Question

Find $X$ and $Y$, if
(i) $X+Y=\left[\begin{array}{cc}7& 0\\ 2& 5\end{array}\right]$ and $X-Y=\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]$
(ii) $2X+3Y=\left[\begin{array}{cc}2& 3\\ 4& 0\end{array}\right]$ and $3X+2Y=\left[\begin{array}{cc}2& -2\\ -1& 5\end{array}\right]$

Answer 1

(i) $X+Y=\left[\begin{array}{cc}7& 0\\ 2& 5\end{array}\right].....\left(1\right)$
$X-Y=\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]......\left(2\right)$
Adding equations (1) and (2), we get:
$2X=\left[\begin{array}{cc}7& 0\\ 2& 5\end{array}\right]+\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}7+3& 0+0\\ 2+0& 5+3\end{array}\right]=\left[\begin{array}{cc}10& 0\\ 2& 8\end{array}\right]$
$\therefore X=\frac{1}{2}\left[\begin{array}{cc}10& 0\\ 2& 8\end{array}\right]=\left[\begin{array}{cc}5& 0\\ 1& 4\end{array}\right]$
Now, $X+Y=\left[\begin{array}{cc}7& 0\\ 2& 5\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}5& 0\\ 1& 4\end{array}\right]+Y=\left[\begin{array}{cc}7& 0\\ 2& 5\end{array}\right]$
$\Rightarrow Y=\left[\begin{array}{cc}7& 0\\ 2& 5\end{array}\right]-\left[\begin{array}{cc}5& 0\\ 1& 4\end{array}\right]$
$\Rightarrow Y=\left[\begin{array}{cc}7-5& 0-0\\ 2-1& 5-4\end{array}\right]$
$\therefore Y=\left[\begin{array}{cc}2& 0\\ 1& 1\end{array}\right]$
(ii) $2X+3Y=\left[\begin{array}{cc}2& 3\\ 4& 0\end{array}\right]....\left(3\right)$
$3X+2Y=\left[\begin{array}{cc}2& -2\\ -1& 5\end{array}\right]....\left(4\right)$
Multiplying equation (3) with (4), we get :
$2(2X+3Y)=2\left[\begin{array}{cc}2& 3\\ 4& 0\end{array}\right]$
$\Rightarrow 4X+6Y=\left[\begin{array}{cc}4& 6\\ 8& 0\end{array}\right]$.....(5)
Multiplying equation (4) with (3), we get :
$3(3X+2Y)=3\left[\begin{array}{cc}2& -2\\ -1& 5\end{array}\right]$
$\Rightarrow 9X+6Y=\left[\begin{array}{cc}6& -6\\ -3& 15\end{array}\right]$.....(6)
From (5) and (6), we have:
$(4X+6Y)-(9X+6Y)=\left[\begin{array}{cc}4& 6\\ 8& 0\end{array}\right]-\left[\begin{array}{cc}6& -6\\ -3& 15\end{array}\right]$
$\Rightarrow -5X=\left[\begin{array}{cc}4-6& 6-(-6)\\ 8-(-3)& 0-15\end{array}\right]=\left[\begin{array}{cc}-2& 12\\ 11& -15\end{array}\right]$
$\therefore X=-\frac{1}{5}\left[\begin{array}{cc}-2& 12\\ 11& -15\end{array}\right]=\left[\begin{array}{cc}\frac{2}{5}& -\frac{12}{5}\\ -\frac{11}{5}& 3\end{array}\right]$
$2X+3Y=\left[\begin{array}{cc}2& 3\\ 4& 0\end{array}\right]$
Now,
$\Rightarrow 2\left[\begin{array}{cc}\frac{2}{5}& -\frac{12}{5}\\ -\frac{11}{5}& 3\end{array}\right]+3Y=\left[\begin{array}{cc}2& 3\\ 4& 0\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}\frac{4}{5}& -\frac{24}{5}\\ -\frac{22}{5}& 6\end{array}\right]+3Y=\left[\begin{array}{cc}2& 3\\ 4& 0\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}\frac{4}{5}& -\frac{24}{5}\\ -\frac{22}{5}& 6\end{array}\right]+3Y=\left[\begin{array}{cc}2& 3\\ 4& 0\end{array}\right]$
$\Rightarrow 3Y=\left[\begin{array}{cc}2& 3\\ 4& 0\end{array}\right]-\left[\begin{array}{cc}\frac{4}{5}& -\frac{24}{5}\\ -\frac{22}{5}& 6\end{array}\right]$
$\Rightarrow 3Y=\left[\begin{array}{cc}2-\frac{4}{5}& 3+\frac{24}{5}\\ 4+\frac{22}{5}& 0-6\end{array}\right]=\left[\begin{array}{cc}\frac{6}{5}& \frac{39}{5}\\ \frac{42}{5}& -2\end{array}\right]$
$\therefore Y=\frac{1}{3}\left[\begin{array}{cc}\frac{6}{5}& \frac{39}{5}\\ \frac{42}{5}& -2\end{array}\right]=\left[\begin{array}{cc}\frac{2}{5}& \frac{13}{5}\\ \frac{14}{5}& -2\end{array}\right]$