Given $A = \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix}$, $B = \begin{pmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{pmatrix}$, find $BA$ and use this to solve the system of equations $y + 2z = 7$, $x - y = 3$, $2x + 3y + 4z = 17$.

Given A = 2 2 -4 \-4 2 -4 \2 -1 5 , B = 1 -1 0 \2 3 4 \0 1 2 , find BA and use this to solve the system of equ

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A = \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix}

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Concept understanding — Matrix Equation Solving

Solving a System of Equations by the Matrix Method

A system of linear equations can be written as a single matrix equation and solved in one clean step using the inverse of a matrix. This is the Class-12 "matrix method" for simultaneous equations.

Writing the system as $AX=B$

Take the system

a_1x+b_1y+c_1z=d_1,\quad a_2x+b_2y+c_2z=d_2,\quad a_3x+b_3y+c_3z=d_3.

Collect the coefficients, the unknowns, and the constants into matrices:

A=\begin{pmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{pmatrix},\quad X=\begin{pmatrix}x\\y\\z\end{pmatrix},\quad B=\begin{pmatrix}d_1\\d_2\\d_3\end{pmatrix}.

Then the whole system is just

AX=B.

Solving when $A$ is invertible

If $\det(A)\neq0$ , then $A^{-1}$ exists, and multiplying both sides on the left by $A^{-1}$ gives

X=A^{-1}B,\qquad\text{where }\;A^{-1}=\frac{1}{\det(A)}\operatorname{adj}(A).

So you compute $\det(A)$ , then $\operatorname{adj}(A)$ , form $A^{-1}$ , and multiply by $B$ . The single column $X=A^{-1}B$ hands you $x$ , $y$ , $z$ at once, and because $A^{-1}$ is unique, the solution is unique.

Watch out

Multiply in the correct order: $X=A^{-1}B$ , not $BA^{-1}$ . Matrix multiplication is not commutative, and $BA^{-1}$ is not even defined here.

When $\det(A)=0$

If $\det(A)=0$ , $A^{-1}$ does not exist and the inverse method fails. The system is then either inconsistent (no solution) or has infinitely many solutions. Decide which by computing $(\operatorname{adj}A)\,B$ :

$(\operatorname{adj}A)\,B\neq O$ → no solution (inconsistent).
$(\operatorname{adj}A)\,B=O$ → infinitely many solutions (consistent, dependent). …

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$BA=6I$ , so $B^{-1}=\tfrac16A$ ; the system is $BX=C$ with $C=(3,17,7)^T$ , giving $X=\tfrac16AC=(2,-1,4)^T$ , i.e. $x=2,\,y=-1,\,z=4$ .

The idea

If the product of two matrices is a scalar times the identity, each is (a scalar multiple of) the other's inverse. Computing $BA$ first hands us $B^{-1}$ for free — and it is $B$ , not $A$ , that is the coefficient matrix of the given system.

Step 1 — Compute $BA$

BA=\begin{pmatrix}1&-1&0\\2&3&4\\0&1&2\end{pmatrix}\begin{pmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{pmatrix}.

Entry by entry (row of $B$ times column of $A$ ):

Row 1: $2+4+0=6$ , $\;2-2+0=0$ , $\;-4+4+0=0$ .
Row 2: $4-12+8=0$ , $\;4+6-4=6$ , $\;-8-12+20=0$ .
Row 3: $0-4+4=0$ , $\;0+2-2=0$ , $\;0-4+10=6$ .

BA=\begin{pmatrix}6&0&0\\0&6&0\\0&0&6\end{pmatrix}=6I\quad\Longrightarrow\quad B^{-1}=\frac16A.

Step 2 — Match the system to $B$

The equations $y+2z=7,\;x-y=3,\;2x+3y+4z=17$ , reordered as

x-y=3,\qquad 2x+3y+4z=17,\qquad y+2z=7,

have coefficient rows equal to the rows of $B$ :

$B\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}3\\17\\7\end{pmatrix}=C.$ …

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

Showing the 12 most recent of 37 on this concept.

CBSE 20241 markMCQ
$\begin{bmatrix} 8 & 14 \\ 9 & 7 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix} X$
›Reveal solutionSolution
We solve the matrix equation $A = BX$ by left-multiplying both sides by $B^{-1}$ , giving $X = B^{-1}A$ . Computing the inverse of $B = \begin{bmatrix}1&2\\3&1\end{bmatrix}$ and multiplying yields $X = \begin{bmatrix}2&0\\3&7\end{bmatrix}$ , which matches option (C).

The core idea here is that a matrix equation like $A = BX$ is solved exactly like the scalar equation $a = bx$ — you isolate $X$ by multiplying both sides by the inverse of $B$ . But because matrix multiplication is not commutative, you must multiply on the left by $B^{-1}$ , not on the right. That single detail is the entire key.

Let’s walk through it.

Set up the equation clearly. We are given

$\begin{bmatrix} 8 & 14 \\ 9 & 7 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix} X.$
Call the left matrix $A$ and the coefficient matrix $B$ , so $A = BX$ . Our job is to find $X$ .

Why left-multiplication by $B^{-1}$ works. If $B$ is invertible, then $B^{-1}B = I$ , the identity matrix. Multiplying both sides of $A = BX$ on the left by $B^{-1}$ gives

$B^{-1}A = B^{-1}(BX) = (B^{-1}B)X = IX = X.$
So $X = B^{-1}A$ . Notice: if we had multiplied on the right instead, we’d get $AB^{-1}$ , which is a completely different (and wrong) matrix.

Watch out

A common mistake is to write $X = A B^{-1}$ by analogy with scalars. But matrix multiplication is not commutative — $B^{-1}A \neq A B^{-1}$ in general. Always multiply on the side where the inverse cancels the original matrix.

Find $B^{-1}$ . For a $2 \times 2$ matrix $B = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , the inverse is

$B^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix},$
provided the determinant $ad - bc \neq 0$ .

Here $a=1$ , $b=2$ , $c=3$ , $d=1$ . The determinant is
$\det(B) = (1)(1) - (2)(3) = 1 - 6 = -5.$
So
$B^{-1} = \frac{1}{-5} \begin{bmatrix} 1 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -\frac{1}{5} & \frac{2}{5} \\ \frac{3}{5} & -\frac{1}{5} \end{bmatrix}.$

Multiply $B^{-1}A$ . Now $A = \begin{bmatrix} 8 & 14 \\ 9 & 7 \end{bmatrix}$ . Compute $X = B^{-1}A$ :

$X = \begin{bmatrix} -\frac{1}{5} & \frac{2}{5} \\ \frac{3}{5} & -\frac{1}{5} \end{bmatrix} \begin{bmatrix} 8 & 14 \\ 9 & 7 \end{bmatrix}.$
Multiply entry by entry: …
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CBSE 2026Set ANNUAL1 markMCQ
$\begin{bmatrix} x+y \\ y+z \\ z+x \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$
(a) 9
(b) 0
(c) 4
(d) 5
›Reveal solutionSolution

Two matrices are equal only if all corresponding entries are equal; adding all three entry-equations gives $x+y+z$ directly.

From $\begin{bmatrix} x+y \\ y+z \\ z+x \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$ , equating corresponding entries:

…

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CBSE 2026Set ANNUAL1 markMCQ
$A = \begin{bmatrix} x & 0 \\ 2 & 3 \end{bmatrix}$
(a) $x = 4$
(b) $x = \pm3$
(c) $x = -3$
(d) $x = -4$
›Reveal solutionSolution

Computing $A^2$ and matching it to $9I$ forces both $x^2=9$ and $2x+6=0$ ; only $x=-3$ satisfies both.

$A=\begin{bmatrix}x & 0\\ 2 & 3\end{bmatrix}$ , so

$A^2=\begin{bmatrix}x & 0\\ 2 & 3\end{bmatrix}\begin{bmatrix}x & 0\\ 2 & 3\end{bmatrix}=\begin{bmatrix}x^2 & 0\\ 2x+6 & 9\end{bmatrix}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If [[x-2y, 0], [5, x]] = [[-3, 0], [5, 3]], then y is equal to:$
(a) 1
(b) 3
(c) 2
(d) 4
›Reveal solutionSolution

Two matrices are equal only if all corresponding entries are equal; comparing the $(2,2)$ entries gives $x=3$ , then the $(1,1)$ entries give $y$ .

Given:

$\begin{bmatrix}x-2y & 0\\ 5 & x\end{bmatrix} = \begin{bmatrix}-3 & 0\\ 5 & 3\end{bmatrix}$

Comparing the $(2,2)$ entries: $x = 3$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$x$
(i) $-2$
(ii) $-1$
(iii) $2$
(iv) $1$
›Reveal solutionSolution

Multiply out the matrices and compare entries.

$\begin{bmatrix}1&3\\2&1\end{bmatrix}\begin{bmatrix}1\\x\end{bmatrix}=\begin{bmatrix}1(1)+3(x)\\2(1)+1(x)\end{bmatrix}=\begin{bmatrix}1+3x\\2+x\end{bmatrix}$

Setting this equal to $\begin{bmatrix}7\\4\end{bmatrix}$ :

…

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CBSE 2025Set ANNUAL1 markMCQ
$x$
(a) $x = -\dfrac{1}{3}, y = 7$
(b) Not possible to find
(c) $x = -\dfrac{2}{3}, y = 7$
(d) $x = -\dfrac{1}{3}, y = -\dfrac{2}{3}$
›Reveal solutionSolution

Equating corresponding entries gives two different equations for x that contradict each other, so no consistent solution exists.

For $\begin{bmatrix} 3x+7 & 5 \\ y+1 & 2-3x \end{bmatrix} = \begin{bmatrix} 0 & y-2 \\ 8 & 4 \end{bmatrix}$ , equating each entry:

$3x+7 = 0 \Rightarrow x = -\dfrac{7}{3}$

$5 = y-2 \Rightarrow y = 7$

$y+1 = 8 \Rightarrow y = 7$ (consistent with above)

$2-3x = 4 \Rightarrow x = -\dfrac{2}{3}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If [[x-2y, 0], [5, x]] = [[-5, 0], [5, 3]], then y is equal to:$
(a) 1
(b) 3
(c) 2
(d) 4
›Reveal solutionSolution

Equal matrices have equal corresponding entries — match the (2,2) entries first to get $x$ , then use the (1,1) entry to get $y$ .

Given $\begin{pmatrix} x-2y & 0 \\ 5 & x\end{pmatrix} = \begin{pmatrix} -5 & 0 \\ 5 & 3\end{pmatrix}$ .

Comparing the (2,2) entries: $x = 3$ .

…

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CBSE 2025Set ANNUAL1 mark
$Q. If [[a+4, 3b], [8, -14]] = [[2a+2, b+4], [8, a-8b]], then find the value of a + b.$
›Reveal solutionSolution

Equate corresponding entries of the two equal matrices to get $a=2$ , $b=2$ , so $a+b=4$ .

Two matrices are equal only if every corresponding entry is equal. Comparing entries of

$\begin{bmatrix} a+4 & 3b \\ 8 & -14 \end{bmatrix} = \begin{bmatrix} 2a+2 & b+4 \\ 8 & a-8b \end{bmatrix}$ :

From the (1,1) entries: $a+4 = 2a+2 \Rightarrow 2 = a \Rightarrow a = 2$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If A = [[2x, 0], [x, x]] and A⁻¹ = [[1, 0], [-1, 2]], then x equals -$
(i) 1
(ii) 2
(iii) 1/2
(iv) −2
›Reveal solutionSolution

Compute $A^{-1}$ from $A=\begin{pmatrix}2x&0\\x&x\end{pmatrix}$ using the $2\times2$ inverse formula and match it to the given $A^{-1}$ .

For $A=\begin{pmatrix}2x&0\\x&x\end{pmatrix}$ , $\det A = (2x)(x)-(0)(x)=2x^2$ .

Using $A^{-1}=\dfrac{1}{\det A}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ for $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ :

$A^{-1}=\dfrac{1}{2x^2}\begin{pmatrix}x&0\\-x&2x\end{pmatrix}=\begin{pmatrix}\tfrac{1}{2x}&0\\-\tfrac{1}{2x}&\tfrac{1}{x}\end{pmatrix}$ .

…

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CBSE 2024Set D1 markMCQ
$2A + B + X = 0$
(a) $\begin{bmatrix} 1 & 2 \\ -7 & -13 \end{bmatrix}$
(b) $\begin{bmatrix} 1 & 2 \\ 7 & 13 \end{bmatrix}$
(c) $\begin{bmatrix} -1 & -2 \\ -7 & -13 \end{bmatrix}$
(d) $\begin{bmatrix} -1 & -2 \\ 7 & 13 \end{bmatrix}$
›Reveal solutionSolution

From $2A + B + X = 0$ , solve $X = -2A - B$ .

$2A = \begin{bmatrix} -2 & 4 \\ 6 & 8 \end{bmatrix}$ , so …

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CBSE 2024Set D1 markMCQ
$[\,x\ \ y\,] = [\,2x - 1\ \ 9\,] \Rightarrow$
(a) $x = 3,\ y = 9$
(b) $x = 1,\ y = 9$
(c) $x = 0,\ y = 9$
(d) $x = 3,\ y = 4$
›Reveal solutionSolution

Equal matrices have equal corresponding entries.

…

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CBSE 2024Set ANNUAL1 markMCQ
$\begin{bmatrix} x+y+z \\ x+z \\ y+z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}$
(a) $5$
(b) $7$
(c) $9$
(d) none of these
›Reveal solutionSolution

Matching the first row of the given matrix equation reads off x+y+z directly, no further algebra needed.

The matrix equation $\begin{bmatrix}x+y+z\\x+z\\y+z\end{bmatrix}=\begin{bmatrix}9\\5\\7\end{bmatrix}$ means corresponding entries are equal:

Row 1: $x+y+z=9$ …

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←Previous questionQ19 — Using matrix method, solve the system of equations 3x + 2y - 2z = 3, x…→Next questionQ21 — If a + b + c ≠ 0 and a b c b c a c a b = 0, then prove that a = b = c.

Q.Given $A = \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix}$ , $B = \begin{pmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{pmatrix}$ , find $BA$ and use this to solve the system of equations $y + 2z = 7$ , $x - y = 3$ , $2x + 3y + 4z = 17$ .

Concept understanding — Matrix Equation Solving

Solving a System of Equations by the Matrix Method

Writing the system as $AX=B$

Solving when $A$ is invertible

When $\det(A)=0$

The idea

Step 1 — Compute $BA$

Step 2 — Match the system to $B$

Method: Using a Given Matrix Product Equal to a Scalar Multiple of $I$ to Find an Inverse

Steps

Common Mistakes

The idea

Step 1 — Compute BABABA

Step 2 — Match the system to BBB

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Step 1 — Compute $BA$

Step 2 — Match the system to $B$

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