Question

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

$\left(ii\right)2X+3Y=\left[\begin{array}{cc}2& 3\\ 4& 0\end{array}\right]and3X+2Y=\left[\begin{array}{cc}2& -2\\ -1& 5\end{array}\right]$

Answer 1

(I) Given, $X+Y=\left[\begin{array}{cc}7& 0\\ 2& 5\end{array}\right].....\left(i\right)$
and $X-Y=\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right].....\left(ii\right)$
Adding Eqs. (i) and (ii), we get
$2X=\left[\begin{array}{cc}7& 0\\ 2& 5\end{array}\right]+\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}10& 0\\ 2& 8\end{array}\right]\Rightarrow X=\frac{1}{2}\left[\begin{array}{cc}10& 0\\ 2& 8\end{array}\right]=\left[\begin{array}{cc}5& 0\\ 1& 4\end{array}\right]$
Subtracting Eq. (ii) from Eq. (i), we get
$2Y=\left[\begin{array}{cc}7& 0\\ 2& 5\end{array}\right]-\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}4& 0\\ 2& 2\end{array}\right]\Rightarrow Y=\frac{1}{2}\left[\begin{array}{cc}4& 0\\ 2& 2\end{array}\right]=\left[\begin{array}{cc}2& 0\\ 1& 1\end{array}\right]$

Given, $2X+3Y=\left[\begin{array}{cc}2& 3\\ 4& 0\end{array}\right]....\left(i\right)$
and $3X+2Y=\left[\begin{array}{cc}2& -2\\ -1& 5\end{array}\right]........\left(ii\right)$
Multiplying Eq. (i) by 2, Eq. by 3and then subtracting, we get
$2(2X+3Y)-3(3X+2Y)=2\left[\begin{array}{cc}2& 3\\ 4& 0\end{array}\right]-3\left[\begin{array}{cc}2& -2\\ -1& 5\end{array}\right]$

$\Rightarrow 4X+6Y-9X-6Y=\left[\begin{array}{cc}4& 6\\ 8& 0\end{array}\right]-\left[\begin{array}{cc}6& -6\\ -3& 15\end{array}\right]\Rightarrow -5X=\left[\begin{array}{cc}4-6& 6+6\\ 8+3& 0-15\end{array}\right]=\left[\begin{array}{cc}-2& 12\\ 11& -15\end{array}\right]\Rightarrow X=-\frac{1}{5}\left[\begin{array}{cc}-2& 12\\ 11& -15\end{array}\right]=\left[\begin{array}{cc}\frac{2}{5}& -\frac{12}{5}\\ -\frac{11}{5}& 3\end{array}\right]$

Then, from Eq. (i), $3Y=\left[\begin{array}{cc}2& 3\\ 4& 0\end{array}\right]-2X=\left[\begin{array}{cc}2& 3\\ 4& 0\end{array}\right]-2\left[\begin{array}{cc}\frac{2}{5}& -\frac{12}{5}\\ -\frac{11}{5}& 3\end{array}\right]$
$=\left[\begin{array}{cc}2-\frac{4}{5}& 3+\frac{24}{5}\\ 4+\frac{22}{5}& 0-6\end{array}\right]=\left[\begin{array}{cc}\frac{6}{5}& \frac{39}{5}\\ \frac{42}{5}& -6\end{array}\right]$
$\therefore Y=\frac{1}{3}\left[\begin{array}{cc}\frac{6}{5}& \frac{39}{5}\\ \frac{42}{5}& -6\end{array}\right]=\left[\begin{array}{cc}\frac{2}{5}& \frac{13}{5}\\ \frac{14}{5}& -2\end{array}\right]$