If $A = \begin{bmatrix} 1 & 3 & 4 \\ 2 & 1 & 2 \\ 5 & 1 & 1 \end{bmatrix}$, find $A^{-1}$. Hence solve the system of equations $x + 3y + 4z = 8$, $2x + y + 2z = 5$ and $5x + y + z = 7$. OR Find the inverse of the following matrix, using elementary transformations: $A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}$.

If A = 1 3 4 \2 1 2 \5 1 1 , find A⁻¹. Hence solve the system of equations x + 3y + 4z = 8, 2x + y + 2z = 5 an

Question

A = \begin{bmatrix} 1 & 3 & 4 \\ 2 & 1 & 2 \\ 5 & 1 & 1 \end{bmatrix}

(OR)

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

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Part (a)Concept understanding — Inverse Matrix Method

The Inverse Matrix Method

Many problems reduce to a system of linear equations, for example

\begin{aligned}2x+3y&=8\\x-\ \ y&=-1.\end{aligned}

The inverse matrix method solves such a system by writing it as a single matrix equation and then undoing the coefficient matrix with its inverse — the matrix analogue of dividing.

Writing the system as $AX=B$

Collect the coefficients, the unknowns and the constants:

A=\begin{pmatrix}2&3\\1&-1\end{pmatrix},\quad X=\begin{pmatrix}x\\y\end{pmatrix},\quad B=\begin{pmatrix}8\\-1\end{pmatrix},

so the whole system becomes $AX=B$ .

The idea

For numbers, $ax=b$ gives $x=a^{-1}b$ provided $a\neq0$ . The same works for matrices: if $A$ is invertible, multiply $AX=B$ on the left by $A^{-1}$ :

A^{-1}(AX)=A^{-1}B\;\Rightarrow\;IX=A^{-1}B\;\Rightarrow\;X=A^{-1}B.

X=A^{-1}B

Multiplying on the left matters — matrix products do not commute, so $BA^{-1}$ would be wrong.

When it works

The inverse $A^{-1}$ exists only when $\det A\neq0$ , so:

$\det A\neq0$ : the system is consistent with the unique solution $X=A^{-1}B$ .
$\det A=0$ : no inverse; the system is either inconsistent (no solution) or has infinitely many — handle it by another method.

Worked steps

For the system above, $\det A=(2)(-1)-(3)(1)=-5\neq0$ , and

A^{-1}=\frac{1}{-5}\begin{pmatrix}-1&-3\\-1&2\end{pmatrix}.

Then …

Part (b)Concept understanding — Inverse Matrix Method

The Inverse Matrix Method

Many problems reduce to a system of linear equations, for example

\begin{aligned}2x+3y&=8\\x-\ \ y&=-1.\end{aligned}

The inverse matrix method solves such a system by writing it as a single matrix equation and then undoing the coefficient matrix with its inverse — the matrix analogue of dividing.

Writing the system as $AX=B$

Collect the coefficients, the unknowns and the constants:

A=\begin{pmatrix}2&3\\1&-1\end{pmatrix},\quad X=\begin{pmatrix}x\\y\end{pmatrix},\quad B=\begin{pmatrix}8\\-1\end{pmatrix},

so the whole system becomes $AX=B$ .

The idea

For numbers, $ax=b$ gives $x=a^{-1}b$ provided $a\neq0$ . The same works for matrices: if $A$ is invertible, multiply $AX=B$ on the left by $A^{-1}$ :

A^{-1}(AX)=A^{-1}B\;\Rightarrow\;IX=A^{-1}B\;\Rightarrow\;X=A^{-1}B.

X=A^{-1}B

Multiplying on the left matters — matrix products do not commute, so $BA^{-1}$ would be wrong.

When it works

The inverse $A^{-1}$ exists only when $\det A\neq0$ , so:

$\det A\neq0$ : the system is consistent with the unique solution $X=A^{-1}B$ .
$\det A=0$ : no inverse; the system is either inconsistent (no solution) or has infinitely many — handle it by another method.

Worked steps

For the system above, $\det A=(2)(-1)-(3)(1)=-5\neq0$ , and

A^{-1}=\frac{1}{-5}\begin{pmatrix}-1&-3\\-1&2\end{pmatrix}.

Then …

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Part (a): $A^{-1}=\dfrac{1}{11}\begin{bmatrix}-1&1&2\\8&-19&6\\-3&14&-5\end{bmatrix}$ and the system has solution $x=y=z=1$ . Part (b): by elementary transformations $A^{-1}=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}$ .

Part (a)

$A=\begin{bmatrix}1&3&4\\2&1&2\\5&1&1\end{bmatrix}$ .

Determinant. Expanding along row 1:

\det A=1\begin{vmatrix}1&2\\1&1\end{vmatrix}-3\begin{vmatrix}2&2\\5&1\end{vmatrix}+4\begin{vmatrix}2&1\\5&1\end{vmatrix}=1(-1)-3(-8)+4(-3)=11\neq0.

Cofactors.

C_{11}=-1,\ C_{12}=8,\ C_{13}=-3,\quad C_{21}=1,\ C_{22}=-19,\ C_{23}=14,\quad C_{31}=2,\ C_{32}=6,\ C_{33}=-5.

Transposing the cofactor matrix gives

\operatorname{adj}A=\begin{bmatrix}-1&1&2\\8&-19&6\\-3&14&-5\end{bmatrix},\qquad A^{-1}=\frac{1}{\det A}\operatorname{adj}A=\frac{1}{11}\begin{bmatrix}-1&1&2\\8&-19&6\\-3&14&-5\end{bmatrix}.

Solve the system. With $B=\begin{bmatrix}8\\5\\7\end{bmatrix}$ , $X=A^{-1}B$ :

$x=\tfrac1{11}(-8+5+14)=1,\quad y=\tfrac1{11}(64-95+42)=1,\quad z=\tfrac1{11}(-24+70-35)=1.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$A$
(a) $\det(A)$
(b) $\frac{1}{\det(A)}$
(c) $1$
(d) $0$
›Reveal solutionSolution

$\det(A^{-1})=\dfrac{1}{\det A}$ .

From $AA^{-1}=I$ , $\det(A)\det(A^{-1})=\det(I)=1$ , so $\det(A^{-1})=\dfrac{1}{\det(A)}$ . (This holds for a …

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CBSE 2026Set ANNUAL1 mark
$AX = B$
›Reveal solutionSolution

The unique solution of $AX=B$ (when $|A| e0$ ) is $X=A^{-1}B$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$
(a) $19$
(b) $\dfrac{1}{19}$
(c) $-19$
(d) $-\dfrac{1}{19}$
›Reveal solutionSolution

Compute |A| and adj(A), form A⁻¹, and compare it entry-by-entry with kA.

$A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$ , $|A| = 2(-2) - 3(5) = -4-15 = -19$

$\text{adj}(A) = \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}$

…

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CBSE 2024Set D1 markMCQ
$A = \begin{bmatrix} 2 & -3 \\ 4 & 6 \end{bmatrix}$
(a) $\begin{bmatrix} \frac{1}{4} & \frac{1}{8} \\ \frac{1}{6} & \frac{1}{12} \end{bmatrix}$
(b) $\begin{bmatrix} \frac{1}{4} & \frac{1}{8} \\ -\frac{1}{6} & \frac{1}{12} \end{bmatrix}$
(c) $\begin{bmatrix} 4 & 8 \\ 6 & 12 \end{bmatrix}$
(d) $\begin{bmatrix} 4 & 8 \\ -6 & 12 \end{bmatrix}$
›Reveal solutionSolution

For a $2\times2$ matrix $A^{-1} = \frac{1}{\det A}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ .

Here $A = \begin{bmatrix} 2 & -3 \\ 4 & 6 \end{bmatrix}$ , so $\det A = (2)(6) - (-3)(4) = 12 + 12 = 24.$

The adjoint (swap diagonal, negate off-diagonal) is $\begin{bmatrix} 6 & 3 \\ -4 & 2 \end{bmatrix}$ . Hence …

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Q.If A=[134212511]A = \begin{bmatrix} 1 & 3 & 4 \\ 2 & 1 & 2 \\ 5 & 1 & 1 \end{bmatrix}A=​125​311​421​​, find A−1A^{-1}A−1. Hence solve the system of equations x+3y+4z=8x + 3y + 4z = 8x+3y+4z=8, 2x+y+2z=52x + y + 2z = 52x+y+2z=5 and 5x+y+z=75x + y + z = 75x+y+z=7.

Part (a)

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Q.If $A = \begin{bmatrix} 1 & 3 & 4 \\ 2 & 1 & 2 \\ 5 & 1 & 1 \end{bmatrix}$ , find $A^{-1}$ . Hence solve the system of equations $x + 3y + 4z = 8$ , $2x + y + 2z = 5$ and $5x + y + z = 7$ .

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