Find X, if

Solution:

(i)

$X+Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right] \quad \ldots(1)$....(1)

$X-Y=\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]$....(2)

Adding equations (1) and (2), we get:

$2 X=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]+\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]=\left[\begin{array}{ll}7+3 & 0+0 \\ 2+0 & 5+3\end{array}\right]=\left[\begin{array}{ll}10 & 0 \\ 2 & 8\end{array}\right]$

$\therefore X=\frac{1}{2}\left[\begin{array}{ll}10 & 0 \\ 2 & 8\end{array}\right]=\left[\begin{array}{ll}5 & 0 \\ 1 & 4\end{array}\right]$

Now, $X+Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]$

$\Rightarrow\left[\begin{array}{ll}5 & 0 \\ 1 & 4\end{array}\right]+Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]$

$\Rightarrow Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]-\left[\begin{array}{ll}5 & 0 \\ 1 & 4\end{array}\right]$

$\Rightarrow Y=\left[\begin{array}{cc}7-5 & 0-0 \\ 2-1 & 5-4\end{array}\right]$

$\therefore Y=\left[\begin{array}{ll}2 & 0 \\ 1 & 1\end{array}\right]$

(ii)

$2 X+3 Y=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]$....(3)

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

Multiplying equation (3) with (2), we get:

$2(2 X+3 Y)=2\left[\begin{array}{lr}2 & 3 \\ 4 & 0\end{array}\right]$

$\Rightarrow 4 X+6 Y=\left[\begin{array}{ll}4 & 6 \\ 8 & 0\end{array}\right]$...(5)

Multiplying equation (4) with (3), we get:

$3(3 X+2 Y)=3\left[\begin{array}{rr}2 & -2 \\ -1 & 5\end{array}\right]$

$\Rightarrow 9 X+6 Y=\left[\begin{array}{rr}6 & -6 \\ -3 & 15\end{array}\right]$...(6)

From (5) and (6), we have:

$(4 X+6 Y)-(9 X+6 Y)=\left[\begin{array}{ll}4 & 6 \\ 8 & 0\end{array}\right]-\left[\begin{array}{rr}6 & -6 \\ -3 & 15\end{array}\right]$

$\Rightarrow-5 X=\left[\begin{array}{ll}4-6 & 6-(-6) \\ 8-(-3) & 0-15\end{array}\right]=\left[\begin{array}{rr}-2 & 12 \\ 11 & -15\end{array}\right]$

$\therefore X=-\frac{1}{5}\left[\begin{array}{lr}-2 & 12 \\ 11 & -15\end{array}\right]=\left[\begin{array}{ll}\frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3\end{array}\right]$

Now, $2 X+3 Y=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]$

$\Rightarrow 2\left[\begin{array}{lr}\frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3\end{array}\right]+3 Y=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]$

$\Rightarrow\left[\begin{array}{ll}\frac{4}{5} & -\frac{24}{5} \\ -\frac{22}{5} & 6\end{array}\right]+3 Y=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]$

$\Rightarrow 3 Y=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]-\left[\begin{array}{ll}\frac{4}{5} & -\frac{24}{5} \\ -\frac{22}{5} & 6\end{array}\right]$

$\Rightarrow 3 Y=\left[\begin{array}{cc}2-\frac{4}{5} & 3+\frac{24}{5} \\ 4+\frac{22}{5} & 0-6\end{array}\right]=\left[\begin{array}{cc}\frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & -6\end{array}\right]$

$\therefore Y=\frac{1}{3}\left[\begin{array}{ll}\frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & -6\end{array}\right]=\left[\begin{array}{lc}\frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2\end{array}\right]$