Answer:
We have, \[x\left[ \begin{matrix} 2x & 2 \\ 3 & x \\ \end{matrix} \right]+2\left[ \begin{matrix} 8 & 5x \\ 4 & 4x \\ \end{matrix} \right]=2\left[ \begin{matrix} {{x}^{2}}+8 & 24 \\ 10 & 6x \\ \end{matrix} \right]\] \[\Rightarrow \] \[\left[ \begin{matrix} 2{{x}^{2}} & 2x \\ 3x & {{x}^{2}} \\ \end{matrix} \right]+\left[ \begin{matrix} 16 & 10x \\ 8 & 8x \\ \end{matrix} \right]=\left[ \begin{matrix} 2{{x}^{2}}+16 & 48 \\ 20 & 12x \\ \end{matrix} \right]\] \[\Rightarrow \] \[\left[ \begin{matrix} 2{{x}^{2}}+16 & 12x \\ 3x+8 & {{x}^{2}}+8x \\ \end{matrix} \right]=\left[ \begin{matrix} 2{{x}^{2}}+16 & 48 \\ 20 & 12x \\ \end{matrix} \right]\] \[\Rightarrow \] 12x = 48 \[\therefore \] x = 4
You need to login to perform this action.
You will be redirected in
3 sec