Find non-zero values of x satisfying the matrix equation:
$x\left[\begin{array}{cc}2 x & 2 \\ 3 & x\end{array}\right]+2\left[\begin{array}{ll}8 & 5 x \\ 4 & 4 x\end{array}\right]=2\left[\begin{array}{cc}\left(x^{2}+8\right) & 24 \\ (10) & 6 x\end{array}\right]$
Given,
$x\left[\begin{array}{cc}2 x & 2 \\ 3 & x\end{array}\right]+2\left[\begin{array}{ll}8 & 5 x \\ 4 & 4 x\end{array}\right]=2\left[\begin{array}{cc}x^{2}+8 & 24 \\ 10 & 6 x\end{array}\right]$
$\Rightarrow$ $\left[\begin{array}{cc}2 x^{2} & 2 x \\ 3 x & x^{2}\end{array}\right]+\left[\begin{array}{cc}16 & 10 x \\ 8 & 8 x\end{array}\right]=\left[\begin{array}{cc}2 x^{2}+16 & 48 \\ 20 & 12 x\end{array}\right]$
$\Rightarrow$ $\left[\begin{array}{cc}2 x^{2}+16 & 2 x+10 x \\ 3 x+8 & x^{2}+8 x\end{array}\right]=\left[\begin{array}{cc}2 x^{2}+16 & 48 \\ 20 & 12 x\end{array}\right]$
On comparing the corresponding elements, we get
2x + 10x = 48
12x = 48
Thus, x = 4
It’s also seen that this value od x also satisfies the equation 3x + 8 = 20 and x2 + 8x = 12x.
Therefore, x = 4 (common) is the solution of the given matrix equation.
$\mathrm{A}=\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]$