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# Find $X$ and $Y$ if $(ii) \quad 2X + 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} \text{ and } 3X + 2Y = \begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix}$

This question has 2 parts and each part has been answered separately here.

Toolbox:
• The scalar multiplication $cA$ of a matrix $A$ and a number $c$ (also called a scalar in the parlance of abstract algebra) is given by multiplying every entry of $A$ by $c$.
• The difference $A-B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A - B)_{i,j} = A_{i,j} - B_{i,j}$ where 1 ≤ i ≤ m and 1 ≤ j ≤ n.
Given $2X + 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} (i)$
Given $3X + 2Y = \begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix} (ii)$
Multiplying (i) by 3 and (ii) by 2 and subtracting the two resulting equations:
$3(2X+3Y) - 2(3X+2Y) = 3 \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} - 2 \begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix}$
i.e., $6X+9Y-6X-4Y = \begin{bmatrix} 3\times2 & 3\times3 \\ 3\times4 & 0 \end{bmatrix} - \begin{bmatrix} 2\times2 & 2\times(-2) \\ 2\times(-1) & 2\times(5) \end{bmatrix}$
i.e., $5Y = \begin{bmatrix} 6 & 9 \\ 12 & 0 \end{bmatrix} - \begin{bmatrix} 4&-4 \\ -2& 10 \end{bmatrix}$
$5Y = \begin{bmatrix} 6-4 & 9-(-4) \\ 12-(-2) & 0-10 \end{bmatrix}$
$5Y = \begin{bmatrix} 2 & 13 \\ 14 & -10 \end{bmatrix}$
$Y = \frac{1}{5} \begin{bmatrix} 2 & 13 \\ 14 & -10 \end{bmatrix}$
$Y = \begin{bmatrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \end{bmatrix}$
Substituting $Y = \begin{bmatrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \end{bmatrix}$ in (i):