Find the matrix $A$ satisfying the matrix equation: $\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

Find the matrix A satisfying the matrix equation: 2 1 \3 2 A -3 2 \5 -3 = 1 0 \0 1 .

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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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The equation is $PAQ=I$ , so $A=P^{-1}Q^{-1}$ . Computing the two inverses and multiplying gives $A=\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$ .

Setting up

Let $P=\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$ and $Q=\begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix}$ , so the equation reads

P\,A\,Q=I.

Because matrix multiplication is not commutative, we must remove $P$ from the left and $Q$ from the right. Multiply on the left by $P^{-1}$ and on the right by $Q^{-1}$ :

P^{-1}(P A Q)Q^{-1}=P^{-1} I Q^{-1}\ \Rightarrow\ A=P^{-1}Q^{-1}.

Checking invertibility and finding the inverses

For $2\times 2$ matrices, $\begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ .

$P^{-1}$ : $\det P = 2\cdot2-1\cdot3 = 1\neq 0$ , so

P^{-1}=\begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}.

$Q^{-1}$ : $\det Q = (-3)(-3)-(2)(5)=9-10=-1\neq 0$ , so

$Q^{-1}=\frac{1}{-1}\begin{bmatrix} -3 & -2 \\ -5 & -3 \end{bmatrix}=\begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix}.$ …

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Q.Find the matrix $A$ satisfying the matrix equation: $\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ .

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 $\det(A)=0$

Setting up

Checking invertibility and finding the inverses

Method: Isolating the middle matrix in $PAQ=I$

Steps

Common Mistakes

Q.Find the matrix AAA satisfying the matrix equation: [2132]A[−325−3]=[1001]\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}[23​12​]A[−35​2−3​]=[10​01​].

Setting up

Checking invertibility and finding the inverses

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Q.Find the matrix $A$ satisfying the matrix equation: $\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ .

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