If $X$ and $Y$ are $2 \times 2$ matrices, then solve the following matrix equations for $X$ and $Y$. $2X + 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix}, 3X + 2Y = \begin{bmatrix} -2 & 2 \\ 1 & -5 \end{bmatrix}$
Answer & explanation
Correct answer: option 1
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
$\begin{bmatrix} -2 & 0 \\ -1 & -3 \end{bmatrix} + Y = \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix}$
$\Rightarrow Y = \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix} - \begin{bmatrix} -2 & 0 \\ -1 & -3 \end{bmatrix} = \begin{bmatrix} 0+2 & 1-0 \\ 1+1 & -1+3 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 2 & 2 \end{bmatrix}$
$∴Y = \begin{bmatrix} 2 & 1 \\ 2 & 2 \end{bmatrix} \text{ and } X = \begin{bmatrix} -2 & 0 \\ -1 & -3 \end{bmatrix}$