If $X$ and $Y$ are $2 \times 2$ matrices, then solve the following matrix equations for $X$ and $Y$. $2X + 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix}, 3X + 2Y = \begin{bmatrix} -2 & 2 \\ 1 & -5 \end{bmatrix}$

$X = \begin{bmatrix} -2 & 0 \\ -1 & -3 \end{bmatrix}, \quad Y = \begin{bmatrix} 2 & 1 \\ 2 & 2 \end{bmatrix}$

The correct answer is Option (1) → $X = \begin{bmatrix} -2 & 0 \\ -1 & -3 \end{bmatrix}, \quad Y = \begin{bmatrix} 2 & 1 \\ 2 & 2 \end{bmatrix}$ ##

We have,

$2X + 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} \quad \dots(i)$

and $3X + 2Y = \begin{bmatrix} -2 & 2 \\ 1 & -5 \end{bmatrix} \quad \dots(ii)$

On subtracting Eq. (i) from Eq. (ii), we get

$∴(3X + 2Y) - (2X + 3Y) = \begin{bmatrix} -2-2 & 2-3 \\ 1-4 & -5-0 \end{bmatrix}$

$(X - Y) = \begin{bmatrix} -4 & -1 \\ -3 & -5 \end{bmatrix} \quad \dots(iii)$

On adding Eqs. (i) and (ii), we get

$(5X + 5Y) = \begin{bmatrix} 0 & 5 \\ 5 & -5 \end{bmatrix}$

$\Rightarrow (X + Y) = \frac{1}{5} \begin{bmatrix} 0 & 5 \\ 5 & -5 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix} \quad \dots(iv)$

On adding Eqs. (iii) and (iv), we get

$(X - Y) + (X + Y) = \begin{bmatrix} -4 & 0 \\ -2 & -6 \end{bmatrix}$

$\Rightarrow 2X = 2 \begin{bmatrix} -2 & 0 \\ -1 & -3 \end{bmatrix}$

$∴X = \begin{bmatrix} -2 & 0 \\ -1 & -3 \end{bmatrix}$

From Eq. (iv), we get

$\begin{bmatrix} -2 & 0 \\ -1 & -3 \end{bmatrix} + Y = \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix}$

$\Rightarrow Y = \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix} - \begin{bmatrix} -2 & 0 \\ -1 & -3 \end{bmatrix} = \begin{bmatrix} 0+2 & 1-0 \\ 1+1 & -1+3 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 2 & 2 \end{bmatrix}$

$∴Y = \begin{bmatrix} 2 & 1 \\ 2 & 2 \end{bmatrix} \text{ and } X = \begin{bmatrix} -2 & 0 \\ -1 & -3 \end{bmatrix}$