Question

Find $x$ and $y$, if $x+y=\left[\begin{array}{cc}7& 0\\ 2& 5\end{array}\right]$ and $x-y=\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]$.

Answer 1

We have,

$x+y=\left[\begin{array}{cc}7& 0\\ 2& 5\end{array}\right]-----\left(1\right)$

$x-y=\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]-----\left(2\right)$

on adding $\left(1\right)$ and $\left(2\right)$ to

$x+y+x-y=\left[\begin{array}{cc}7& 0\\ 2& 5\end{array}\right]+\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]$

$2x=\left[\begin{array}{cc}10& 0\\ 2& 8\end{array}\right]$

$x=\frac{1}{2}\left[\begin{array}{cc}10& 0\\ 2& 8\end{array}\right]$

$x=\left[\begin{array}{cc}5& 0\\ 1& 4\end{array}\right]$

on subtracting $\left(1\right)$ and $\left(2\right)$

$x+y-x+y=\left[\begin{array}{cc}7& 0\\ 2& 5\end{array}\right]-\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]$

$2y=\left[\begin{array}{cc}4& 0\\ 2& 2\end{array}\right]$

$y=\frac{1}{2}\left[\begin{array}{cc}4& 0\\ 2& 2\end{array}\right]$

$y=\left[\begin{array}{cc}2& 0\\ 1& 1\end{array}\right]$

Hence, this is the answer.