Solution:

The given equation is

$2\left[ \begin{matrix} 2 & 4 \\ -3 & a \\ \end{matrix} \right]+3\left[ \begin{matrix} 1 & b \\ 8 & -4 \\ \end{matrix} \right]=\left[ \begin{matrix} 7 & 10 \\ 18 & 1 \\ \end{matrix} \right]$

 

Taking Left Hand Side

= $2\left[ \begin{matrix} 2 & 4 \\ -3 & a \\ \end{matrix} \right]+3\left[ \begin{matrix} 1 & b \\ 8 & -4 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2\times 2 & 2\times 4 \\ 2\times \left( -3 \right) & 2\times a \\ \end{matrix} \right]+\left[ \begin{matrix} 3\times 1 & 3\times b \\ 3\times 8 & 3\times \left( -4 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4 & 8 \\ -6 & 2a \\ \end{matrix} \right]+\left[ \begin{matrix} 3 & 3b \\ 24 & -12 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4+3 & 8+3b \\ \left( -6 \right)+24 & 2a+\left( -12 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 7 & 8+3b \\ 18 & 2a-12 \\ \end{matrix} \right]$

 

By equating it with the Right Hand Side, we have:

$\left[ \begin{matrix} 7 & 8+3b \\ 18 & 2a-12 \\ \end{matrix} \right]=~\left[ \begin{matrix} 7 & 10 \\ 18 & 1 \\ \end{matrix} \right]$

 

By comparing corresponding elements:

$8+3b=10$

$3b=10-8$

$3b=2$

$b=\frac{2}{3}\quad\quad\quad\quad\quad ...~(1)$

 

$2a-12=1$

$2a=1+12$

$2a=13$

$a=\frac{13}{2} \quad\quad\quad\quad\quad ...~(2)$

 

From equations $(1)$ and $(2)$, we get $a=\frac{13}{2}$ and $b=\frac{2}{3}$