Find non-zero values of $x$ satisfying the matrix equation: $x\begin{bmatrix} 2x & 2 \\ 3 & x \end{bmatrix} + 2\begin{bmatrix} 8 & 5x \\ 4 & 4x \end{bmatrix} = 2\begin{bmatrix} x^2+8 & 24 \\ 10 & 6x \end{bmatrix}$.

Find non-zero values of x satisfying the matrix equation: x 2x 2 \3 x + 2 8 5x \4 4x = 2 x²+8 24 \10 6x .

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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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Multiply the scalars in, add, and equate entries. Every constraint gives $x=4$ (the (2,2) entry also allows $x=0$ , which is excluded). So the non-zero value is $x=4$ .

Turning the matrix equation into scalar equations

Two matrices are equal iff their corresponding entries are equal, so we simplify each side and compare positions.

Left side. Distribute the scalars:

x\begin{bmatrix} 2x & 2 \\ 3 & x \end{bmatrix}=\begin{bmatrix} 2x^2 & 2x \\ 3x & x^2 \end{bmatrix},\qquad 2\begin{bmatrix} 8 & 5x \\ 4 & 4x \end{bmatrix}=\begin{bmatrix} 16 & 10x \\ 8 & 8x \end{bmatrix}.

Adding:

\text{LHS}=\begin{bmatrix} 2x^2+16 & 2x+10x \\ 3x+8 & x^2+8x \end{bmatrix}=\begin{bmatrix} 2x^2+16 & 12x \\ 3x+8 & x^2+8x \end{bmatrix}.

Right side.

\text{RHS}=2\begin{bmatrix} x^2+8 & 24 \\ 10 & 6x \end{bmatrix}=\begin{bmatrix} 2x^2+16 & 48 \\ 20 & 12x \end{bmatrix}.

Equating entries

$(1,1)$ : $2x^2+16=2x^2+16$ — an identity, no restriction.
$(1,2)$ : $12x=48\ \Rightarrow\ x=4$ . …

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

KEAM 2026Set eng-2026-04194 marksMCQ
$(x \quad 3 \quad -1)\begin{pmatrix} 1 & 1 & 1 \\ -1 & 0 & 1 \\ 1 & 0 & -1 \end{pmatrix}\begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} = 0$
›Reveal solutionSolution

Do the matrix multiplication in stages (right pair first), reduce the row-vector $\times$ column-vector to a scalar, set it to $0$ , and solve for $x$ .

Concept. The product of a $1\times3$ row, a $3\times3$ matrix, and a $3\times1$ column is a $1\times1$ scalar. Multiplication is associative, so evaluate the $3\times3$ times the column first.

Step 1 — multiply the matrix by the column $(2,3,1)^{T}$ .

$\begin{pmatrix}1&1&1\\-1&0&1\\1&0&-1\end{pmatrix}\begin{pmatrix}2\\3\\1\end{pmatrix}=\begin{pmatrix}1\cdot2+1\cdot3+1\cdot1\\-1\cdot2+0\cdot3+1\cdot1\\1\cdot2+0\cdot3-1\cdot1\end{pmatrix}=\begin{pmatrix}6\\-1\\1\end{pmatrix}.$ …
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KEAM 2025Set eng-2025-04274 marksMCQ
$A = \begin{pmatrix} 0 & 2 \\ 3 & 4 \end{pmatrix}$
›Reveal solutionSolution

Compute $(I+A)\begin{pmatrix}4&-3\\2&-1\end{pmatrix}$ ; the bottom-right entry gives $x=-14$ .

First,
$I+A=\begin{pmatrix}1&0\\0&1\end{pmatrix}+\begin{pmatrix}0&2\\3&4\end{pmatrix}=\begin{pmatrix}1&2\\3&5\end{pmatrix}.$
Multiply: …
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KEAM 2025Set eng-2025-04294 marksMCQ
$A = \begin{bmatrix} X & 0 \\ 1 & 1 \end{bmatrix}$
›Reveal solutionSolution

Squaring $A$ gives entries $X^2$ and $X+1$ ; matching to $B$ gives $X^2=16$ and $X+1=5$ , both satisfied by $X=4$ .

Compute $A^2$ with $A=\begin{bmatrix}X&0\\1&1\end{bmatrix}$ :

$A^2=\begin{bmatrix}X&0\\1&1\end{bmatrix}\begin{bmatrix}X&0\\1&1\end{bmatrix}=\begin{bmatrix}X^2&0\\X+1&1\end{bmatrix}.$ …
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KEAM 2024Set eng-2024-06054 marksMCQ
$A=\begin{pmatrix}0 & 1\\ -1 & 2\end{pmatrix}$
›Reveal solutionSolution

From $XA=B$ , $X=BA^{-1}$ ; computing gives $\begin{pmatrix}3&-1\\-3&1\end{pmatrix}$ .

$\det A=0\cdot2-1\cdot(-1)=1$ , so $A^{-1}=\begin{pmatrix}2&-1\\1&0\end{pmatrix}$ . …
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KEAM 2026Set eng-2026-04224 marksMCQ
$A$
›Reveal solutionSolution

Multiply the series by $A$ , subtract, to collapse it to $A^{11}=I$ .

Given $I+A+A^2+\cdots+A^{10}=O$ .

Multiply by $A$ : $A+A^2+\cdots+A^{11}=O$ .

Subtract the original: $A^{11}-I=O\Rightarrow A^{11}=I$ . …
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KEAM 2024Set eng-2024-06064 marksMCQ
$A=\begin{pmatrix}3 & -2 & 1\\ -1 & 3 & -1\end{pmatrix}$
›Reveal solutionSolution

Multiply $A$ (2×3) by $B$ (3×1) and match to $\begin{pmatrix}-2\\6\end{pmatrix}$ .

First entry of $AB$ : $3(1)+(-2)(\alpha)+1(-1)=2-2\alpha$ . Setting $2-2\alpha=-2$ gives $\alpha=2$ . …
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KEAM 2026Set eng-2026-04194 marksMCQ
$A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$
›Reveal solutionSolution

Squaring $\alpha I+\beta A$ and matching with $A$ makes the diagonal entries $\alpha^2-\beta^2$ equal $0$ .

Note $A^2=\begin{pmatrix}0&1\\-1&0\end{pmatrix}^2=-I$ . Then
$(\alpha I+\beta A)^2=\alpha^2 I+2\alpha\beta A+\beta^2 A^2=(\alpha^2-\beta^2)I+2\alpha\beta A.$
In matrix form this is $\begin{pmatrix}\alpha^2-\beta^2&2\alpha\beta\\-2\alpha\beta&\alpha^2-\beta^2\end{pmatrix}$ . …
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KEAM 2026Set eng-2026-04184 marksMCQ
$A=\begin{bmatrix}3 & 5\\ -2 & -3\end{bmatrix}$
›Reveal solutionSolution

$B=A A^{-2}=A^{-1}$ , and $A^{-1}=\begin{bmatrix}-3&-5\\2&3\end{bmatrix}$ .

From $BA^{2}=A$ , multiply on the right by $A^{-2}$ :
$B=A\,A^{-2}=A^{-1}.$
For $A=\begin{bmatrix}3&5\\-2&-3\end{bmatrix}$ , $\det A=(3)(-3)-(5)(-2)=-9+10=1$ , so …
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Q.Find non-zero values of $x$ satisfying the matrix equation: $x\begin{bmatrix} 2x & 2 \\ 3 & x \end{bmatrix} + 2\begin{bmatrix} 8 & 5x \\ 4 & 4x \end{bmatrix} = 2\begin{bmatrix} x^2+8 & 24 \\ 10 & 6x \end{bmatrix}$ .

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 $\det(A)=0$

Turning the matrix equation into scalar equations

Equating entries

Method: Solving a scalar unknown from a matrix equation

Steps

Common Mistakes

Q.Find non-zero values of xxx satisfying the matrix equation: x[2x23x]+2[85x44x]=2[x2+824106x]x\begin{bmatrix} 2x & 2 \\ 3 & x \end{bmatrix} + 2\begin{bmatrix} 8 & 5x \\ 4 & 4x \end{bmatrix} = 2\begin{bmatrix} x^2+8 & 24 \\ 10 & 6x \end{bmatrix}x[2x3​2x​]+2[84​5x4x​]=2[x2+810​246x​].

Turning the matrix equation into scalar equations

Equating entries

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Q.Find non-zero values of $x$ satisfying the matrix equation: $x\begin{bmatrix} 2x & 2 \\ 3 & x \end{bmatrix} + 2\begin{bmatrix} 8 & 5x \\ 4 & 4x \end{bmatrix} = 2\begin{bmatrix} x^2+8 & 24 \\ 10 & 6x \end{bmatrix}$ .

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