NCERT Class 12 Determinants Ex 4.5 Q16 Solution

📘 Concept & Theory

Concept/Theory

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A system of simultaneous linear equations is said to be consistent if it has at least one solution. For a system of two linear equations in two variables,

\[ a_1x+b_1y=c_1 \]

\[ a_2x+b_2y=c_2, \]

the equations can be represented in matrix form as

\[ AX=B, \]

where

\[ A= \begin{bmatrix} a_1&b_1\\ a_2&b_2 \end{bmatrix}, \quad X= \begin{bmatrix} x\\ y \end{bmatrix}, \quad B= \begin{bmatrix} c_1\\ c_2 \end{bmatrix}. \]

If

\[ |A|\neq 0, \]

then the coefficient matrix \(A\) is non-singular and the system has a unique solution. Therefore, the system is consistent.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given equations in matrix form \(AX=B\).
Identify the coefficient matrix \(A\).
Calculate the determinant \(|A|\).
Check whether \(|A|\) is zero or non-zero.
Since a non-zero determinant implies a unique solution, conclude whether the system is consistent.

✏️ Solution

Complete Solution

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Step-by-step Solution · 10 steps

Given — system of equations:
\[ \begin{align} x+2y&=2 \tag{1}\\ 2x+3y&=3 \tag{2} \end{align} \]
The coefficients of \(x\) and \(y\) form the coefficient matrix. Thus,
\[A=\begin{bmatrix}1&2\\2&3\end{bmatrix}\]
The variables form the column matrix
\[X=\begin{bmatrix}x\\y\end{bmatrix},\]
and the constant terms form
\[B=\begin{bmatrix}2\\3\end{bmatrix}\]
Therefore, the given system can be written in matrix form as
\[\begin{bmatrix}1&2\\2&3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\3\end{bmatrix}\]
For a system \(AX=B\), if
\[|A|\neq0,\]
then \(A\) is non-singular and the system has a unique solution. Hence, it is consistent.
Now calculate the determinant of the coefficient matrix:
\[|A|=\begin{vmatrix}1&2\\2&3\end{vmatrix}\]
For a determinant of order \(2\),
\[\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc\]
Therefore,
\[|A|=(1)(3)-(2)(2)\]
Simplifying
\[|A|=3-4\]
Hence,
\[|A|=-1\]
Since
\[|A|=-1\neq0,\]
the coefficient matrix \(A\) is non-singular.
Therefore, the system has a unique solution.
A system having a unique solution is consistent.
Hence, the given system of equations is consistent.

🎯 Exam Significance

Exam Significance

›

This question tests an important application of determinants: determining the consistency of a system of linear equations. In a CBSE board examination, students should clearly identify the coefficient matrix and calculate its determinant without skipping the intermediate steps. The key conclusion is

\[ |A|\neq0\Rightarrow\text{unique solution}\Rightarrow\text{consistent system}. \]

Writing this logical chain makes the answer complete and mathematically justified.

Significance for Competitive Entrance Examinations

The determinant criterion is frequently useful in questions involving systems of linear equations, matrices, inverse matrices, and Cramer's Rule. For a two-variable system, recognizing immediately that a non-zero determinant of the coefficient matrix guarantees a unique solution can significantly reduce calculation time. The same idea extends naturally to higher-order systems.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 7 points

A system of linear equations can be expressed in matrix form as \(AX=B\).
The coefficient matrix contains the coefficients of the variables.
For a two-variable system, calculate the determinant using
\[ \begin{vmatrix} a&b\\ c&d \end{vmatrix}=ad-bc. \]
If
\[ |A|\neq0, \]
the coefficient matrix is non-singular.
A non-singular coefficient matrix gives a unique solution.
A system with a unique solution is consistent.
For the present system,
\[ |A|=-1\neq0, \]
so the system is consistent.

📘 Concept & Theory

Concept/Theory

›

A system of simultaneous linear equations is consistent if it has at least one solution. For a system represented in matrix form as

\[ AX=B, \]

if the determinant of the coefficient matrix satisfies

\[ |A|\neq0, \]

then \(A\) is a non-singular matrix. Consequently, the system has a unique solution and is therefore consistent.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given system in matrix form \(AX=B\).
Identify the coefficient matrix \(A\).
Calculate the determinant \(|A|\).
Check whether \(|A|\) is zero or non-zero.
Use the determinant criterion to conclude whether the system is consistent.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 11 steps

Given — system of equations
\[\begin{aligned}2x-y&=5\\x+y&=4\end{aligned}\]
Comparing these equations with the standard form
\[a_1x+b_1y=c_1,\]
\[a_2x+b_2y=c_2,\]
we have
\[a_1=2,\quad b_1=-1,\quad c_1=5,\]
and
\[a_2=1,\quad b_2=1,\quad c_2=4\]
Therefore, the coefficient matrix is
\[A=\begin{bmatrix}2&-1\\1&1\end{bmatrix}\]
The variable matrix is
\[X=\begin{bmatrix}x\\y\end{bmatrix},\]
and the constant matrix is
\[B=\begin{bmatrix}5\\4\end{bmatrix}\]
Hence, the given system can be written in matrix form as
\[AX=B,\]
that is,
\[\begin{bmatrix}2&-1\\1&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}5\\4\end{bmatrix}\]
For the system to have a unique solution, and hence to be consistent, we require
\[|A|\neq0.\]
Now,
\[|A|=\begin{vmatrix}2&-1\\1&1\end{vmatrix}\]
Using the determinant formula
\[\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc,\]
we get
\[|A|=(2)(1)-(-1)(1)\]
Therefore,
\[|A|=2+1\]
Hence,
\[|A|=3\]
Since
\[|A|=3\neq0,\]
the coefficient matrix \(A\) is non-singular.
Therefore, the system has a unique solution.
Hence, the given system of equations is consistent.

🎯 Exam Significance

Exam Significance

›

This problem reinforces the determinant-based criterion for examining the consistency of a system of linear equations. In board examinations, students should explicitly form the coefficient matrix and show the determinant calculation. The decisive step is

\[ |A|\neq0, \]

which establishes that the system has a unique solution and is therefore consistent.

Significance for Competitive Entrance Examinations

For competitive examinations, the determinant criterion provides a quick way to classify a system without necessarily solving for \(x\) and \(y\). Recognising that

\[ |A|\neq0 \]

immediately implies a unique solution can save substantial calculation time in questions involving matrices, determinants, inverse matrices, and systems of linear equations.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 6 points

The system can be represented as \(AX=B\).
The coefficient matrix is
\[ A= \begin{bmatrix} 2&-1\\ 1&1 \end{bmatrix} \]
Its determinant is
\[ |A|=(2)(1)-(-1)(1)=3. \]
Since
\[ |A|=3\neq0, \]
the matrix \(A\) is non-singular.
A non-singular coefficient matrix gives a unique solution.
A system having a unique solution is consistent.

📘 Concept & Theory

Concept/Theory

›

A system of simultaneous linear equations is consistent if it has at least one solution. For a system

\[ AX=B, \]

where \(A\) is the coefficient matrix, the condition

\[ |A|\neq0 \]

guarantees a unique solution and hence consistency.

However, when

\[ |A|=0, \]

the coefficient matrix is singular. In this case, the system may either have infinitely many solutions or no solution. Therefore, the determinant being zero alone is not sufficient to conclude that the system is inconsistent. The equations must be examined further.

For two linear equations

\[ a_1x+b_1y=c_1 \]

and

\[ a_2x+b_2y=c_2, \]

the system is inconsistent when

\[ \frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}. \]

Geometrically, this represents two distinct parallel lines.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given equations in matrix form \(AX=B\).
Identify the coefficient matrix \(A\).
Calculate \(|A|\).
Since \(|A|=0\), do not conclude inconsistency immediately.
Compare the ratios of the coefficients and constant terms.
Use the consistency criterion to determine whether the system has a solution.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 11 steps

Given — system of equations
\[\begin{aligned}x+3y&=52x+6y&=8\end{aligned}\]
The coefficient matrix is
\[A=\begin{bmatrix}1&3\\2&6\end{bmatrix}\]
The variable matrix is
\[X=\begin{bmatrix}x\\y\end{bmatrix},\]
and the constant matrix is
\[B=\begin{bmatrix}5\\8\end{bmatrix}\]
Therefore, the system can be written as
\[AX=B,\]
that is,
\[\begin{bmatrix}1&3\\2&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}5\\8\end{bmatrix}\]
Now calculate the determinant of the coefficient matrix:
\[|A|=\begin{vmatrix}1&3\\2&6\end{vmatrix}\]
Using
\[\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc,\]
we obtain
\[|A|=(1)(6)-(3)(2)\]
Hence,
\[|A|=6-6=0.\]
Thus, the coefficient matrix \(A\) is singular. Therefore, we must examine the equations further to determine whether the system is consistent or inconsistent.
Comparing the coefficients of the two equations, we have
\[a_1=1,\quad b_1=3,\quad c_1=5,\]
and
\[a_2=2,\quad b_2=6,\quad c_2=8.\]
Now calculate the required ratios:
\[\frac{a_1}{a_2}=\frac{1}{2},\]
\[\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2},\]
and
\[\frac{c_1}{c_2}=\frac{5}{8}\]
Therefore,
\[\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{1}{2},\]
but
\[\frac{c_1}{c_2}=\frac{5}{8}\]
Since
\[\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2},\]

the two equations represent distinct parallel lines.
Hence, the two equations have no common solution.
Therefore, the given system of equations is inconsistent.

🎯 Exam Significance

Exam Significance

›

This question highlights an important point that is frequently tested in board examinations: when

\[ |A|=0, \]

the system cannot be classified as inconsistent merely from the determinant. Students must apply the complete consistency criterion. The correct conclusion follows from

\[ \frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}. \]

Showing these ratios explicitly makes the reasoning complete and avoids the common error of treating \(|A|=0\) as an automatic indication of inconsistency.

Significance for Competitive Entrance Examinations

This distinction is particularly important in competitive examinations. A zero determinant indicates that the coefficient matrix is singular, but it does not by itself distinguish between infinitely many solutions and no solution. Recognising the ratio criterion quickly allows students to classify the system without solving the equations completely.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 7 points

The coefficient matrix is
\[A=\begin{bmatrix}1&3\\2&6\end{bmatrix}.\]
\[ |A|=(1)(6)-(3)(2)=0. \]
A zero determinant means that the coefficient matrix is singular; it does not automatically mean that the system is inconsistent.
The relevant ratios are
\[ \frac{a_1}{a_2}=\frac{1}{2},\quad \frac{b_1}{b_2}=\frac{1}{2},\quad \frac{c_1}{c_2}=\frac{5}{8}. \]
Since
\[ \frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}, \]
the system has no solution.
The two equations represent distinct parallel lines.
Hence, the given system is inconsistent.

📘 Concept & Theory

Concept/Theory

›

A system of three simultaneous linear equations can be expressed in the matrix form

\[ AX=B, \]

where \(A\) is the coefficient matrix, \(X\) is the column matrix of variables, and \(B\) is the column matrix of constants.

If

\[ |A|\neq0, \]

then \(A\) is non-singular and the system has a unique solution. Therefore, the system is consistent.

If \(|A|=0\), the coefficient matrix is singular, and further analysis is required to determine whether the system has infinitely many solutions or no solution.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given system in the form \(AX=B\).
Identify the coefficient matrix \(A\).
Calculate the determinant \(|A|\) by expanding along the first row.
Simplify the determinant to obtain a condition on the parameter \(a\).
Use the condition \(|A|\neq0\) to determine when the system is guaranteed to be consistent.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 22 steps

Given — system of equations
\[\begin{aligned}x+y+z&=1\\2x+3y+2z&=2\\ax+ay+2az&=4\end{aligned}\]
Comparing the equations with the standard matrix equation
\[AX=B,\]
the coefficients of \(x,y,z\) form the coefficient matrix
\[A=\begin{bmatrix}1&1&1\\2&3&2\\a&a&2a\end{bmatrix}\]
The variable matrix is
\[X=\begin{bmatrix}x\\y\\z\end{bmatrix},\]
and the constant matrix is
\[B=\begin{bmatrix}1\\2\\4\end{bmatrix}\]
Therefore, the system can be written in matrix form as
\[\boxed{\bbox[5pt]{\begin{bmatrix}1&1&1\\2&3&2\\a&a&2a\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\4\end{bmatrix}}}\]
For the system to have a unique solution, and hence to be consistent, we require
\[|A|\neq0.\]
Now,
\[|A|=\begin{vmatrix}1&1&1\\2&3&2\\a&a&2a\end{vmatrix}\]
Expansion Along the First Row
Expanding the determinant along the first row, we obtain
\[\begin{aligned}|A|&=1\begin{vmatrix}3&2\\a&2a\end{vmatrix}-1\begin{vmatrix}2&2\\a&2a\end{vmatrix}+1\begin{vmatrix}2&3\\a&a\end{vmatrix}\end{aligned}\]
Evaluate each determinant separately:
\[\begin{vmatrix}3&2\\a&2a\end{vmatrix}=(3)(2a)-(2)(a)\]
Hence,
\[\begin{vmatrix}3&2\\a&2a\end{vmatrix}=6a-2a=4a\]
Similarly,
\[\begin{vmatrix}2&2\\a&2a\end{vmatrix}=(2)(2a)-(2)(a)\]
Therefore,
\[\begin{vmatrix}2&2\\a&2a\end{vmatrix}=4a-2a=2a\]
Finally,
\[\begin{vmatrix}2&3\\a&a\end{vmatrix}=(2)(a)-(3)(a)\]
Thus,
\[\begin{vmatrix}2&3\\a&a\end{vmatrix}=2a-3a=-a\]
Substituting these values into the expansion,
\[\begin{aligned}|A|&=4a-2a+(-a)\\&=4a-2a-a\\&=2a-a\\&=a\end{aligned}\]
Hence,
\[\boxed{\bbox[5pt]{|A|=a}}\]
For the coefficient matrix to be non-singular, we require
\[|A|\neq0\]
Since
\[|A|=a,\]
we get
\[a\neq0\]
Therefore, for
\[\boxed{\bbox[5pt]{a\neq0}},\]
the coefficient matrix is non-singular and the system has a unique solution.
Hence, the system is consistent for \(a\neq0\).
Special Case: \(a=0\)
It is important not to conclude that the system is inconsistent when \(a=0\) merely because \(|A|=0\). The case \(a=0\) requires separate examination.
Substituting \(a=0\) into the third equation gives
\[0x+0y+0z=4,\]
or
\[0=4,\]
which is impossible.
Therefore, when
\[a=0,\]
the system has no solution and is inconsistent.
Thus, the complete conclusion is
\[ \boxed{\bbox[5pt]{ \begin{cases} a\neq0 &\Rightarrow \text{system is consistent and has a unique solution},\\ a=0 &\Rightarrow \text{system is inconsistent}. \end{cases} }} \]

🎯 Exam Significance

Exam Significance

›

This question combines two important Class 12 topics: matrix representation of a system of linear equations and the determinant criterion for consistency. It also tests the student's ability to handle a parameter \(a\). In a board examination, every stage of the determinant expansion should be shown clearly, especially the signs \(+,-,+\) while expanding along the first row.

A particularly important examination point is that

\[ |A|=0 \]

does not automatically mean that the system is inconsistent. The special case \(a=0\) must be checked separately. Here, it produces the contradiction \(0=4\), which establishes inconsistency.

Significance for Competitive Entrance Examinations

Parameter-dependent systems are common in competitive entrance examinations. The fastest reliable approach is to first evaluate the determinant of the coefficient matrix. If it is non-zero, the system immediately has a unique solution. If it vanishes, the exceptional parameter value must be tested separately.

This problem therefore reinforces the important distinction:

\[ |A|\neq0 \Rightarrow \text{unique solution}, \]

whereas

\[ |A|=0 \Rightarrow \text{further investigation required}. \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

A system of three linear equations can be written as \(AX=B\).
The coefficient matrix is
\[ A= \begin{bmatrix} 1&1&1\\ 2&3&2\\ a&a&2a \end{bmatrix}. \]
Expanding along the first row gives
\[ |A|=a. \]
For \(a\neq0\),
\[ |A|\neq0, \]
so the system has a unique solution and is consistent.
For \(a=0\), the third equation becomes
\[ 0=4, \]
which is impossible.
Therefore, \(a=0\) gives an inconsistent system.
The complete classification is
\[ \boxed{\bbox[5pt]{a\neq0\Rightarrow\text{consistent},\qquad a=0\Rightarrow\text{inconsistent}}} \]

📘 Concept & Theory

Concept/Theory

›

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

\[ AX=B, \]

where \(A\) is the coefficient matrix.

If

\[ |A|\neq0, \]

then the coefficient matrix is non-singular and the system has a unique solution. Hence, the system is consistent.

If

\[ |A|=0, \]

the coefficient matrix is singular. In that case, we cannot conclude immediately that the system is inconsistent. Further examination is necessary to determine whether the system has infinitely many solutions or no solution.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given equations in matrix form \(AX=B\).
Identify the coefficient matrix \(A\).
Calculate \(|A|\), choosing a convenient row for expansion.
If \(|A|\neq0\), conclude that the system is consistent with a unique solution.
If \(|A|=0\), substitute or simplify the equations further to determine whether a solution exists.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 25 steps

Given — system of equations
\[\begin{align} 3x-y-2z&=2\tag{1}\\ 2y-z&=-1\tag{2}\\ 3x-5y&=3\tag{3} \end{align}\]
To write the system in matrix form, insert the coefficient of every variable in every equation. A variable that is absent from an equation has coefficient \(0\).
Thus, equation (1) gives the coefficient row
\[\begin{bmatrix}3&-1&-2\end{bmatrix},\]
equation (2) gives
\[\begin{bmatrix}0&2&-1\end{bmatrix},\]
and equation (3) gives
\[\begin{bmatrix}3&-5&0\end{bmatrix}\]
Therefore, the coefficient matrix is
\[A=\begin{bmatrix}3&-1&-2\\0&2&-1\\3&-5&0\end{bmatrix}\]
The variable matrix is
\[X=\begin{bmatrix}x\\y\\z\end{bmatrix},\]
while the constant matrix is
\[B=\begin{bmatrix}2\\-1\\3\end{bmatrix}\]
Hence, the system can be written as
\[AX=B,\]
that is,
\[\boxed{\bbox[5pt]{\begin{bmatrix}3&-1&-2\\0&2&-1\\3&-5&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\-1\\3\end{bmatrix}}}\]
Now calculate the determinant of the coefficient matrix:
\[|A|=\begin{vmatrix}3&-1&-2\\0&2&-1\\3&-5&0\end{vmatrix}\]
Expansion Along the Second Row
The second row contains a zero, so expanding along this row reduces the calculation.
The second row is
\[\begin{bmatrix}0&2&-1\end{bmatrix}.\]
Therefore, using the cofactor signs \(+,-,+\) for the second row positions \(21,22,23\), we get
\[\begin{aligned}|A|&=0\begin{vmatrix}-1&-2\\-5&0\end{vmatrix}-2\begin{vmatrix}3&-2\\3&0\end{vmatrix}+(-1)\begin{vmatrix}3&-1\\3&-5\end{vmatrix}.\end{aligned}\]
The first term is zero:
\[0\begin{vmatrix}-1&-2\\-5&0\end{vmatrix}=0\]
Now evaluate the second minor:
\[\begin{vmatrix}3&-2\\3&0\end{vmatrix}=(3)(0)-(-2)(3)\]
Hence,
\[\begin{vmatrix}3&-2\\3&0\end{vmatrix}=0+6=6\]
Next, evaluate the third minor:
\[\begin{vmatrix}3&-1\\3&-5\end{vmatrix}=(3)(-5)-(-1)(3)\]
Therefore,
\[\begin{vmatrix}3&-1\\3&-5\end{vmatrix}=-15+3=-12\]
Substituting these values,
\[\begin{aligned}|A|&=0-2(6)+(-1)(-12)\\&=-12+12\\&=0\end{aligned}\]
Thus,
\[\boxed{|A|=0}\]
Since the determinant of the coefficient matrix is zero, the coefficient matrix is singular. Therefore, we must examine the original equations further before deciding consistency.
Further Examination of the System
From equation (2),
\[2y-z=-1\]
Rearranging,
\[z=2y+1 \]
Substitute this value of \(z\) into equation (1):
\[3x-y-2(2y+1)=2\]
Expanding,
\[3x-y-4y-2=2\]
Combining like terms,
\[3x-5y-2=2\]
Therefore,
\[3x-5y=4\]
But equation (3) gives
\[3x-5y=3\]
Hence, the same expression \(3x-5y\) would have to be simultaneously equal to \(4\) and \(3\):
\[3x-5y=4\]
and
\[3x-5y=3\]
This leads to the contradiction
\[4=3,\]
which is impossible.

🎯 Exam Significance

Exam Significance

›

This problem is particularly important because it demonstrates the correct interpretation of a zero determinant. The result

\[ |A|=0 \]

alone does not prove inconsistency. Students must perform an additional check. Here, substitution produces the contradiction

\[ 4=3, \]

which conclusively establishes that the system has no solution.

For board examinations, students should show the determinant calculation carefully and then explicitly examine the system when the determinant is zero. This avoids the common mistake of writing “\(|A|=0\), hence inconsistent” without justification.

Significance for Competitive Entrance Examinations

This question develops an important speed-and-accuracy skill for competitive examinations: identifying what information a determinant actually provides. A non-zero determinant immediately gives a unique solution, whereas a zero determinant signals a singular system and requires further classification.

In this problem, equation (2) gives \(z\) immediately, making substitution an efficient way to expose the contradiction. Recognising such a convenient equation can reduce calculation time considerably.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 8 points

The coefficient matrix is
\[ A= \begin{bmatrix} 3&-1&-2\\ 0&2&-1\\ 3&-5&0 \end{bmatrix}. \]
The determinant is
\[ |A|=0. \]
A zero determinant means that the coefficient matrix is singular; it does not by itself establish inconsistency.
From the second equation,
\[ z=2y+1. \]
Substitution into the first equation gives
\[ 3x-5y=4. \]
The third equation gives
\[ 3x-5y=3. \]
The resulting contradiction
\[ 4=3 \]
shows that the system has no solution.
Therefore, the given system is inconsistent.

📘 Concept & Theory

Concept/Theory

›

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

\[ AX=B, \]

where \(A\) is the coefficient matrix, \(X\) is the column matrix of variables, and \(B\) is the column matrix of constants.

For a square system, if the determinant of the coefficient matrix satisfies

\[ |A|\neq0, \]

then \(A\) is non-singular. Consequently, the system has a unique solution and is therefore consistent.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given system in matrix form \(AX=B\).
Identify the coefficient matrix \(A\).
Calculate the determinant \(|A|\).
Expand the determinant along a convenient row or column.
Check whether the resulting determinant is non-zero.
If \(|A|\neq0\), conclude that the system has a unique solution and is consistent.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 15 steps

Given — system of equation
\[ \begin{align} 5x-y+4z&=5\tag{1} \\ 2x+3y+5z&=2\tag{2} \\ 5x-2y+6z&=-1\tag{3} \end{align} \]
The coefficients of \(x\), \(y\), and \(z\) in equations (1), (2), and (3) form the coefficient matrix:
\[A=\begin{bmatrix}5&-1&4\\2&3&5\\5&-2&6\end{bmatrix}\]
The variable matrix is
\[X=\begin{bmatrix}x\\y\\z\end{bmatrix},\]
and the constant matrix is
\[B=\begin{bmatrix}5\\2\\-1\end{bmatrix}\]
Therefore, the given system can be written in matrix form as
\[AX=B,\]
that is,
\[\boxed{\bbox[5pt]{\begin{bmatrix}5&-1&4\\2&3&5\\5&-2&6\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\2\\-1\end{bmatrix}}}\]
To examine the consistency of the system, calculate the determinant of the coefficient matrix:
\[|A|=\begin{vmatrix}5&-1&4\\2&3&5\\5&-2&6\end{vmatrix}\]
Expansion Along the First Row
Expanding along the first row, the cofactor signs are
\[+,\,-,\,+.\]
Therefore,
\[\begin{aligned}|A|&=5\begin{vmatrix}3&5\\-2&6\end{vmatrix}-(-1) \begin{vmatrix}2&5\\5&6\end{vmatrix}+4\begin{vmatrix}2&3\\5&-2\end{vmatrix}\end{aligned}\]
Since the second entry of the first row is \(-1\), its contribution is
\[-(-1)\begin{vmatrix}2&5\\5&6\end{vmatrix}=+\begin{vmatrix}2&5\\5&6\end{vmatrix}\]
Now evaluate the first minor:
\[\begin{vmatrix}3&5\\-2&6\end{vmatrix}=(3)(6)-(5)(-2)\]
Hence,
\[\begin{vmatrix}3&5\\-2&6\end{vmatrix}=18+10=28\]
Next, evaluate the second minor:
\[\begin{vmatrix}2&5\\5&6\end{vmatrix}=(2)(6)-(5)(5)\]
Therefore,
\[\begin{vmatrix}2&5\\5&6\end{vmatrix}=12-25=-13\]
Finally, evaluate the third minor:
\[\begin{vmatrix}2&3\\5&-2\end{vmatrix}=(2)(-2)-(3)(5)\]
Thus,
\[\begin{vmatrix}2&3\\5&-2\end{vmatrix}=-4-15=-19\]
Substituting the three minors into the expansion,
\[\begin{aligned}|A|&=5(28)+(-13)+4(-19)\\&=140-13-76\\&=127-76\\&=51\end{aligned}\]
Hence,
\[\boxed{|A|=51}\]
Since
\[|A|=51\neq0,\]
the coefficient matrix \(A\) is non-singular.
A system having a unique solution is consistent.
Hence, the given system of equations is consistent.

🎯 Exam Significance

Exam Significance

›

This problem tests the application of determinants to systems of three simultaneous linear equations. The essential board-examination argument is to form the coefficient matrix correctly, calculate its determinant, and establish that

\[ |A|\neq0. \]

Students should be particularly careful with the negative coefficient \(-1\) and the corresponding cofactor sign during expansion.

A complete solution should show the evaluation of each \(2\times2\) minor rather than jumping directly to the final determinant. This makes the sign handling transparent and reduces the possibility of calculation errors.

Significance for Competitive Entrance Examinations

The determinant test provides a rapid method for deciding whether a square system has a unique solution. Here,

\[ |A|=51\neq0 \]

immediately establishes uniqueness and consistency without requiring the values of \(x\), \(y\), and \(z\) to be calculated.

In time-bound entrance examinations, this distinction is useful: if the question asks only about consistency, solving the entire system is unnecessary once a non-zero determinant has been established.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

The coefficient matrix is
\[A=\begin{bmatrix}5&-1&4\\2&3&5\\5&-2&6\end{bmatrix}\]
The system can be represented as \(AX=B\).
Expanding \(|A|\) along the first row gives
\[|A|=5\begin{vmatrix}3&5\\-2&6\end{vmatrix}+\begin{vmatrix}2&5\\5&6\end{vmatrix}+4\begin{vmatrix}2&3\\5&-2\end{vmatrix}\]
The determinant evaluates to
\[ |A|=51. \]
Since
\[ |A|=51\neq0, \]
the coefficient matrix is non-singular.
The system therefore has a unique solution.
Hence, the given system is consistent.

📘 Concept & Theory

Concept/Theory

›

The matrix method is used to solve a system of linear equations by expressing it in the form

\[ AX=B. \]

If the coefficient matrix \(A\) is non-singular, that is,

\[ |A|\neq0, \]

then \(A^{-1}\) exists. Multiplying \(AX=B\) by \(A^{-1}\) on the left gives

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

Since

\[ A^{-1}A=I, \]

we obtain

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

For a \(2\times2\) matrix

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

its inverse is

\[A^{-1}=\frac{1}{|A|}\begin{bmatrix}d&-b\\-c&a\end{bmatrix},\]

provided

\[ |A|=ad-bc\neq0 \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the system in matrix form \(AX=B\).
Identify the coefficient matrix \(A\), variable matrix \(X\), and constant matrix \(B\).
Calculate \(|A|\) to verify that \(A^{-1}\) exists.
Find \(\operatorname{adj}(A)\).
Use
\[ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A). \]
Use \(X=A^{-1}B\) to obtain \(x\) and \(y\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 28 steps

Given — system
\[ \begin{align} 5x+2y&=4 \tag{1}\\ 7x+3y&=5 \tag{2} \end{align} \]
Write the System in Matrix Form
The coefficients of \(x\) and \(y\) form the coefficient matrix:
\[ A= \begin{bmatrix} 5&2\\ 7&3 \end{bmatrix} \]
The variable matrix is
\[X=\begin{bmatrix}x\\y\end{bmatrix},\]
and the constant matrix is
\[B=\begin{bmatrix}4\\5\end{bmatrix}\]
Therefore,
\[\boxed{\bbox[5pt]{\begin{bmatrix}5&2\\7&3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}4\\5\end{bmatrix}}}\]
Thus,
\[AX=B\]
Find the Determinant of \(A\)
Before finding \(A^{-1}\), we must verify that the determinant of \(A\) is non-zero.
\[|A|=\begin{vmatrix}5&2\\7&3\end{vmatrix}\]
Using
\[\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc,\]
we get
\[|A|=(5)(3)-(2)(7)\]
Therefore,
\[|A|=15-14\]
Hence,
\[\boxed{|A|=1}\]
Since
\[|A|=1\neq0,\]
the matrix \(A\) is non-singular and \(A^{-1}\) exists. Therefore, the system has a unique solution.
Use the Matrix Method
We have
\[AX=B\]
Multiplying both sides on the left by \(A^{-1}\),
\[A^{-1}(AX)=A^{-1}B\]
Using
\[A^{-1}A=I,\]
we obtain
\[IX=A^{-1}B\]
Since
\[IX=X,\]
we get
\[\boxed{\bbox[5pt]{X=A^{-1}B}}\]
Find \(\operatorname{adj}(A)\)
We have
\[A=\begin{bmatrix}5&2\\7&3\end{bmatrix}\]
For
\[A=\begin{bmatrix}a&b\\c&d\end{bmatrix},\]
the adjoint is
\[\operatorname{adj}(A)=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\]
Therefore,
\[\operatorname{adj}(A)=\begin{bmatrix}3&-2\\-7&5\end{bmatrix}\]
Find \(A^{-1}\)
The inverse of \(A\) is given by
\[A^{-1}=\frac{1}{|A|}\operatorname{adj}(A)\]
Since
\[|A|=1\]
and
\[\operatorname{adj}(A)=\begin{bmatrix}3&-2\\-7&5\end{bmatrix},\]
we have
\[A^{-1}=\frac{1}{1}\begin{bmatrix}3&-2\\-7&5\end{bmatrix}\]
Hence,
\[\boxed{\bbox[5pt]{A^{-1}=\begin{bmatrix}3&-2\\-7&5\end{bmatrix}}}\]
Calculate \(X=A^{-1}B\)
We have
\[X=A^{-1}B\]
Substituting the values of \(A^{-1}\) and \(B\),
\[\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3&-2\\-7&5\end{bmatrix}\begin{bmatrix}4\\5\end{bmatrix}\]
Performing the matrix multiplication,
\[\begin{aligned}\begin{bmatrix}x\\y\end{bmatrix}&=\begin{bmatrix}(3)(4)+(-2)(5)\\(-7)(4)+(5)(5)\end{bmatrix}\end{aligned}\]
Therefore,
\[\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}12-10\\-28+25\end{bmatrix}\]
Hence,
\[\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\-3\end{bmatrix}\]
Comparing the corresponding entries,
\[\boxed{\bbox[5pt]{x=2}}\]
and
\[\boxed{\bbox[5pt]{y=-3}}\]

🎯 Exam Significance

Exam Significance

›

This is a standard application of the inverse-matrix method and is important for the board examination. Students should clearly show the sequence

\[ AX=B \Rightarrow X=A^{-1}B, \]

followed by the calculation of \(|A|\), \(\operatorname{adj}(A)\), \(A^{-1}\), and finally the matrix multiplication.

A frequent error is to write

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

The correct formula is

\[ \boxed{\bbox[5pt]{ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A) }}\]

Here, because \(|A|=1\), the numerical value of the inverse happens to equal \(\operatorname{adj}(A)\).

Significance for Competitive Entrance Examinations

The matrix method provides a systematic approach to solving simultaneous equations. In competitive examinations, the determinant can first be checked to determine whether the inverse exists. If the question asks only for the solution, the identity

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

provides a direct route.

For larger systems, the same principle applies, although calculating an inverse may be more laborious. The determinant, adjoint, inverse matrix, and matrix multiplication therefore form an interconnected set of concepts that are useful in both direct and application-based questions.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

The system is represented as
\[ AX=B. \]
The coefficient matrix is
\[ A= \begin{bmatrix} 5&2\\ 7&3 \end{bmatrix}. \]
The determinant is
\[ |A|=15-14=1\neq0. \]
Since \(|A|\neq0\), the inverse \(A^{-1}\) exists and the system has a unique solution.
The adjoint is
\[ \operatorname{adj}(A)= \begin{bmatrix} 3&-2\\ -7&5 \end{bmatrix}. \]
The inverse is
\[ A^{-1}= \begin{bmatrix} 3&-2\\ -7&5 \end{bmatrix}. \]
Using
\[ X=A^{-1}B, \]
we obtain
\[ X= \begin{bmatrix} 2\\ -3 \end{bmatrix}. \]
Therefore,
\[ \boxed{\bbox[5pt]{x=2,\quad y=-3}}. \]

📘 Concept & Theory

Concept/Theory

›

The matrix method expresses a system of simultaneous linear equations in the form

\[ AX=B. \]

If

\[ |A|\neq0, \]

then \(A\) is non-singular and its inverse exists. Multiplying \(AX=B\) by \(A^{-1}\) on the left gives

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

Since

\[ A^{-1}A=I, \]

we obtain

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

Thus, the unknowns can be found by calculating the inverse of the coefficient matrix.

For a \(2\times2\) matrix

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

the inverse is

\[A^{-1}=\frac{1}{|A|}\begin{bmatrix}d&-b\\-c&a\end{bmatrix},\]

provided

\[|A|=ad-bc\neq0\]

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given equations in matrix form \(AX=B\).
Identify the coefficient matrix \(A\), variable matrix \(X\), and constant matrix \(B\).
Calculate \(|A|\) to verify that \(A^{-1}\) exists.
Find \(\operatorname{adj}(A)\).
Calculate \(A^{-1}\) using
\[ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A). \]
Use \(X=A^{-1}B\) to determine \(x\) and \(y\).
Verify the solution in the original equations.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 23 steps

Given — system:
\[ \begin{align} 2x-y&=-2 \tag{1}\\ 3x+4y&=3 \tag{2} \end{align} \]
Write the System in Matrix Form
The coefficients of \(x\) and \(y\) form the coefficient matrix:
\[A=\begin{bmatrix}2&-1\\3&4\end{bmatrix}\]
The variable matrix is
\[X=\begin{bmatrix}x\\y\end{bmatrix}\]
and the constant matrix is
\[B=\begin{bmatrix}-2\\3\end{bmatrix}\]
Therefore, the system can be written as
\[\boxed{\bbox[5pt]{\begin{bmatrix}2&-1\\3&4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-2\\3\end{bmatrix}}}\]
Hence,
\[AX=B\]
Find the Determinant of \(A\)
To Find — \(A^{-1}\), first calculate the determinant of \(A\):
\[|A|=\begin{vmatrix}2&-1\\3&4\end{vmatrix}\]
Using
\[\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc,\]
we get
\[A|=(2)(4)-(-1)(3)\]
Therefore,
\[|A|=8+3\]
Hence,
\[\boxed{|A|=11}\]
Since
\[|A|=11\neq0,\]
the matrix \(A\) is non-singular and \(A^{-1}\) exists.
Therefore, the system has a unique solution and is consistent.
Find \(\operatorname{adj}(A)\)
We have
\[A=\begin{bmatrix}2&-1\\3&4\end{bmatrix}\]
For
\[A=\begin{bmatrix}a&b\\c&d\end{bmatrix},\]
the adjoint is
\[\operatorname{adj}(A)=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\]
Therefore,
\[\operatorname{adj}(A)=\begin{bmatrix}4&1\\-3&2\end{bmatrix}.\]
Find \(A^{-1}\)
Using
\[A^{-1}=\frac{1}{|A|}\operatorname{adj}(A),\]
we obtain
\[A^{-1}=\frac{1}{11}\begin{bmatrix}4&1\\-3&2\end{bmatrix}.\]
Hence,
\[\boxed{\bbox[5pt]{A^{-1}=\begin{bmatrix}\frac{4}{11}&\frac{1}{11}\\-\frac{3}{11}&\frac{2}{11}\end{bmatrix}}}\]
Calculate \(X=A^{-1}B\)
From
\[AX=B,\]
multiplying by \(A^{-1}\) on the left gives
\[A^{-1}(AX)=A^{-1}B.\]
Since
\[A^{-1}A=I,\]
we get
\[X=A^{-1}B\]
Substituting the values,
\[\begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{11}\begin{bmatrix}4&1\\-3&2\end{bmatrix}\begin{bmatrix}-2\\3\end{bmatrix}\]
Performing the matrix multiplication,
\[\begin{aligned}\begin{bmatrix}x\\y\end{bmatrix}&=\frac{1}{11}\begin{bmatrix}(4)(-2)+(1)(3)\\(-3)(-2)+(2)(3)\end{bmatrix}\end{aligned}\]
Simplifying the first entry,
\[(4)(-2)+(1)(3)=-8+3=-5\]
Simplifying the second entry,
\[ (-3)(-2)+(2)(3)=6+6=12 \]
Therefore,
\[\begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{11}\begin{bmatrix}-5\\12\end{bmatrix}\]
Hence,
\[\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-\frac{5}{11}\\\frac{12}{11}\end{bmatrix}\]
Therefore,
\[\boxed{x=-\frac{5}{11}}\]
and
\[\boxed{y=\frac{12}{11}}\]

🎯 Exam Significance

Exam Significance

›

The determinant test allows the existence of a unique solution to be established immediately without solving the equations. Once

\[ |A|\neq0, \]

the matrix inverse method can be applied directly.

The problem also reinforces efficient matrix multiplication. In a time-bound examination, keeping the common factor \(1/11\) outside the matrix multiplication avoids unnecessary fractions during intermediate calculations.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

The system is written as
\[ AX=B. \]
The coefficient matrix is
\[ A= \begin{bmatrix} 2&-1\\ 3&4 \end{bmatrix}. \]
The determinant is
\[ |A|=(2)(4)-(-1)(3)=11. \]
Since
\[ |A|=11\neq0, \]
\(A^{-1}\) exists and the system has a unique solution.
The adjoint is
\[ \operatorname{adj}(A)= \begin{bmatrix} 4&1\\ -3&2 \end{bmatrix}. \]
The inverse is
\[ A^{-1} = \frac{1}{11} \begin{bmatrix} 4&1\\ -3&2 \end{bmatrix}. \]
Using
\[ X=A^{-1}B, \]
we obtain
\[ X= \begin{bmatrix} -\frac{5}{11}\\ \frac{12}{11} \end{bmatrix}. \]
Therefore,
\[ \boxed{x=-\frac{5}{11},\quad y=\frac{12}{11}}. \]

📘 Concept & Theory

Concept/Theory

›

A system of simultaneous linear equations can be expressed in the matrix form

\[ AX=B, \]

where \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constant matrix.

If

\[ |A|\neq0, \]

then \(A\) is non-singular and its inverse exists. Multiplying

\[ AX=B \]

by \(A^{-1}\) on the left gives

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

Since

\[ A^{-1}A=I, \]

we obtain

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

For a \(2\times2\) matrix

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

its inverse is

\[ A^{-1} = \frac{1}{|A|} \begin{bmatrix} d&-b\\ -c&a \end{bmatrix}, \]

provided

\[ |A|=ad-bc\neq0. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given system in the form \(AX=B\).
Identify the coefficient matrix \(A\), variable matrix \(X\), and constant matrix \(B\).
Calculate \(|A|\) to check whether \(A^{-1}\) exists.
Find \(\operatorname{adj}(A)\).
Calculate \(A^{-1}\) using
\[ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A). \]
Use
\[ X=A^{-1}B \]
to obtain the values of \(x\) and \(y\).
Verify the obtained values in the original equations.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 23 steps

Given — system:
\[ \begin{align} 4x-3y&=3\tag{1}\\ 3x-5y&=7\tag{3} \end{align} \]
Write the System in Matrix Form
The coefficients of \(x\) and \(y\) form the coefficient matrix:
\[A=\begin{bmatrix}4&-3\\3&-5\end{bmatrix}\]
The variable matrix is
\[X=\begin{bmatrix}x\\y\end{bmatrix},\]
and the constant matrix is
\[B=\begin{bmatrix}3\\7\end{bmatrix}\]
Therefore,
\[\boxed{\begin{bmatrix}4&-3\\3&-5\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3\\7\end{bmatrix}}\]
Hence,
\[AX=B\]
Find the Determinant of \(A\)
We have
\[|A|=\begin{vmatrix}4&-3\\3&-5\end{vmatrix}\]
Using
\[\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc,\]
we get
\[\begin{aligned}|A|&=(4)(-5)-(-3)(3)\\&=-20+9\\&=-11\end{aligned}\]
Since
\[|A|=-11\neq0,\]
the matrix \(A\) is non-singular and \(A^{-1}\) exists. Thus, the system has a unique solution.
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,
\[A=\begin{bmatrix}4&-3\\3&-5\end{bmatrix}.\]
Therefore,
\[\operatorname{adj}(A)=\begin{bmatrix}-5&3\\-3&4\end{bmatrix}\]
Find \(A^{-1}\)
Using the inverse formula,
\[A^{-1}=\frac{1}{|A|}\operatorname{adj}(A)\]
Substituting
\[|A|=-11\]
and
\[\operatorname{adj}(A)=\begin{bmatrix}-5&3\\-3&4\end{bmatrix},\]
we get
\[A^{-1}=-\frac{1}{11}\begin{bmatrix}-5&3\\-3&4\end{bmatrix}\]
Hence,
\[\boxed{\bbox[5pt]{A^{-1}=-\frac{1}{11}\begin{bmatrix}-5&3\\-3&4\end{bmatrix}}}\]
Calculate \(X=A^{-1}B\)
From
\[AX=B,\]
multiplying both sides by \(A^{-1}\) on the left,
\[A^{-1}AX=A^{-1}B\]
Since
\[A^{-1}A=I,\]
we obtain
\[X=A^{-1}B\]
Therefore,
\[\begin{bmatrix}x\\y\end{bmatrix}=-\frac{1}{11}\begin{bmatrix}-5&3\\-3&4\end{bmatrix}\begin{bmatrix}3\\7\end{bmatrix}\]
Performing the matrix multiplication,
\[\begin{aligned}\begin{bmatrix}x\\y\end{bmatrix}&=-\frac{1}{11}\begin{bmatrix}(-5)(3)+(3)(7)\\(-3)(3)+(4)(7)\end{bmatrix}\end{aligned}\]
Simplifying the first entry,
\[(-5)(3)+(3)(7)=-15+21=6\]
Simplifying the second entry,
\[(-3)(3)+(4)(7)=-9+28=19\]
Hence,
\[\begin{bmatrix}x\\y\end{bmatrix}=-\frac{1}{11}\begin{bmatrix}6\\19\end{bmatrix}\]
Therefore,
\[\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-\frac{6}{11}\\-\frac{19}{11}\end{bmatrix}\]
Thus,
\[\boxed{\bbox[5pt]{x=-\frac{6}{11}}}\]
and
\[\boxed{\bbox[5pt]{y=-\frac{19}{11}}}\]

🎯 Exam Significance

Exam Significance

›

This problem is an important application of the inverse-matrix method for solving simultaneous linear equations. A complete board-examination solution should demonstrate the matrix representation, calculation of the determinant, construction of the adjoint, calculation of the inverse, and multiplication by the constant matrix.

The determinant

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

establishes that \(A^{-1}\) exists and that the system has a unique solution. Students should take particular care with the signs while calculating both the determinant and the adjoint.

Significance for Competitive Entrance Examinations

This question reinforces the direct relationship between the determinant and the inverse matrix. Once

\[ |A|\neq0 \]

is established, there is no need to investigate consistency further: the system has a unique solution.

For objective and time-bound examinations, the inverse can be constructed immediately using the \(2\times2\) formula. Keeping the factor

\[ -\frac{1}{11} \]

outside the matrix multiplication also makes the arithmetic faster and reduces the risk of errors involving fractions.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

The system is represented as
\[ AX=B. \]
The coefficient matrix is
\[ A= \begin{bmatrix} 4&-3\\ 3&-5 \end{bmatrix}. \]
The determinant is
\[ |A|=(4)(-5)-(-3)(3)=-11. \]
Since
\[ |A|=-11\neq0, \]
\(A^{-1}\) exists and the system has a unique solution.
The adjoint is
\[ \operatorname{adj}(A)= \begin{bmatrix} -5&3\\ -3&4 \end{bmatrix}. \]
The inverse matrix is
\[ A^{-1} = -\frac{1}{11} \begin{bmatrix} -5&3\\ -3&4 \end{bmatrix}. \]
Using
\[ X=A^{-1}B, \]
we obtain
\[ X= \begin{bmatrix} -\frac{6}{11}\\ -\frac{19}{11} \end{bmatrix}. \]
Therefore,
\[ \boxed{x=-\frac{6}{11},\quad y=-\frac{19}{11}}. \]

📘 Concept & Theory

Concept/Theory

›

A system of simultaneous linear equations can be represented in matrix form as

\[ AX=B, \]

where \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constant matrix.

If

\[ |A|\neq0, \]

then \(A\) is non-singular and its inverse exists. In that case,

\[ AX=B. \]

Multiplying both sides by \(A^{-1}\) on the left,

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

Since

\[ A^{-1}A=I, \]

we obtain

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

For a \(2\times2\) matrix

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

the inverse is

\[ A^{-1} = \frac{1}{|A|} \begin{bmatrix} d&-b\\ -c&a \end{bmatrix}, \]

provided

\[ |A|=ad-bc\neq0. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given equations in the form \(AX=B\).
Identify the coefficient matrix \(A\), variable matrix \(X\), and constant matrix \(B\).
Calculate \(|A|\) to verify that the inverse of \(A\) exists.
Find \(\operatorname{adj}(A)\).
Calculate \(A^{-1}\) using
\[ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A). \]
Use
\[ X=A^{-1}B \]
to determine \(x\) and \(y\).
Verify the obtained values in the original equations.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 23 steps

Given — system of equations
\[ \begin{align} 5x+y&=3\tag{1}\\ 3x+2y&=5\tag{2} \end{align} \]
Write the System in Matrix Form
The coefficients of \(x\) and \(y\) form the coefficient matrix:
\[A=\begin{bmatrix}5&1\\3&2\end{bmatrix}\]
The variable matrix is
\[X=\begin{bmatrix}x\\y\end{bmatrix},\]
and the constant matrix is
\[B=\begin{bmatrix}3\\5\end{bmatrix} \]
Therefore, the system can be written as
\[\boxed{\begin{bmatrix}5&1\\3&2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3\\5\end{bmatrix}}\]
Hence,
\[AX=B\]
Find the Determinant of \(A\)
We have
\[|A|=\begin{vmatrix}5&1\\3&2\end{vmatrix}\]
Using
\[\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc,\]
we get
\[|A|=(5)(2)-(1)(3)\]
Therefore,
\[\begin{aligned}|A|=10-3&=7\\&=7\end{aligned}\]
Since
\[|A|=7\neq0,\]
the matrix \(A\) is non-singular and \(A^{-1}\) exists. Therefore, the system has a unique solution.
Find \(\operatorname{adj}(A)\)
We have
\[A=\begin{bmatrix}5&1\\3&2\end{bmatrix}\]
For a \(2\times2\) matrix,
\[A=\begin{bmatrix}a&b\\c&d \end{bmatrix},\]
the adjoint is
\[\operatorname{adj}(A)=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\]
Therefore,
\[\operatorname{adj}(A)=\begin{bmatrix}2&-1\\-3&5\end{bmatrix}\]
Find \(A^{-1}\)
Using
\[A^{-1}=\frac{1}{|A|}\operatorname{adj}(A),\]
we obtain
\[A^{-1}=\frac{1}{7}\begin{bmatrix}2&-1\\-3&5\end{bmatrix}\]
Hence,
\[\boxed{A^{-1}=\begin{bmatrix}\frac{2}{7}&-\frac{1}{7}\\-\frac{3}{7}&\frac{5}{7}\end{bmatrix}}\]
Calculate \(X=A^{-1}B\)
From
\[AX=B,\]
multiplying both sides by \(A^{-1}\) on the left gives
\[A^{-1}(AX)=A^{-1}B\]
Since
\[A^{-1}A=I,\]
we get
\[X=A^{-1}B\]
Therefore,
\[\begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{7}\begin{bmatrix}2&-1\\-3&5\end{bmatrix}\begin{bmatrix}3\\5\end{bmatrix}\]
Performing the matrix multiplication,
\[\begin{aligned}\begin{bmatrix}x\\y\end{bmatrix}&=\frac{1}{7}\begin{bmatrix}(2)(3)+(-1)(5)\\(-3)(3)+(5)(5)\end{bmatrix}\end{aligned}\]
Simplifying the first entry,
\[ (2)(3)+(-1)(5)=6-5=1. \]
Simplifying the second entry,
\[ (-3)(3)+(5)(5)=-9+25=16. \]
Hence,
\[\begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{7}\begin{bmatrix}1\\16\end{bmatrix}\]
Therefore,
\[\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}\frac{1}{7}\\\frac{16}{7}\end{bmatrix}\]
Thus,
\[ \boxed{x=\frac{1}{7}} \]
and
\[ \boxed{y=\frac{16}{7}}\]

🎯 Exam Significance

Exam Significance

›

This question is a standard application of the inverse-matrix method for solving simultaneous linear equations. The complete procedure should be presented in the logical order

\[AX=B\rightarrow |A|\rightarrow\operatorname{adj}(A)\rightarrow A^{-1}\rightarrow X=A^{-1}B\]

An important point is to construct the coefficient matrix directly from the coefficients in the original equations. Here,

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

not a matrix obtained by accidentally interchanging any coefficient.

Also, the correct determinant is

\[ |A|=(5)(2)-(1)(3)=7, \]

so the inverse exists. Showing these intermediate steps makes the answer complete and mathematically rigorous for board evaluation.

Significance for Competitive Entrance Examinations

For entrance examinations, the determinant provides an immediate test for the existence of a unique solution. Since

\[ |A|=7\neq0, \]

the system has a unique solution, and there is no need to solve the equations by elimination merely to establish uniqueness.

The \(2\times2\) inverse formula also allows rapid calculation:

\[A^{-1}=\frac{1}{7}\begin{bmatrix}2&-1\\-3&5\end{bmatrix}\]

Maintaining the common factor \(1/7\) until the final multiplication keeps the calculation compact and reduces arithmetic errors.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

The system is written as
\[ AX=B \]
The coefficient matrix is
\[A=\begin{bmatrix}5&1\\3&2\end{bmatrix}\]
The determinant is
\[ |A|=(5)(2)-(1)(3)=7 \]
Since
\[ |A|=7\neq0, \]
\(A^{-1}\) exists and the system has a unique solution.
The adjoint is
\[\operatorname{adj}(A)=\begin{bmatrix}2&-1\\-3&5\end{bmatrix}\]
The inverse matrix is
\[A^{-1}=\frac{1}{7}\begin{bmatrix}2&-1\\-3&5\end{bmatrix}\]
Using
\[ X=A^{-1}B, \]
we obtain
\[X=\begin{bmatrix}\frac{1}{7}\\\frac{16}{7}\end{bmatrix}\]
Therefore,
\[ \boxed{x=\frac{1}{7},\quad y=\frac{16}{7}}. \]

📘 Concept & Theory

Concept/Theory

›

A system of simultaneous linear equations can be expressed in the matrix form

\[ AX=B, \]

where \(A\) is the coefficient matrix, \(X\) is the column matrix of variables, and \(B\) is the column matrix of constants.

If

\[ |A|\neq0, \]

then \(A\) is a non-singular matrix and its inverse exists. Starting with

\[ AX=B, \]

multiply both sides by \(A^{-1}\):

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

Since

\[ A^{-1}A=I, \]

we obtain

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

For a \(3\times3\) matrix, the inverse is obtained using

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

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

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given equations in the matrix form \(AX=B\).
Identify the coefficient matrix \(A\), variable matrix \(X\), and constant matrix \(B\).
Calculate \(|A|\) and check whether it is non-zero.
Find all the cofactors of \(A\).
Construct the cofactor matrix.
Transpose the cofactor matrix to obtain \(\operatorname{adj}(A)\).
Calculate \(A^{-1}\) using
\[ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A). \]
Use
\[ X=A^{-1}B \]
to determine \(x\), \(y\), and \(z\).
Verify the obtained values in the original equations.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 37 steps

Given — system of equations
\[ \begin{align} 2x+y+z&=1\tag{1}\\ x-2y-z&=\frac{3}{2}\tag{2}\\ 3y-5z&=9\tag{3} \end{align} \]
Notice that the coefficient of \(x\) in the third equation is \(0\). Therefore, equation (3) is written as
\[0x+3y-5z=9\]
Write the System in Matrix Form
The coefficient matrix is
\[A=\begin{bmatrix}2&1&1\\1&-2&-1\\0&3&-5\end{bmatrix}\]
The variable matrix is
\[X=\begin{bmatrix}x\\y\\z\end{bmatrix}\]
The constant matrix is
\[B=\begin{bmatrix}1\\\frac{3}{2}\\9\end{bmatrix}\]
Therefore,
\[\boxed{\bbox[5pt]{\begin{bmatrix}2&1&1\\1&-2&-1\\0&3&-5\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\\frac{3}{2}\\9\end{bmatrix}}}\]
Hence,
\[AX=B\]
Find the Determinant of \(A\)
We have
\[|A|=\begin{vmatrix}2&1&1\\1&-2&-1\\0&3&-5\end{vmatrix}\]
Expanding along the third row is convenient because its first element is \(0\):
\[\begin{aligned}|A|&=0\begin{vmatrix}1&1\\-2&-1\end{vmatrix}-3\begin{vmatrix}2&1\\1&-1\end{vmatrix}-5\begin{vmatrix}2&1\\1&-2\end{vmatrix}\end{aligned}\]
Now evaluate the first \(2\times2\) determinant:
\[\begin{vmatrix}1&1\\-2&-1\end{vmatrix}=(1)(-1)-(1)(-2)=-1+2=1\]
Its contribution is
\[0(1)=0\]
Next,
\[\begin{vmatrix}2&1\\1&-1\end{vmatrix}=(2)(-1)-(1)(1)=-2-1=-3\]
Therefore, its contribution is
\[-3(-3)=9\]
Finally,
\[\begin{vmatrix}2&1\\1&-2\end{vmatrix}=(2)(-2)-(1)(1)=-4-1=-5\]
Therefore, its contribution is
\[-5(-5)=25\]
Hence,
\[\begin{aligned}|A|&=0+9+25\\&=34\end{aligned}\]
Thus,
\[\boxed{|A|=34}\]
Since
\[|A|=34\neq0,\]
the matrix \(A\) is non-singular and \(A^{-1}\) exists. Therefore, the system is consistent and has a unique solution.
Find the Cofactors of \(A\)
Let \(A_{ij}\) denote the cofactor corresponding to the element in the \(i\)-th row and \(j\)-th column. Recall that
\[ A_{ij}=(-1)^{i+j}M_{ij}, \]
where \(M_{ij}\) is the corresponding minor.
Cofactors of the first row:
\[\begin{aligned}A_{11}&=(+1)\begin{vmatrix}-2&-1\\3&-5\end{vmatrix}\\&=(-2)(-5)-(-1)(3)\\&=10+3\\&=13\end{aligned}\]

\[\begin{aligned}A_{12}&=(-1)\begin{vmatrix}1&-1\\0&-5\end{vmatrix}\\&=-\left[(1)(-5)-(-1)(0)\right]\\&=-(-5)\\&=5\end{aligned}\]

\[\begin{aligned}A_{13}&=(+1)\begin{vmatrix}1&-2\\0&3\end{vmatrix}\\&=(1)(3)-(-2)(0)\\&=3\end{aligned}\]
Cofactors of the second row:
\[\begin{aligned}A_{21}&=(-1)\begin{vmatrix}1&1\\3&-5\end{vmatrix}\\&=-\left[(1)(-5)-(1)(3)\right]\\&=-(-5-3)\\&=8\end{aligned}\]

\[\begin{aligned}A_{22}&=(+1)\begin{vmatrix}2&1\\0&-5\end{vmatrix}\\&=(2)(-5)-(1)(0)\\&=-10\end{aligned}\]

\[\begin{aligned}A_{23}&=(-1)\begin{vmatrix}2&1\\0&3\end{vmatrix}\\&=-\left[(2)(3)-(1)(0)\right]\\&=-6\end{aligned}\]
Cofactors of the third row:
\[\begin{aligned}A_{31}&=(+1)\begin{vmatrix}1&1\\-2&-1\end{vmatrix}\\&=(1)(-1)-(1)(-2)\\&=-1+2\\&=1\end{aligned}\]

\[\begin{aligned}A_{32}&=(-1)\begin{vmatrix}2&1\\1&-1\end{vmatrix}\\&=-\left[(2)(-1)-(1)(1)\right]\\&=-(-2-1)\\&=3\end{aligned}\]

\[\begin{aligned}A_{33}&=(+1)\begin{vmatrix}2&1\\1&-2\end{vmatrix}\\&=(2)(-2)-(1)(1)\\&=-4-1\\&=-5\end{aligned}\]
Therefore, the cofactor matrix is
\[ C=\begin{bmatrix} 13&5&3\\ 8&-10&-6\\ 1&3&-5 \end{bmatrix} \]
Find the Adjoint of \(A\)
The adjoint of a matrix is the transpose of its cofactor matrix:
\[ \operatorname{adj}(A)=C^{T}. \]
Therefore,
\[\operatorname{adj}(A)=\begin{bmatrix}13&8&1\\5&-10&3\\3&-6&-5\end{bmatrix}\]
Hence,
\[\boxed{\bbox[5pt]{\operatorname{adj}(A)=\begin{bmatrix}13&8&1\\5&-10&3\\3&-6&-5\end{bmatrix}}}\]
Find \(A^{-1}\)
We use the formula
\[A^{-1}=\frac{1}{|A|}\operatorname{adj}(A)\]
Since
\[|A|=34,\]
we get
\[A^{-1}=\frac{1}{34}\begin{bmatrix}13&8&1\\5&-10&3\\3&-6&-5\end{bmatrix}\]
Thus,
\[\boxed{\bbox[5pt]{A^{-1}=\frac{1}{34}\begin{bmatrix}13&8&1\\5&-10&3\\3&-6&-5\end{bmatrix}}}\]
Calculate \(X=A^{-1}B\)
From
\[AX=B,\]
multiplying both sides by \(A^{-1}\) on the left gives
\[A^{-1}AX=A^{-1}B\]
Since
\[A^{-1}A=I,\]
we obtain
\[X=A^{-1}B\]
Therefore,
\[\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{34}\begin{bmatrix}13&8&1\\5&-10&3\\3&-6&-5\end{bmatrix}\begin{bmatrix}1\\\frac{3}{2}\\9\end{bmatrix}\]
Performing the matrix multiplication, the first component is
\[\begin{aligned}x&=\frac{1}{34}\left[(13)(1)+(8)\left(\frac{3}{2}\right)+(1)(9)\right]\\&=\frac{1}{34}\left[13+12+9\right]\\&=\frac{34}{34}\\&=1\end{aligned}\]
The second component is
\[\begin{aligned}y&=\frac{1}{34}\left[(5)(1) +(-10)\left(\frac{3}{2}\right)+(3)(9)\right]\\ &=\frac{1}{34}\left[5-15+27\right]\\ &=\frac{17}{34}\\&=\frac{1}{2}\end{aligned}\]
The third component is
\[\begin{aligned}z&=\frac{1}{34}\left[(3)(1) +(-6)\left(\frac{3}{2}\right)+(-5)(9)\right]\\ &=\frac{1}{34}\left[3-9-45\right]\\ &=\frac{-51}{34}\\&=-\frac{3}{2}\end{aligned}\]
Therefore,
\[\boxed{\bbox[5pt]{\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\\frac{1}{2}\\-\frac{3}{2}\end{bmatrix}}}\]
Hence,
\[ \boxed{x=1,\qquad y=\frac{1}{2},\qquad z=-\frac{3}{2}}. \]

🎯 Exam Significance

Exam Significance

›

This problem is a comprehensive application of the inverse-matrix method for a \(3\times3\) system. It tests several important concepts from the chapter: matrix representation of a system, determinant, minors, cofactors, adjoint, inverse matrix, and matrix multiplication.

A particularly important point is the zero coefficient of \(x\) in the third equation. It must be entered as \(0\) in the coefficient matrix:

\[A=\begin{bmatrix}2&1&1\\1&-2&-1\\0&3&-5\end{bmatrix}\]

Omitting this zero changes the matrix and leads to an incorrect solution.

The calculation

\[ |A|=34\neq0 \]

proves that \(A^{-1}\) exists. Therefore, the system is consistent and has a unique solution. In a board examination, students should clearly show the cofactor calculations and the transposition required to obtain the adjoint.

Significance for Competitive Entrance Examinations

This problem is useful for mastering the relationship between determinants, cofactors, adjoints, and inverse matrices. The determinant provides the quickest structural test:

\[ |A|\neq0 \quad\Longrightarrow\quad A^{-1}\text{ exists} \quad\Longrightarrow\quad \text{unique solution}. \]

For time-bound examinations, expanding the determinant along the third row is efficient because its first element is zero. This eliminates one minor immediately and reduces the amount of arithmetic.

Another important examination skill is maintaining the factor

\[ \frac{1}{34} \]

outside the matrix multiplication until the final stage. This makes the calculation cleaner and minimizes errors involving fractions.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

A system of three linear equations can be represented as
\[ AX=B. \]
A missing variable in an equation must be represented by a zero coefficient. Here, the third equation contributes the row
\[ [0\quad3\quad-5]. \]
The determinant of the coefficient matrix is
\[ |A|=34. \]
Since
\[ |A|=34\neq0, \]
the coefficient matrix is non-singular and its inverse exists.
The cofactor matrix is
\[ \begin{bmatrix} 13&5&3\\ 8&-10&-6\\ 1&3&-5 \end{bmatrix}. \]
The adjoint is the transpose of the cofactor matrix:
\[ \operatorname{adj}(A)= \begin{bmatrix} 13&8&1\\ 5&-10&3\\ 3&-6&-5 \end{bmatrix}. \]
The inverse is
\[ A^{-1} = \frac{1}{34} \begin{bmatrix} 13&8&1\\ 5&-10&3\\ 3&-6&-5 \end{bmatrix}. \]
Using
\[ X=A^{-1}B, \]
the solution is
\[ \boxed{x=1,\quad y=\frac{1}{2},\quad z=-\frac{3}{2}}. \]
Substitution into all three original equations confirms the solution.

📘 Concept & Theory

Concept/Theory

›

A system of simultaneous linear equations can be expressed compactly in matrix form as

\[ AX=B, \]

where \(A\) is the coefficient matrix, \(X\) contains the unknown variables, and \(B\) contains the constant terms.

If

\[ |A|\neq0, \]

then \(A\) is non-singular, so its inverse \(A^{-1}\) exists. Starting from

\[ AX=B, \]

multiply both sides on the left by \(A^{-1}\):

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

Since

\[ A^{-1}A=I, \]

we get

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

For a \(3\times3\) matrix, its inverse is obtained from

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

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

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given equations in the form \(AX=B\).
Identify the coefficient matrix \(A\), variable matrix \(X\), and constant matrix \(B\).
Calculate \(|A|\) and check whether \(A^{-1}\) exists.
Find all the cofactors of \(A\).
Form the cofactor matrix and transpose it to obtain \(\operatorname{adj}(A)\).
Use
\[ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A) \]
to calculate \(A^{-1}\).
Use
\[ X=A^{-1}B \]
to obtain \(x\), \(y\), and \(z\).
Finally, substitute the values back into the original equations to verify the result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 33 steps

Given — system of equations
\[ \begin{align} x-y+z&=4\tag{1}\\ 2x+y-3z&=0\tag{2}\\ x+y+z&=2\tag{3} \end{align} \]
Write the System in Matrix Form
Reading the coefficients of \(x\), \(y\), and \(z\) from the three equations gives
\[ A= \begin{bmatrix} 1&-1&1\\ 2&1&-3\\ 1&1&1 \end{bmatrix}\]
The variable matrix is
\[X=\begin{bmatrix}x\\y\\z\end{bmatrix},\]
while the constant matrix is
\[B=\begin{bmatrix}4\\0\\2\end{bmatrix}\]
Therefore, the given system can be written as
\[\boxed{\bbox[5pt]{\begin{bmatrix}1&-1&1\\2&1&-3\\1&1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}4\\0\\2\end{bmatrix}}}\]
Thus,
\[AX=B\]
Find the Determinant of \(A\)
We now calculate the determinant of the coefficient matrix:
\[|A|=\begin{vmatrix}1&-1&1\\2&1&-3\\1&1&1\end{vmatrix}\]
Expanding along the first row,
\[\begin{aligned}|A|&=1\begin{vmatrix}1&-3\\1&1\end{vmatrix}-(-1)\begin{vmatrix}2&-3\\1&1\end{vmatrix}+1\begin{vmatrix}2&1\\1&1\end{vmatrix}\end{aligned}\]
First minor:
\[\begin{aligned}\begin{vmatrix}1&-3\\1&1\end{vmatrix}&=(1)(1)-(-3)(1)\\&=1+3\\&=4\end{aligned}\]
Second minor:
\[\begin{aligned}\begin{vmatrix}2&-3\\1&1\end{vmatrix}&=(2)(1)-(-3)(1)\\&=2+3\\&=5\end{aligned}\]
Third minor:
\[\begin{aligned}\begin{vmatrix}2&1\\1&1\end{vmatrix}&=(2)(1)-(1)(1)\\&=2-1\\&=1\end{aligned}\]
Therefore,
\[\begin{aligned}|A|&=4+5+1\\&=10\end{aligned}\]
Hence,
\[\boxed{|A|=10}\]
Since
\[|A|=10\neq0,\]
\(A\) is non-singular and \(A^{-1}\) exists. Therefore, the given system is consistent and has a unique solution.
Find the Cofactors of \(A\)
The cofactor corresponding to \(a_{ij}\) is
\[A_{ij}=(-1)^{i+j}M_{ij},\]
where \(M_{ij}\) denotes the corresponding minor.
The signs of the cofactors follow the pattern
\[\begin{bmatrix}+&-&+\\-&+&-\\+&-&+\end{bmatrix}\]
Cofactors of the first row:
\[\begin{aligned}A_{11}&=(+1)\begin{vmatrix}1&-3\\1&1\end{vmatrix}\\&=(1)(1)-(-3)(1)\\&=1+3\\&=4\end{aligned}\]

\[\begin{aligned}A_{12}&=(-1)\begin{vmatrix}2&-3\\1&1\end{vmatrix}\\&=-\left[(2)(1)-(-3)(1)\right]\\&=-(2+3)\\&=-5\end{aligned}\]
\[\begin{aligned}A_{13}&=(+1)\begin{vmatrix}2&1\\1&1\end{vmatrix}\\&=(2)(1)-(1)(1)\\&=2-1\\&=1\end{aligned}\]
Cofactors of the second row:
\[\begin{aligned}A_{21}&=(-1)\begin{vmatrix}-1&1\\1&1\end{vmatrix}\\&=-\left[(-1)(1)-(1)(1)\right]\\&=-(-1-1)\\&=2\end{aligned}\]

\[\begin{aligned}A_{22}&=(+1)\begin{vmatrix}1&1\\1&1\end{vmatrix}\\&=(1)(1)-(1)(1)\\&=0\end{aligned}\]

\[\begin{aligned}A_{23}&=(-1)\begin{vmatrix}1&-1\\1&1\end{vmatrix}\\&=-\left[(1)(1)-(-1)(1)\right]\\&=-(1+1)\\&=-2\end{aligned}\]
Cofactors of the third row:
\[\begin{aligned}A_{31}&=(+1)\begin{vmatrix}-1&1\\1&-3\end{vmatrix}\\&=(-1)(-3)-(1)(1)\\&=3-1\\&=2\end{aligned}\]

\[\begin{aligned}A_{32}&=(-1)\begin{vmatrix}1&1\\2&-3\end{vmatrix}\\&=-\left[(1)(-3)-(1)(2)\right]\\&=-(-3-2)\\&=5\end{aligned}\]

\[\begin{aligned}A_{33}&=(+1)\begin{vmatrix}1&-1\\2&1\end{vmatrix}\\&=(1)(1)-(-1)(2)\\&=1+2\\&=3\end{aligned}\]
Therefore, the cofactor matrix is
\[C=\begin{bmatrix}4&-5&1\\2&0&-2\\2&5&3\end{bmatrix}\]
Find the Adjoint of \(A\)
Remember that the adjoint is obtained by taking the transpose of the cofactor matrix:
\[ \operatorname{adj}(A)=C^T. \]
Hence,
\[\operatorname{adj}(A)=\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}\]
Therefore,
\[\boxed{\bbox[5pt]{\operatorname{adj}(A)=\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}}}\]
Find \(A^{-1}\)
Using
\[A^{-1}=\frac{1}{|A|}\operatorname{adj}(A),\]
we get
\[A^{-1}=\frac{1}{10}\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}\]
Thus,
\[\boxed{\bbox[5pt]{A^{-1}=\frac{1}{10}\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}}}\]
Calculate \(X=A^{-1}B\)
From
\[AX=B,\]
multiply both sides on the left by \(A^{-1}\):
\[A^{-1}AX=A^{-1}B\]
Since
\[A^{-1}A=I,\]
we obtain
\[X=A^{-1}B\]
Therefore,
\[\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{10}\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}\begin{bmatrix}4\\0\\2\end{bmatrix}\]
For the first component,
\[\begin{aligned}x&=\frac{1}{10}\left[(4)(4)+(2)(0)+(2)(2)\right]\\&=\frac{1}{10}[16+0+4]\\&=\frac{20}{10}\\&=2\end{aligned}\]
For the second component,
\[\begin{aligned}y&=\frac{1}{10}\left[(-5)(4)+(0)(0)+(5)(2)\right]\\&=\frac{1}{10}[-20+0+10]\\&=\frac{-10}{10}\\&=-1\end{aligned}\]
For the third component,
\[\begin{aligned}z&=\frac{1}{10}\left[(1)(4)+(-2)(0)+(3)(2)\right]\\&=\frac{1}{10}[4+0+6]\\&=\frac{10}{10}\\&=1\end{aligned}\]
Therefore,
\[\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\-1\\1\end{bmatrix}\]
Hence,
\[\boxed{x=2,\quad y=-1,\quad z=1}\]

🎯 Exam Significance

Exam Significance

›

This question is a useful application of the inverse-matrix method and combines several important ideas from the chapter: matrix representation, determinants, minors, cofactors, adjoint, inverse of a matrix, and matrix multiplication.

For a board examination, it's important to show the logical sequence clearly. In particular, calculate \(|A|\) before attempting to find \(A^{-1}\). Here,

\[ |A|=10\neq0, \]

so the inverse exists and the system has a unique solution.

Pay special attention to the cofactor signs:

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

A sign error at this stage can change the adjoint and consequently the final answer.

Another frequent mistake is confusing the cofactor matrix with the adjoint. The cofactor matrix here is

\[ C= \begin{bmatrix} 4&-5&1\\ 2&0&-2\\ 2&5&3 \end{bmatrix}, \]

while its transpose gives

\[ \operatorname{adj}(A)= \begin{bmatrix} 4&2&2\\ -5&0&5\\ 1&-2&3 \end{bmatrix} \]

Writing the intermediate steps makes the solution easier to follow and helps ensure that marks are not lost for an unexplained jump between the cofactor calculation, adjoint, inverse, and final solution.

Significance for Competitive Entrance Examinations

In competitive examinations, the first useful observation is often the determinant:

\[ |A|=10\neq0. \]

This immediately tells you that \(A\) is invertible and that the system has a unique solution.

When calculating a determinant, choose an expansion that keeps the arithmetic simple. For this matrix, expanding along the first row is convenient:

\[ |A|=4+5+1=10. \]

The inverse-matrix method is particularly useful when a question explicitly asks for the matrix method or when the same coefficient matrix occurs with different constant matrices. Once \(A^{-1}\) is available, the solution follows directly from

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

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

The system is represented in matrix form as
\[ AX=B. \]
The coefficient matrix is
\[ A= \begin{bmatrix} 1&-1&1\\ 2&1&-3\\ 1&1&1 \end{bmatrix} \]
The determinant is
\[ |A|=10. \]
Since
\[ |A|=10\neq0, \]
\(A^{-1}\) exists and the system has a unique solution.
The cofactor matrix is
\[ C= \begin{bmatrix} 4&-5&1\\ 2&0&-2\\ 2&5&3 \end{bmatrix}. \]
The adjoint is
\[ \operatorname{adj}(A)= \begin{bmatrix} 4&2&2\\ -5&0&5\\ 1&-2&3 \end{bmatrix} \]
The inverse is
\[ A^{-1} = \frac{1}{10} \begin{bmatrix} 4&2&2\\ -5&0&5\\ 1&-2&3 \end{bmatrix} \]
Using
\[ X=A^{-1}B, \]
we obtain
\[ \boxed{x=2,\quad y=-1,\quad z=1}. \]
Substitution into all three original equations confirms that the solution is correct.

📘 Concept & Theory

Concept/Theory

›

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

\[ AX=B, \]

where \(A\) is the coefficient matrix, \(X\) is the column matrix of unknowns, and \(B\) is the column matrix of constants.

If

\[ |A|\neq0, \]

then \(A\) is non-singular and its inverse exists. Multiplying

\[ AX=B \]

by \(A^{-1}\) on the left gives

\[ A^{-1}AX=A^{-1}B, \]

and hence

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

The inverse of a non-singular square matrix is calculated using

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

Therefore, this problem requires the calculation of the determinant, cofactors, adjoint, inverse, and finally the product \(A^{-1}B\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given equations in matrix form \(AX=B\).
Identify the coefficient matrix \(A\), variable matrix \(X\), and constant matrix \(B\).
Calculate \(|A|\) to determine whether \(A^{-1}\) exists.
Find all nine cofactors of \(A\).
Form the cofactor matrix.
Transpose the cofactor matrix to obtain \(\operatorname{adj}(A)\).
Calculate \(A^{-1}\) using
\[ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A). \]
Use
\[ X=A^{-1}B \]
to determine \(x\), \(y\), and \(z\).
Verify the values in the original equations.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 33 steps

Given — equations are
\[ \begin{aligned} 2x+3y+3z&=5\\ x-2y+z&=-4\\ 3x-y-2z&=3 \end{aligned} \]
Write the System in Matrix Form
Comparing the coefficients of \(x\), \(y\), and \(z\), the coefficient matrix is
\[ A= \begin{bmatrix} 2&3&3\\ 1&-2&1\\ 3&-1&-2 \end{bmatrix} \]
The variable matrix is
\[X=\begin{bmatrix}x\\y\\z\end{bmatrix},\]
and the constant matrix is
\[B=\begin{bmatrix}5\\-4\\3\end{bmatrix}\]
Hence,
\[\boxed{\bbox[5pt]{\begin{bmatrix}2&3&3\\1&-2&1\\3&-1&-2\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}5\\-4\\ 3\end{bmatrix}}}\]
Thus,
\[AX=B\]
Find the Determinant of \(A\)
We have
\[|A|=\begin{vmatrix}2&3&3\\1&-2&1\\3&-1&-2\end{vmatrix}\]
Expanding along the first row,
\[\begin{aligned}|A|&=2\begin{vmatrix}-2&1\\-1&-2\end{vmatrix}-3 \begin{vmatrix}1&1\\3&-2\end{vmatrix}+3\begin{vmatrix}1&-2\\3&-1\end{vmatrix}\end{aligned}\]
Evaluating the first minor,
\[\begin{aligned}\begin{vmatrix}-2&1\\-1&-2\end{vmatrix} &=(-2)(-2)-(1)(-1)\\&=4+1\\&=5\end{aligned}\]
Evaluating the second minor,
\[\begin{aligned}\begin{vmatrix}1&1\\3&-2\end{vmatrix} &=(1)(-2)-(1)(3)\\&=-2-3\\&=-5\end{aligned}\]
Evaluating the third minor,
\[\begin{aligned}\begin{vmatrix}1&-2\\3&-1\end{vmatrix} &=(1)(-1)-(-2)(3)\\&=-1+6\\&=5\end{aligned}\]
Therefore,
\[\begin{aligned}|A|&=2(5)-3(-5)+3(5)\\&=10+15+15\\&=40\end{aligned}\]
Hence,
\[\boxed{|A|=40}\]
the matrix \(A\) is non-singular. Therefore, \(A^{-1}\) exists and the given system has a unique solution.
Find the Cofactors of \(A\)
The cofactor of an element \(a_{ij}\) is given by
\[A_{ij}=(-1)^{i+j}M_{ij},\]
where \(M_{ij}\) is the corresponding minor.
The cofactor sign pattern is
\[\begin{bmatrix}+&-&+\\-&+&-\\+&-&+\end{bmatrix}\]
Cofactors of the first row:
\[\begin{aligned}A_{11}&=(+1)\begin{vmatrix}-2&1\\-1&-2\end{vmatrix}\\&=(-2)(-2)-(1)(-1)\\&=4+1\\&=5\end{aligned}\]

\[\begin{aligned}A_{12}&=(-1)\begin{vmatrix}1&1\\3&-2\end{vmatrix}\\&=-\left[(1)(-2)-(1)(3)\right]\\&=-(-2-3)\\&=5\end{aligned}\]

\[\begin{aligned}A_{13}&=(+1)\begin{vmatrix}1&-2\\3&-1\end{vmatrix}\\&=(1)(-1)-(-2)(3)\\&=-1+6\\&=5\end{aligned}\]
Cofactors of the second row:
\[\begin{aligned}A_{21}&=(-1)\begin{vmatrix}3&3\\-1&-2\end{vmatrix}\\&=-\left[(3)(-2)-(3)(-1)\right]\\&=-(-6+3)\\&=3\end{aligned}\]

\[\begin{aligned}A_{22}&=(+1)\begin{vmatrix}2&3\\3&-2\end{vmatrix}\\&=(2)(-2)-(3)(3)\\&=-4-9\\&=-13\end{aligned}\]

\[\begin{aligned}A_{23}&=(-1)\begin{vmatrix}2&3\\3&-1\end{vmatrix}\\&=-\left[(2)(-1)-(3)(3)\right]\\&=-(-2-9)\\&=11\end{aligned}\]
Cofactors of the third row:
\[\begin{aligned}A_{31}&=(+1)\begin{vmatrix}3&3\\-2&1\end{vmatrix}\\&=(3)(1)-(3)(-2)\\&=3+6\\&=9\end{aligned}\]

\[\begin{aligned}A_{32}&=(-1)\begin{vmatrix}2&3\\1&1\end{vmatrix}\\&=-\left[(2)(1)-(3)(1)\right]\\&=-(2-3)\\&=1\end{aligned}\]

\[\begin{aligned}A_{33}&=(+1)\begin{vmatrix}2&3\\1&-2\end{vmatrix}\\&=(2)(-2)-(3)(1)\\&=-4-3\\&=-7\end{aligned}\]
Therefore, the cofactor matrix is
\[C=\begin{bmatrix}5&5&5\\3&-13&11\\9&1&-7\end{bmatrix}\]
Find the Adjoint of \(A\)
The adjoint of \(A\) is the transpose of its cofactor matrix:
\[ \operatorname{adj}(A)=C^T. \]
Therefore,
\[\operatorname{adj}(A)=\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}\]
Hence,
\[\boxed{\bbox[5pt]{\operatorname{adj}(A)=\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}}}\]
Find the Inverse of \(A\)
We use
\[A^{-1}=\frac{1}{|A|}\operatorname{adj}(A)\]
Since
\[|A|=40,\]
we obtain
\[\begin{aligned}A^{-1}&=\frac{1}{40}\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}\end{aligned}\]
Thus,
\[\boxed{\bbox[5pt]{A^{-1}=\frac{1}{40}\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}}}\]
Find \(X=A^{-1}B\)
We have
\[AX=B\]
Multiplying both sides by \(A^{-1}\),
\[A^{-1}AX=A^{-1}B\]
Since
\[A^{-1}A=I,\]
we get
\[X=A^{-1}B\]
Therefore,
\[\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{40}\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}\begin{bmatrix}5\\-4\\3\end{bmatrix}\]
Calculating the first component,
\[\begin{aligned}x&=\frac{1}{40}\left[(5)(5)+(3)(-4)+(9)(3)\right]\\ &=\frac{1}{40}\left[25-12+27\right]\\ &=\frac{40}{40}\\ &=1\end{aligned} \]
Calculating the second component,
\[\begin{aligned}y&=\frac{1}{40}\left[(5)(5)+(-13)(-4)+(1)(3)\right]\\ &=\frac{1}{40}\left[25+52+3\right]\\ &=\frac{80}{40}\\ &=2\end{aligned} \]
Calculating the third component,
\[\begin{aligned}z&=\frac{1}{40}\left[(5)(5)+(11)(-4)+(-7)(3)\right]\\ &=\frac{1}{40}\left[25-44-21\right]\\ &=\frac{-40}{40}\\ &=-1\end{aligned} \]
Hence,
\[\begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}1\\2\\-1\end{bmatrix}. \]
Therefore,
\[\boxed{x=1,\qquad y=2,\qquad z=-1}\]

🎯 Exam Significance

Exam Significance

›

This problem is an important application of the inverse-matrix method in Class 12 Mathematics. It tests several connected concepts from matrices and determinants: representation of a system as \(AX=B\), evaluation of a determinant, calculation of cofactors, formation of the adjoint, calculation of the inverse, and matrix multiplication.

The determinant must be checked before finding the inverse:

\[ |A|=40\neq0. \]

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

For board examinations, every major intermediate step should be shown. In particular, do not directly write the adjoint without establishing the cofactor matrix. Also remember that the adjoint is the transpose of the cofactor matrix, not the cofactor matrix itself.

The cofactor matrix is

\[\begin{bmatrix}5&5&5\\3&-13&11\\9&1&-7\end{bmatrix},\]

whereas the adjoint is

\[\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}\]

Correct signs while calculating cofactors are especially important because one incorrect sign changes the adjoint and therefore changes the final solution.

Significance for Competitive Entrance Examinations

For competitive examinations, the determinant provides an immediate structural check:

\[ |A|=40\neq0. \]

Therefore, the coefficient matrix is invertible and the system has exactly one solution.

The cofactor and determinant calculations also provide useful practice in sign management and minor evaluation. In a time-bound examination, careful selection of the row or column for determinant expansion can reduce arithmetic considerably.

Once the inverse has been obtained, the entire system is reduced to the compact relation

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

This is particularly useful when several systems share the same coefficient matrix \(A\), because the inverse can be calculated once and reused with different constant matrices.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

The system is first written as
\[ AX=B. \]
The coefficient matrix is
\[A=\begin{bmatrix}2&3&3\\1&-2&1\\3&-1&-2\end{bmatrix}\]
The determinant is
\[|A|=40\neq0.\]
Therefore, \(A\) is non-singular, \(A^{-1}\) exists, and the system has a unique solution.
The cofactor matrix is
\[C=\begin{bmatrix}5&5&5\\3&-13&11\\9&1&-7\end{bmatrix}\]
The adjoint is
\[\operatorname{adj}(A)=\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}\]
The inverse is
\[A^{-1}=\frac{1}{40}\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}\]
Using
\[ X=A^{-1}B, \]
we obtain
\[ \boxed{x=1,\quad y=2,\quad z=-1}. \]
Substitution into all three original equations confirms the solution.

📘 Concept & Theory

Concept/Theory

›

A system of three simultaneous linear equations can be expressed in the matrix form

\[ AX=B, \]

where \(A\) is the coefficient matrix, \(X\) is the column matrix containing the variables, and \(B\) is the column matrix of constants.

If

\[ |A|\neq0, \]

then \(A\) is non-singular and its inverse exists. Multiplying \(AX=B\) by \(A^{-1}\) on the left gives

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

Since

\[ A^{-1}A=I, \]

we obtain

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

The inverse of a non-singular matrix is calculated using

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

Thus, we first calculate the determinant and cofactors of \(A\), form \(\operatorname{adj}(A)\), calculate \(A^{-1}\), and finally determine \(X\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given equations in matrix form \(AX=B\).
Identify the coefficient matrix \(A\) and constant matrix \(B\).
Calculate \(|A|\).
Since \(|A|\neq0\), conclude that \(A^{-1}\) exists and the system has a unique solution.
Calculate all nine cofactors of \(A\).
Form the cofactor matrix.
Transpose the cofactor matrix to obtain \(\operatorname{adj}(A)\).
Calculate \(A^{-1}\).
Use \(X=A^{-1}B\) to obtain \(x,y,z\).
Verify the solution in the original equations.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 32 steps

Write the System in Matrix Form
The given equations are
\[ \begin{aligned} x-y+2z&=7\\ 3x+4y-5z&=-5\\ 2x-y+3z&=12 \end{aligned} \]
Comparing the coefficients of \(x\), \(y\), and \(z\), the coefficient matrix is
\[A=\begin{bmatrix}1&-1&2\\3&4&-5\\2&-1&3\end{bmatrix}\]
The variable matrix is
\[X=\begin{bmatrix}x\\y\\z\end{bmatrix},\]
and the constant matrix is
\[B=\begin{bmatrix}7\\-5\\12\end{bmatrix}\]
Therefore,
\[\boxed{\bbox[5pt]{\begin{bmatrix}1&-1&2\\3&4&-5\\2&-1&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}7\\-5\\12\end{bmatrix}}}\]
Thus,
\[AX=B\]
Calculate the Determinant of \(A\)
We have
\[|A|=\begin{vmatrix}1&-1&2\\3&4&-5\\2&-1&3\end{vmatrix}\]
Expanding along the first row,
\[\begin{aligned}|A|&=1\begin{vmatrix}4&-5\\-1&3\end{vmatrix} -(-1)\begin{vmatrix}3&-5\\2&3\end{vmatrix} +2\begin{vmatrix}3&4\\2&-1\end{vmatrix}\end{aligned}\]
Evaluate the first minor:
\[\begin{aligned}\begin{vmatrix}4&-5\\-1&3\end{vmatrix}&=(4)(3)-(-5)(-1)\\&=12-5\\&=7\end{aligned}\]
Evaluate the second minor:
\[\begin{aligned}\begin{vmatrix}3&-5\\2&3\end{vmatrix}&=(3)(3)-(-5)(2)\\&=9+10\\&=19\end{aligned}\]
Evaluate the third minor:
\[\begin{aligned}\begin{vmatrix}3&4\\2&-1\end{vmatrix} &=(3)(-1)-(4)(2)\\ &=-3-8\\ &=-11 \end{aligned} \]
Hence,
\[ \begin{aligned} |A| &=1(7)-(-1)(19)+2(-11)\\ &=7+19-22\\ &=26-22\\ &=4 \end{aligned} \]
Therefore,
\[ \boxed{|A|=4}. \]
Since
\[|A|=4\neq0,\]
matrix \(A\) is non-singular. Hence \(A^{-1}\) exists and the system has a unique solution.
Find the Cofactors of \(A\)
The cofactor of \(a_{ij}\) is
\[ A_{ij}=(-1)^{i+j}M_{ij}, \]
where \(M_{ij}\) is the corresponding minor.
The cofactor sign pattern is
\[ \begin{bmatrix} +&-&+\ -&+&-\ +&-&+ \end{bmatrix}. \]
Cofactors of the first row:
\[\begin{aligned}A_{11}&=(+1)\begin{vmatrix}4&-5\\-1&3\end{vmatrix}\\&=(4)(3)-(-5)(-1)\\&=12-5\\&=7\end{aligned}\]

\[\begin{aligned}A_{12}&=(-1)\begin{vmatrix}3&-5\\2&3\end{vmatrix}\\&=-\left[(3)(3)-(-5)(2)\right]\\&=-(9+10)\\&=-19\end{aligned}\]

\[\begin{aligned}A_{13}&=(+1)\begin{vmatrix}3&4\\2&-1\end{vmatrix}\\&=(3)(-1)-(4)(2)\\&=-3-8\\&=-11\end{aligned}\]
Cofactors of the second row:
\[\begin{aligned}A_{21}&=(-1)\begin{vmatrix}-1&2\\-1&3\end{vmatrix}\\&=-\left[(-1)(3)-(2)(-1)\right]\\&=-(-3+2)\\&=1\end{aligned}\]

\[\begin{aligned}A_{22}&=(+1)\begin{vmatrix}1&2\\2&3\end{vmatrix}\\&=(1)(3)-(2)(2)\\&=3-4\\&=-1\end{aligned}\]

\[\begin{aligned}A_{23}&=(-1)\begin{vmatrix}1&-1\\\end{vmatrix}\\&=-\left[(1)(-1)-(-1)(2)\right]\\&=-(-1+2)\\&=-1\end{aligned}\]
Cofactors of the third row:
\[\begin{aligned}A_{31}&=(+1)\begin{vmatrix}-1&2\\4&-5\end{vmatrix}\\&=(-1)(-5)-(2)(4)\\&=5-8\\&=-3\end{aligned}\]

\[\begin{aligned}A_{32}&=(-1)\begin{vmatrix}1&2\\3&-5\end{vmatrix}\\&=-\left[(1)(-5)-(2)(3)\right]\\&=-(-5-6)\\&=11\end{aligned}\]

\[\begin{aligned}A_{33}&=(+1)\begin{vmatrix}1&-1\\3&4\end{vmatrix}\\&=(1)(4)-(-1)(3)\\&=4+3\\&=7\end{aligned}\]
Therefore, the cofactor matrix is
\[C=\begin{bmatrix}7&-19&-11\\1&-1&-1\\-3&11&7\end{bmatrix}\]
Find the Adjoint of \(A\)
The adjoint of a matrix is the transpose of its cofactor matrix:
\[ \operatorname{adj}(A)=C^T. \]
Hence,
\[\begin{aligned}\operatorname{adj}(A)&=\begin{bmatrix}7&-19&-11\\1&-1&11\\-3&-1&7\end{bmatrix}\end{aligned}\]
Therefore,
\[\boxed{\operatorname{adj}(A)=\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}}\]
Find the Inverse of \(A\)
We use the formula
\[A^{-1}=\frac{1}{|A|}\operatorname{adj}(A)\]
Since \(|A|=4,\) we get
\[\begin{aligned}A^{-1}&=\frac{1}{4}\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}\end{aligned}\]
Thus,
\[\boxed{A^{-1}=\frac{1}{4}\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}}\]
Calculate \(X=A^{-1}B\)
From
\[ AX=B, \]
multiplying both sides by \(A^{-1}\), we obtain
\[ A^{-1}AX=A^{-1}B. \]
Since
\[ A^{-1}A=I, \]
it follows that
\[ X=A^{-1}B. \]
Therefore,
\[\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{4}\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}\begin{bmatrix}7\\-5\\12\end{bmatrix}\]
For the first component,
\[\begin{aligned}x&=\frac{1}{4}\left[(7)(7)+(1)(-5)+(-3)(12)\right]\\ &=\frac{1}{4}\left[49-5-36\right]\\ &=\frac{8}{4}\\ &=2\end{aligned} \]
For the second component,
\[\begin{aligned}y&=\frac{1}{4}\left[(-19)(7)+(-1)(-5)+(11)(12)\right]\\&=\frac{1}{4}\left[-133+5+132\right]\\&=\frac{4}{4}\\&=1\end{aligned}\]
For the third component,
\[\begin{aligned}z&=\frac{1}{4}\left[(-11)(7)+(-1)(-5)+(7)(12)\right]\\ &=\frac{1}{4}\left[-77+5+84\right]\\ &=\frac{12}{4}\\ &=3\end{aligned}\]
Hence,
\[\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\1\\3\end{bmatrix}\]
Therfore,
\[\boxed{x=2,\qquad y=1,\qquad z=3}\]

🎯 Exam Significance

Exam Significance

›

This problem is a comprehensive application of the inverse-matrix method for solving simultaneous linear equations. It combines important concepts from both matrices and determinants, including determinant evaluation, cofactors, adjoint, inverse of a matrix, and matrix multiplication.

A crucial first check is

\[ |A|=4\neq0. \]

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

For board examinations, the intermediate calculations should be shown clearly. In particular, the cofactor matrix must be distinguished from the adjoint. The adjoint is obtained by transposing the cofactor matrix.

Here,

\[C=\begin{bmatrix}7&-19&-11\\1&-1&-1\\-3&11&7\end{bmatrix}\]

whereas

\[\operatorname{adj}(A)=\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}\]

Writing the cofactors systematically and maintaining the correct \(+,-,+\) sign pattern helps avoid errors in the adjoint and inverse.

Significance for Competitive Entrance Examinations

This problem reinforces the relationship between the determinant and the solvability of a system. Since

\[ |A|\neq0, \]

the coefficient matrix is invertible and the system has exactly one solution.

For time-bound examinations, determinant expansion should be performed strategically. Once the inverse is available, the solution follows directly from

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

The problem also provides useful practice in cofactor signs, minor evaluation, transposition, and matrix multiplication, all of which are frequently tested either directly or as components of more advanced matrix and determinant questions.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 10 points

The system is represented as
\[ AX=B. \]
The coefficient matrix is
\[ A= \begin{bmatrix} 1&-1&2\\ 3&4&-5\\ 2&-1&3 \end{bmatrix} \]
The determinant is
\[ |A|=4\neq0. \]
Therefore, \(A\) is non-singular and \(A^{-1}\) exists.
The system consequently has a unique solution.
The cofactor matrix is
\[ C= \begin{bmatrix} 7&-19&-11\\ 1&-1&-1\\ -3&11&7 \end{bmatrix} \]
The adjoint is
\[ \operatorname{adj}(A)= \begin{bmatrix} 7&1&-3\\ -19&-1&11\\ -11&-1&7 \end{bmatrix} \]
The inverse is
\[ A^{-1} = \frac{1}{4} \begin{bmatrix} 7&1&-3\\ -19&-1&11\\ -11&-1&7 \end{bmatrix} \]
Using
\[ X=A^{-1}B, \]
the solution is
\[ \boxed{x=2,\quad y=1,\quad z=3}. \]
Substitution into all three original equations confirms the solution.

📘 Concept & Theory

Concept/Theory

›

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

\[ |A|\neq0. \]

When the inverse exists,

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

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

A system of linear equations can be written in the matrix form

\[ AX=B. \]

If \(A^{-1}\) exists, then

\[ A^{-1}AX=A^{-1}B, \]

and therefore

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

This gives the values of the unknowns directly.

🗺️ Solution Roadmap

Step-by-step Plan

›

Calculate all the cofactors of the given matrix \(A\).
Calculate \(|A|\) and verify that it is non-zero.
Form the cofactor matrix.
Transpose the cofactor matrix to obtain \(\operatorname{adj}(A)\).
Use \(A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A)\) to find \(A^{-1}\).
Write the given system as \(AX=B\).
Use \(X=A^{-1}B\) to calculate \(x,y,z\).
Verify the obtained values in the original equations.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 33 steps

Finding \(A^{-1}\)
Given — Matrix
\[A=\begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix}\]
Find the Cofactors of \(A\)
The cofactor of the element \(a_{ij}\) is
\[ A_{ij}=(-1)^{i+j}M_{ij}, \]
where \(M_{ij}\) is the corresponding minor.
The cofactor sign pattern is
\[\begin{bmatrix}+&-&+\\-&+&-\\+&-&+\end{bmatrix}\]
Cofactor \(A_{11}\):
\[\begin{aligned}A_{11}&=(+1)\begin{vmatrix}2&-4\\1&-2\end{vmatrix}\\&=(2)(-2)-(-4)(1)\\&=-4+4\\&=0\end{aligned}\]
Cofactor \(A_{12}\):
\[\begin{aligned}A_{12}&=(-1)\begin{vmatrix}3&-4\\1&-2\end{vmatrix}\\&=-\left[(3)(-2)-(-4)(1)\right]\\&=-(-6+4)\\&=2\end{aligned}\]
Cofactor \(A_{13}\):
\[\begin{aligned}A_{13}&=(+1)\begin{vmatrix}3&2\\1&1\end{vmatrix}\\&=(3)(1)-(2)(1)\\&=3-2\\&=1\end{aligned}\]
Cofactor \(A_{21}\):
\[\begin{aligned}A_{21}&=(-1)\begin{vmatrix}-3&5\\1&-2\end{vmatrix}\\&=-\left[(-3)(-2)-(5)(1)\right]\\&=-(6-5)\\&=-1\end{aligned}\]
Cofactor \(A_{22}\):
\[\begin{aligned}A_{22}&=(+1)\begin{vmatrix}2&5\\1&-2\end{vmatrix}\\&=(2)(-2)-(5)(1)\\&=-4-5\\&=-9\end{aligned}\]
Cofactor \(A_{23}\):
\[\begin{aligned}A_{23}&=(-1)\begin{vmatrix}2&-3\\1&1\end{vmatrix}\\&=-\left[(2)(1)-(-3)(1)\right]\\&=-(2+3)\\&=-5\end{aligned}\]
Cofactor \(A_{31}\):
\[\begin{aligned}A_{31}&=(+1)\begin{vmatrix}-3&5\\2&-4\end{vmatrix}\\&=(-3)(-4)-(5)(2)\\&=12-10\\&=2\end{aligned}\]
Cofactor \(A_{32}\):
\[\begin{aligned}A_{32}&=(-1)\begin{vmatrix}2&5\\3&-4\end{vmatrix}\\&=-\left[(2)(-4)-(5)(3)\right]\\&=-(-8-15)\\&=23\end{aligned}\]
Cofactor \(A_{33}\):
\[\begin{aligned}A_{33}&=(+1)\begin{vmatrix}2&-3\\3&2\end{vmatrix}\\&=(2)(2)-(-3)(3)\\&=4+9\\&=13\end{aligned}\]
Hence, the cofactor matrix is
\[C=\begin{bmatrix}0&2&1\\-1&-9&-5\\2&23&13\end{bmatrix}\]
Find the Determinant of \(A\)
Expanding \(|A|\) along the first row,
\[\begin{aligned}|A|&=2\begin{vmatrix}2&-4\\1&-2\end{vmatrix} -(-3)\begin{vmatrix}3&-4\\1&-2\end{vmatrix} +5\begin{vmatrix}3&2\\1&1\end{vmatrix}\end{aligned}\]
Using the minors already calculated,
\[\begin{aligned}|A|&=2(0)+3(-2)+5(1)\\&=0-6+5\\&=-1\end{aligned}\]
Therefore
\[\boxed{|A|=-1}\]
Since
\[ |A|=-1\neq0, \]
the matrix \(A\) is non-singular and \(A^{-1}\) exists.
Find the Adjoint of \(A\)
The adjoint is the transpose of the cofactor matrix:
\[ \operatorname{adj}(A)=C^T. \]
Therefore,
\[\begin{aligned}\operatorname{adj}(A)&=\begin{bmatrix}0&2&1\\2&-9&23\\1&-5&13\end{bmatrix}\end{aligned}\]
Thus,
\[\boxed{\operatorname{adj}(A)=\begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix}}.\]
Find \(A^{-1}\)
We use
\[ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A). \]
Since
\[|A|=-1,\]
we get
\[\begin{aligned}A^{-1}&=-\begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix}\\ &=\begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}\end{aligned}\]
Hence,
\[\boxed{A^{-1}=\begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}}\]
Solving the System Using \(A^{-1}\)
Given — system is
\[\begin{aligned}2x-3y+5z&=11\\3x+2y-4z&=-5\\x+y-2z&=-3\end{aligned}\]
Comparing coefficients of \(x,y,z\), we obtain
\[A=\begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix},\]

\[X=\begin{bmatrix}x\\y\\z\end{bmatrix},\]
and
\[B=\begin{bmatrix}11\\-5\\-3\end{bmatrix}\]
Therefore,
\[\boxed{\begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}11\\-5\\-3\end{bmatrix}}\]
Hence
\[AX=B\]
Use \(X=A^{-1}B\)
Since \(A^{-1}\) exists,
\[ A^{-1}AX=A^{-1}B \]
Using
\[ A^{-1}A=I, \]
we obtain
\[ X=A^{-1}B. \]
Substituting the values of \(A^{-1}\) and \(B\),
\[\begin{aligned}\begin{bmatrix}x\\y\\z\end{bmatrix} &=\begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix} \begin{bmatrix}11\\-5\\-3\end{bmatrix}\end{aligned}\]
Calculate the first component:
\[\begin{aligned}x&=0(11)+1(-5)+(-2)(-3)\\&=0-5+6\\&=1\end{aligned}\]
Calculate the second component:
\[\begin{aligned}y&=(-2)(11)+9(-5)+(-23)(-3)\\&=-22-45+69\\&=2\end{aligned}\]
Calculate the third component:
\[\begin{aligned}z&=(-1)(11)+5(-5)+(-13)(-3)\\&=-11-25+39\\&=3\end{aligned}\]
Therefore,
\[\begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}1\\2\\3\end{bmatrix}\]
Hence,
\[\boxed{x=1,\quad y=2,\quad z=3}\]

🎯 Exam Significance

Exam Significance

›

This is an important application of the inverse-matrix method prescribed in the Class 12 Mathematics syllabus. The question tests several connected concepts: cofactors, determinant, adjoint, inverse of a matrix, matrix representation of a system of linear equations, and matrix multiplication.

The determinant check

\[ |A|=-1\neq0 \]

is particularly important because it establishes that \(A^{-1}\) exists and that the system has a unique solution.

In a board examination, students should not jump directly from the matrix \(A\) to its inverse. The cofactor matrix and its transpose should be shown clearly when the question requires the inverse by the adjoint method.

A common source of error is confusing the cofactor matrix with the adjoint matrix. The adjoint is obtained by transposing the complete cofactor matrix:

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

Significance for Competitive Entrance Examinations

This problem is useful for developing speed and accuracy in determinant and matrix calculations. Once

\[ |A|\neq0 \]

is established, the existence of \(A^{-1}\) immediately implies a unique solution for the corresponding system.

For objective and time-bound examinations, careful use of the cofactor sign pattern

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

can significantly reduce sign errors. The relationship

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

is also a fundamental identity that can be used in more advanced problems involving matrix equations, determinants, and systems of linear equations.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

For a square matrix \(A\), the inverse exists when
\[ |A|\neq0. \]
For the given matrix,
\[ |A|=-1\neq0. \]
The cofactor matrix is
\[ C= \begin{bmatrix} 0&2&1\\ -1&-9&-5\\ 2&23&13 \end{bmatrix}. \]
The adjoint is the transpose of the cofactor matrix:
\[ \operatorname{adj}(A)= \begin{bmatrix} 0&-1&2\\ 2&-9&23\\ 1&-5&13 \end{bmatrix}. \]
The inverse is
\[ A^{-1}= \begin{bmatrix} 0&1&-2\\ -2&9&-23\\ -1&5&-13 \end{bmatrix}. \]
A system of linear equations can be expressed as
\[ AX=B. \]
When \(A^{-1}\) exists, the solution is obtained from
\[ X=A^{-1}B. \]
For this system,
\[ \boxed{x=1,\quad y=2,\quad z=3}. \]
Substitution into the original equations provides a direct verification of the result.

📘 Concept & Theory

Concept/Theory

›

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

\[ AX=B. \]

If

\[ |A|\neq0, \]

then \(A\) is non-singular and its inverse exists. The solution is then obtained using

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

In this problem, the three unknowns represent the cost per kilogram of onion, wheat and rice. We therefore form a \(3\times3\) coefficient matrix from the quantities of the three items and a column matrix from their total costs.

🗺️ Solution Roadmap

Step-by-step Plan

›

Let the costs per kg of onion, wheat and rice be \(x,y,z\), respectively.
Form the three linear equations from the given information.
Write the equations in the matrix form \(AX=B\).
Calculate \(|A|\) to determine whether \(A^{-1}\) exists.
Find the cofactors of \(A\).
Form \(\operatorname{adj}(A)\) by transposing the cofactor matrix.
Calculate \(A^{-1}\).
Use \(X=A^{-1}B\) to obtain the three costs.
Verify the answer in the original equations.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 42 steps

Define the Unknown Costs
Let the cost of
\[ 1\text{ kg onion ₹}=x, \]

\[ 1\text{ kg wheat ₹}=y, \]
and
\[ 1\text{ kg rice ₹}=z. \]
Form the Linear Equations
According to the first condition, the cost of \(4\) kg onion, \(3\) kg wheat and \(2\) kg rice is ₹\(60\). Therefore,
\[ 4x+3y+2z=60. \]
According to the second condition, the cost of \(2\) kg onion, \(4\) kg wheat and \(6\) kg rice is ₹\(90\). Therefore,
\[ 2x+4y+6z=90. \]
According to the third condition, the cost of \(6\) kg onion, \(2\) kg wheat and \(3\) kg rice is ₹\(70\). Therefore,
\[ 6x+2y+3z=70. \]
Thus, the required system is
\[ \boxed{\bbox[5pt]{ \begin{aligned} 4x+3y+2z&=60\\ 2x+4y+6z&=90\\ 6x+2y+3z&=70 \end{aligned}}} \]
Write the System in Matrix Form
Comparing the coefficients of \(x,y,z\), we have
\[A=\begin{bmatrix}4&3&2\\2&4&6\\6&2&3\end{bmatrix}\]

\[X=\begin{bmatrix}x\\y\\z\end{bmatrix},\]
and
\[B=\begin{bmatrix}60\\90\\70\end{bmatrix}\]
Hence,
\[\boxed{\bbox[5pt]{\begin{bmatrix}4&3&2\\2&4&6\\6&2&3\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}60\\90\\70\end{bmatrix}}}\]
Therefore,
\[AX=B\]
Find the Determinant of \(A\)
We have
\[|A|=\begin{vmatrix}4&3&2\\2&4&6\\6&2&3\end{vmatrix}\]
Expanding along the first row,
\[\begin{aligned}|A|&=4\begin{vmatrix}4&6\\2&3\end{vmatrix} -3\begin{vmatrix}2&6\\6&3\end{vmatrix} +2\begin{vmatrix}2&4\\6&2\end{vmatrix}\end{aligned}\]
Evaluate the first minor:
\[\begin{aligned}\begin{vmatrix}4&6\\2&3\end{vmatrix}&=(4)(3)-(6)(2)\\&=12-12\\&=0\end{aligned}\]
Evaluate the second minor:
\[\begin{aligned}\begin{vmatrix}2&6\\6&3\end{vmatrix}&=(2)(3)-(6)(6)\\&=6-36\\&=-30\end{aligned}\]
Evaluate the third minor:
\[\begin{aligned}\begin{vmatrix}2&4\\6&2\end{vmatrix}&=(2)(2)-(4)(6)\\&=4-24\\&=-20\end{aligned}\]
Therefore,
\[\begin{aligned}|A|&=4(0)-3(-30)+2(-20)\\&=0+90-40\\&=50\end{aligned}\]
Hence,
\[\boxed{|A|=50}\]
Since
\[ |A|=50\neq0, \]
the matrix \(A\) is non-singular. Therefore, \(A^{-1}\) exists and the system has a unique solution.
Find the Cofactors of \(A\)
The cofactor sign pattern is
\[ \begin{bmatrix} +&-&+\\ -&+&-\\ +&-&+ \end{bmatrix} \]
The cofactor \(A_{11}\) is
\[\begin{aligned}A_{11}&=\begin{vmatrix}4&6\\2&3\end{vmatrix}\\&=12-12\\&=0\end{aligned}\]
The cofactor \(A_{12}\) is
\[\begin{aligned}A_{12}&=-\begin{vmatrix}2&6\\6&3\end{vmatrix}\\&=-(6-36)\\&=30\end{aligned}\]
The cofactor \(A_{13}\) is
\[\begin{aligned}A_{13}&=\begin{vmatrix}2&4\\6&2\end{vmatrix}\\&=4-24\\&=-20\end{aligned}\]
The cofactor \(A_{21}\) is
\[\begin{aligned}A_{21}&=-\begin{vmatrix}3&2\\2&3\end{vmatrix}\\&=-\left[(3)(3)-(2)(2)\right]\\&=-(9-4)\\&=-5\end{aligned}\]
The cofactor \(A_{22}\) is
\[\begin{aligned}A_{22}&=\begin{vmatrix}4&2\\6&3\end{vmatrix}\\&=(4)(3)-(2)(6)\\&=12-12\\&=0\end{aligned}\]
The cofactor \(A_{23}\) is
\[\begin{aligned}A_{23}&=-\begin{vmatrix}4&3\\6&2\end{vmatrix}\\&=-\left[(4)(2)-(3)(6)\right]\\&=-(8-18)\\&=10\end{aligned}\]
The cofactor \(A_{31}\) is
\[\begin{aligned}A_{31}&=\begin{vmatrix}3&2\\4&6\end{vmatrix}\\&=(3)(6)-(2)(4)\\&=18-8\\&=10\end{aligned}\]
The cofactor \(A_{32}\) is
\[\begin{aligned}A_{32}&=-\begin{vmatrix}4&2\\2&6\end{vmatrix}\\&=-\left[(4)(6)-(2)(2)\right]\\&=-(24-4)\\&=-20\end{aligned}\]
The cofactor \(A_{33}\) is
\[\begin{aligned}A_{33}&=\begin{vmatrix}4&3\\2&4\end{vmatrix}\\&=(4)(4)-(3)(2)\\&=16-6\\&=10\end{aligned}\]
Therefore, the cofactor matrix is
\[C=\begin{bmatrix}0&30&-20\\-5&0&10\\10&-20&10\end{bmatrix}\]
Find the Adjoint of \(A\)
The adjoint is the transpose of the cofactor matrix:
\[ \operatorname{adj}(A)=C^T. \]
Therefore,
\[\begin{aligned}\operatorname{adj}(A)&=\begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}\end{aligned}\]
Hence,
\[\boxed{\bbox[5pt]{\operatorname{adj}(A)=\begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}}}\]
Find \(A^{-1}\)
Use
\[A^{-1}=\frac{1}{|A|}\operatorname{adj}(A)\]
Since \(|A|=50,\) we obtain
\[\begin{aligned}A^{-1}&=\frac{1}{50}\begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}\end{aligned}\]
Taking the common factor \(5\) outside,
\[\begin{aligned}A^{-1}&=\frac{1}{10}\begin{bmatrix}0&-1&2\\6&0&-4\\-4&2&2\end{bmatrix}\end{aligned}\]
Thus,
\[\boxed{\bbox[5pt]{A^{-1}=\frac{1}{10}\begin{bmatrix}0&-1&2\\6&0&-4\\-4&2&2\end{bmatrix}}}\]
Calculate \(X=A^{-1}B\)
From
\[ AX=B, \]
multiplying both sides by \(A^{-1}\), we get
\[ A^{-1}AX=A^{-1}B. \]
Since \(A^{-1}A=I,\) we obtain
\[ X=A^{-1}B. \]
Therefore,
\[\begin{aligned}\begin{bmatrix}x\\y\\z\end{bmatrix}&=\frac{1}{10}\begin{bmatrix}0&-1&2\\6&0&-4\\-4&2&2\end{bmatrix}\begin{bmatrix}60\\90\\70\end{bmatrix}\end{aligned}\]
Calculate the first component:
\[\begin{aligned}x&=\frac{1}{10}\left[0(60)-1(90)+2(70)\right]\\&=\frac{1}{10}\left[0-90+140\right]\\&=\frac{50}{10}\\&=5\end{aligned}\]
Calculate the second component:
\[\begin{aligned}y&=\frac{1}{10}\left[6(60)+0(90)-4(70)\right]\\&=\frac{1}{10}\left[360-280\right]\\&=\frac{80}{10}\\&=8\end{aligned}\]
Calculate the third component:
\[\begin{aligned}z&=\frac{1}{10}\left[-4(60)+2(90)+2(70)\right]\\&=\frac{1}{10}\left[-240+180+140\right]\\&=\frac{80}{10}\\&=8\end{aligned}\]
Hence,
\[\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\8\\8\end{bmatrix}\]
Therefore,
\[\boxed{x=5,\qquad y=8,\qquad z=8}\]
Interpret the Answer
Since \(x,y,z\) represent the costs per kilogram of onion, wheat and rice respectively,
\[ \boxed{\text{Cost of onion = ₹}5\text{ per kg}} \]

\[ \boxed{\text{Cost of wheat = ₹}8\text{ per kg}} \]

\[ \boxed{\text{Cost of rice = ₹}8\text{ per kg}} \]

🎯 Exam Significance

Exam Significance

›

This is an important application-based problem from the inverse-matrix method. It converts a real-life cost problem into a system of three simultaneous linear equations and then solves the system using

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

For board examinations, the important scoring steps are the correct identification of \(A\), \(X\), and \(B\), calculation of \(|A|\), formation of the cofactor matrix, obtaining the adjoint by transposition, and finally calculating \(A^{-1}B\).

The determinant

\[ |A|=50\neq0 \]

is especially significant because it confirms that \(A^{-1}\) exists and that the system has a unique solution.

Significance for Competitive Entrance Examinations

This problem is useful for strengthening the connection between determinants, inverse matrices and simultaneous equations. In competitive examinations, recognizing the matrix structure quickly can reduce a lengthy word problem to the compact relation

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

The calculation also reinforces the importance of the cofactor sign pattern and the distinction between the cofactor matrix and its transpose. These details are frequent sources of errors in matrix-inverse questions.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

Assign variables to the unit costs before constructing the equations.
The given information produces the system
\[ 4x+3y+2z=60,\quad 2x+4y+6z=90,\quad 6x+2y+3z=70. \]
The matrix equation is
\[ AX=B. \]
The determinant of the coefficient matrix is
\[ |A|=50\neq0. \]
Therefore, \(A^{-1}\) exists and the system has a unique solution.
The inverse is
\[ A^{-1} = \frac{1}{10} \begin{bmatrix} 0&-1&2\\ 6&0&-4\\ -4&2&2 \end{bmatrix} \]
The solution is obtained from
\[ X=A^{-1}B. \]
The costs are
\[ \boxed{\text{Onion = ₹}5/\text{kg},\quad \text{Wheat = ₹}8/\text{kg},\quad \text{Rice = ₹}8/\text{kg}}. \]
Substitution into all three original equations confirms the answer.

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

Chapter Complete!

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Determinants — Learning Resources

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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

Chapter Complete!

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Determinants — Learning Resources

Frequently Asked Questions

Q 01 What is the answer to NCERT Class 12 Determinants Exercise 4.5 Question 16?

Q 02 How is Question 16 of Exercise 4.5 solved?

Q 03 What are the equations in Exercise 4.5 Question 16?

Q 04 What matrix is used as the coefficient matrix in Question 16?

Q 05 What is the determinant of the coefficient matrix?

Q 06 Why is the determinant important in this problem?

Q 07 What is the inverse of matrix A in Question 16?

Q 08 What formula is used to solve the system by matrix method?

Q 09 What are the costs of onion, wheat and rice per kg?

Q 10 Is Exercise 4.5 Question 16 important for CBSE Class 12 exams?

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