Find matrices

Solution:

Given : $(2 X-Y)=\left[\begin{array}{ccc}6 & -6 & 0 \\ -4 & 2 & 1\end{array}\right]$                 ...(1)

$(X+2 Y)=\left[\begin{array}{ccc}3 & 2 & 5 \\ -2 & 1 & -7\end{array}\right]$                            ...(2)

Multiplying eq. (1) by eq. (2), we get

$2(2 X-Y)=2\left[\begin{array}{ccc}6 & -6 & 0 \\ -4 & 2 & 1\end{array}\right]$

$\Rightarrow 4 X-2 Y=\left[\begin{array}{ccc}12 & -12 & 0 \\ -8 & 4 & 2\end{array}\right]$       ...(3)

From eq. (3) and eq. (4), we get

$(4 X-2 Y)+(X+2 Y)=\left[\begin{array}{ccc}12 & -12 & 0 \\ -8 & 4 & 2\end{array}\right]+\left[\begin{array}{ccc}3 & 2 & 5 \\ -2 & 1 & -7\end{array}\right]$

$\Rightarrow 5 X=\left[\begin{array}{ccc}12+3 & -12+2 & 0+5 \\ -8-2 & 4+1 & 2-7\end{array}\right]$

$\Rightarrow 5 X=\left[\begin{array}{ccc}15 & -10 & 5 \\ -10 & 5 & -5\end{array}\right]$

$\Rightarrow X=\frac{1}{5}\left[\begin{array}{ccc}15 & -10 & 5 \\ -10 & 5 & -5\end{array}\right]$

$\Rightarrow X=\left[\begin{array}{ccc}3 & -2 & 1 \\ -2 & 1 & -1\end{array}\right]$

Putting the value of $X$ in eq. (2), we get

$(X+2 Y)=\left[\begin{array}{ccc}3 & 2 & 5 \\ -2 & 1 & -7\end{array}\right]$

$\Rightarrow\left[\begin{array}{ccc}3 & -2 & 1 \\ -2 & 1 & -1\end{array}\right]+2 Y=\left[\begin{array}{ccc}3 & 2 & 5 \\ -2 & 1 & -7\end{array}\right]$

$\Rightarrow 2 Y=\left[\begin{array}{ccc}3 & 2 & 5 \\ -2 & 1 & -7\end{array}\right]-\left[\begin{array}{ccc}3 & -2 & 1 \\ -2 & 1 & -1\end{array}\right]$

$\Rightarrow 2 Y=\left[\begin{array}{cc}3-3 & 2+2 & 5-1 \\ -2+2 & 1-1 & -7+1\end{array}\right]$

$\Rightarrow Y=\left[\begin{array}{ccc}0 & 2 & 2 \\ 0 & 0 & -3\end{array}\right]$