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 .

NCERT Exemplar · Q8

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Manipur CohsemShort· 3mImportance★★★★★

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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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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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