
\[\begin{aligned}A_{23}&=(-1)\begin{vmatrix}x&0\\0&0\end{vmatrix}\\&=-\left(0-0\right)\\&=0\end{aligned}\]
Cofactors of the Third Row
\[\begin{aligned}A_{31}&=(+1)\begin{vmatrix}0&0\\y&0\end{vmatrix}\\&=0-0\\&=0\end{aligned}\]
\[\begin{aligned}A_{32}&=(-1)\begin{vmatrix}x&0\\0&0\end{vmatrix}\\&=-\left(0-0\right)\\&=0\end{aligned}\]
\[\begin{aligned}A_{33}&=(+1)\begin{vmatrix}x&0\\0&y\end{vmatrix}\\&=xy-0\\&=xy\end{aligned}\]
Form the Cofactor Matrix
Thus, the cofactor matrix is
\[C=\begin{bmatrix}yz&0&0\\0&xz&0\\0&0&xy\end{bmatrix}\]
Find the Adjoint of \(A\)
The adjoint is the transpose of the cofactor matrix:
\[ \operatorname{adj}(A)=C^T. \]
Since \(C\) is a diagonal matrix, its transpose is the same matrix. Hence,
\[ \boxed{\bbox[5pt]{ \operatorname{adj}(A)= \begin{bmatrix} yz&0&0\\ 0&xz&0\\ 0&0&xy \end{bmatrix}} }\]
Calculate \(A^{-1}\)
Using
\[A^{-1}=\frac{1}{|A|}\operatorname{adj}(A),\]
and \(|A|=xyz\), we get
\[\begin{aligned}A^{-1}&=\frac{1}{xyz}\begin{bmatrix}yz&0&0\\0&xz&0\\0&0&xy\end{bmatrix}\end{aligned}\]
Dividing each entry by \(xyz\)
\[\begin{aligned}A^{-1}&=\begin{bmatrix}\dfrac{yz}{xyz}&0&0\\0&\dfrac{xz}{xyz}&0\\0&0&\dfrac{xy}{xyz}\end{bmatrix}\\ &=\begin{bmatrix}\dfrac1x&0&0\\0&\dfrac1y&0\\0&0&\dfrac1z\end{bmatrix}\end{aligned}\]
Equivalently,
\[\boxed{\bbox[5pt]{A^{-1}=\begin{bmatrix}x^{-1}&0&0\\0&y^{-1}&0\\0&0&z^{-1}\end{bmatrix}}}\]

🎯 Exam Significance

Exam Significance

›

This is an important standard result for diagonal matrices. The question tests the complete procedure for finding an inverse using determinants, cofactors and adjoints. It is particularly useful for reinforcing the condition

\[ |A|\neq0 \]

for the existence of an inverse.

The question also provides a useful opportunity to distinguish between the cofactor matrix and the adjoint. Here the cofactor matrix is diagonal, so its transpose is unchanged:

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

In a Board examination, explicitly mentioning this step makes the solution mathematically complete.

Significance for Competitive Entrance Examinations

For competitive examinations, this problem illustrates an important shortcut: the inverse of a non-singular diagonal matrix is obtained simply by taking the reciprocal of each non-zero diagonal element.

\[ \operatorname{diag}(x,y,z)^{-1} = \operatorname{diag}\left(\frac1x,\frac1y,\frac1z\right). \]

Thus, although the cofactor method is required for demonstrating the general theory, a multiple-choice or numerical question can usually be solved immediately by recognizing the diagonal structure.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

A matrix is invertible if and only if its determinant is non-zero.
For the given matrix,
\[ |A|=xyz. \]
Since \(x,y,z\neq0\),
\[ xyz\neq0, \]
so \(A^{-1}\) exists.
The cofactor matrix is
\[ \begin{bmatrix} yz&0&0\\ 0&xz&0\\ 0&0&xy \end{bmatrix}. \]
Because the cofactor matrix is diagonal, its transpose is unchanged.
The inverse is
\[ A^{-1}= \begin{bmatrix} 1/x&0&0\\ 0&1/y&0\\ 0&0&1/z \end{bmatrix}. \]
For any non-singular diagonal matrix, its inverse is obtained by taking the reciprocal of each diagonal element.
The result can be verified through
\[ AA^{-1}=A^{-1}A=I. \]

Q9

NUMERIC3 marks

Let

\[A=\begin{bmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{bmatrix},\qquad0\leq\theta\leq2\pi\]

Then the determinant \(\Delta=|A|\) belongs to which of the following intervals?

\[\begin{aligned}(A)\quad&\Delta=0\\(B)\quad&\Delta\in(2,\infty)\\(C)\quad&\Delta\in(2,4)\\(D)\quad&\Delta\in[2,4]\end{aligned}\]

📘 Concept & Theory

Concept/Theory

›

To determine the range of a determinant containing trigonometric expressions, we first evaluate the determinant algebraically and then use the known range of the trigonometric function.

Here, the determinant contains only \(\sin\theta\). Therefore, after simplifying the determinant, we can put

\[ s=\sin\theta \]

Since

\[ -1\leq\sin\theta\leq1, \]

we have

\[ -1\leq s\leq1 \]

A useful point in this problem is that the determinant simplifies to an expression involving \(\sin^2\theta\). Since

\[ 0\leq\sin^2\theta\leq1, \]

its range can then be determined directly.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the determinant \(\Delta=|A|\).
Expand the determinant along the first row.
Evaluate each \(2\times2\) minor carefully.
Simplify the determinant in terms of \(\sin^2\theta\).
Use
\[ 0\leq\sin^2\theta\leq1. \]
Determine the minimum and maximum possible values of \(\Delta\).
Identify the correct option.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 21 steps

Given —
\[A=\begin{bmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{bmatrix}\]
Let
\[\Delta=|A|\]
Therefore,
\[\Delta=\begin{vmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix}\]
Expand Along the First Row
The cofactor signs along the first row are
\[ +,-,+. \]
Hence,
\[\begin{aligned}\Delta&=1\begin{vmatrix}1&\sin\theta\\-\sin\theta&1\end{vmatrix}- \sin\theta\begin{vmatrix}-\sin\theta&\sin\theta\\-1&1\end{vmatrix}+ 1\begin{vmatrix}-\sin\theta&1\\-1&-\sin\theta\end{vmatrix}\end{aligned}\]
Evaluate the First Minor
\[\begin{aligned}\begin{vmatrix}1&\sin\theta\\-\sin\theta&1\end{vmatrix} &=(1)(1)-(\sin\theta)(-\sin\theta)\\&=1+\sin^2\theta\end{aligned}\]
Therefore, the first contribution is
\[\boxed{1+\sin^2\theta}\]
Evaluate the Second Minor
\[\begin{aligned}\begin{vmatrix}-\sin\theta&\sin\theta\\-1&1\end{vmatrix} &=(-\sin\theta)(1)-(\sin\theta)(-1)\\&=-\sin\theta+\sin\theta\\&=0\end{aligned}\]
Hence, the entire second contribution is
\[-\sin\theta(0)=0\]
Thus, the second contribution is
\[\boxed{0}\]
Evaluate the Third Minor
\[\begin{aligned}\begin{vmatrix}-\sin\theta&1\\-1&-\sin\theta\end{vmatrix} &=(-\sin\theta)(-\sin\theta)-(1)(-1)\\&=\sin^2\theta+1\end{aligned}\]
Therefore, the third contribution is
\[\boxed{1+\sin^2\theta}\]
Simplify the Determinant
Combining all three contributions,
\[\begin{aligned}\Delta&=(1+\sin^2\theta)-0+(1+\sin^2\theta)\\&=2+2\sin^2\theta\\&=2(1+\sin^2\theta)\end{aligned}\]
Thus,
\[\boxed{\Delta=2(1+\sin^2\theta)}\]
Determine the Range
For every real value of \(\theta\),
\[-1\leq\sin\theta\leq1\]
Squaring throughout gives
\[0\leq\sin^2\theta\leq1\]
Adding \(1\) throughout,
\[1\leq1+\sin^2\theta\leq2\]
Multiplying throughout by \(2\),
\[2\leq2(1+\sin^2\theta)\leq4\]
Since
\[\Delta=2(1+\sin^2\theta)\]
we obtain
\[\boxed{2\leq\Delta\leq4}\]
Check That Both Endpoints Are Attainable
The lower endpoint \(2\) is attained when
\[ \sin\theta=0. \]
For example, \(\theta=0\) lies in the given interval \(0\leq\theta\leq2\pi\). Hence,
\[ \Delta=2. \]
The upper endpoint \(4\) is attained when
\[ \sin^2\theta=1, \]
or
\[ \sin\theta=\pm1. \]
For example, \(\theta=\frac{\pi}{2}\) lies in the given interval. Hence,
\[ \Delta=4. \]
Therefore, both endpoints are included, and the range is the closed interval
\[ [2,4]. \]

💡 Answer

Final Answer

›

Correct option is

\[\boxed{\text{(D)}\ \Delta\in[2,4]}\]

🎯 Exam Significance

Exam Significance

›

This question tests two important skills: expansion of a third-order determinant and determination of the range of a trigonometric expression. It is important to evaluate every \(2\times2\) minor carefully because an incorrect minor can completely alter the range.

For Board examinations, the safest approach is to write the complete first-row expansion before simplifying. In particular, the cofactor signs

\[ +,-,+ \]

should be explicitly maintained.

The final range should also be written as a closed interval because both \(2\) and \(4\) are actually attained within

\[ 0\leq\theta\leq2\pi. \]

Significance for Competitive Entrance Examinations

This is a useful determinant-range problem because the determinant simplifies to

\[ \Delta=2(1+\sin^2\theta). \]

Once this form is recognized, the answer follows immediately from

\[ 0\leq\sin^2\theta\leq1. \]

In a time-bound examination, there is no need to solve for individual values of \(\theta\). The key observation is that \(\sin^2\theta\) has the exact range \([0,1]\), which directly gives

\[ \Delta\in[2,4]. \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

For expansion along the first row, the cofactor signs are
\[ +,-,+. \]
The correct second minor is zero:
\[ \begin{vmatrix} -\sin\theta&\sin\theta\\ -1&1 \end{vmatrix}=0. \]
The first and third minors are both
\[ 1+\sin^2\theta. \]
Therefore,
\[ \Delta=2(1+\sin^2\theta). \]
Since
\[ 0\leq\sin^2\theta\leq1, \]
we have
\[ 2\leq\Delta\leq4. \]
The value \(\Delta=2\) is attained when \(\sin\theta=0\).
The value \(\Delta=4\) is attained when \(\sin^2\theta=1\).
Both endpoints are included, so the interval is closed.
The correct answer is
\[ \boxed{\text{(D)}\ \Delta\in[2,4]}. \]

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

Chapter Complete!

Recent posts

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

Chapter Complete!

Recent posts

Determinants — Learning Resources

Frequently Asked Questions

Q 01 What is covered in the NCERT Class 12 Determinants Miscellaneous Exercise?

Q 02 Are these NCERT Class 12 Determinants solutions step by step?

Q 03 How can I find the inverse of a diagonal matrix?

Q 04 When does the inverse of a matrix exist?

Q 05 How are determinants useful in solving linear equations?

Q 06 Are these Determinants solutions useful for CBSE Board exams?

Q 07 Are these solutions useful for JEE and other entrance exams?

Q 08 How can I avoid mistakes while expanding a determinant?

Q 09 What is the importance of the range of sin²? in determinant problems?

Q 10 What are the key topics to revise from the Determinants chapter?

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