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# Find $X$ and $Y$ if: $(i) \quad X + Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} \text{ and } X - Y = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$

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

Toolbox:
• 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.
• 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 $c$.
Given $X + Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} \text{ and } X - Y = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$:
Adding the two, $X+Y + X-Y = 2X = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$
Therefore, $2X = \begin{bmatrix} 7+3 & 0+0 \\ 2+0 & 5+3 \end{bmatrix} = \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix}$
$X = \frac{1}{2} \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix}$
$X = \begin{bmatrix} \frac{1}{2}10 & 0 \\ \frac{1}{2}2 & \frac{1}{2}8 \end{bmatrix}$
$X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix}$
Given $X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix}$ and $X - Y = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$:
$Y = X - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$
Substituting for $X$, $Y = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} -\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$
$Y = \begin{bmatrix} 5-3 & 0-0 \\ 1-0 & 4-3 \end{bmatrix}$
$Y=\begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$