If $A = \left[ {\begin{array}{cccccccccccccccccccc}1&5\\7&{12}\end{array}} \right]$ and $B = \left[ {\begin{array}{llllllllllllllllllll}9&1\\7&8\end{array}} \right]$, then find a matrix C such that $3A + 5B + 2C$ is a null matrix.

We have, $A = \left[ {\begin{array}{cccccccccccccccccccc}1&5\\7&{12}\end{array}} \right]$ and $B = \left[ {\begin{array}{llllllllllllllllllll}9&1\\7&8\end{array}} \right]$

Let $C = \left[ {\begin{array}{llllllllllllllllllll}a&b\\C&d\end{array}} \right]$

$\therefore$ $3A + 5B + 2C = 0$
$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}3&{15}\\{21}&{36}\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}{45}&5\\{35}&{40}\end{array}} \right] + \left[ {\begin{array}{llllllllllllllllllll}{2a}&{2b}\\{2c}&{2d}\end{array}} \right]$

$= \left[ {\begin{array}{llllllllllllllllllll}0&0\\0&0\end{array}} \right]$

$\Rightarrow$ $\left[ {\begin{array}{llllllllllllllllllll}{48 + 2a}&{20 + 2b}\\{56 + 2c}&{76 + 2d}\end{array}} \right]$

$= \left[ {\begin{array}{llllllllllllllllllll}0&0\\0&0\end{array}} \right]$

$\Rightarrow$ $2a + 48 = 0 \Rightarrow a = - 24$

Hence, $20 + 2b = 0 \Rightarrow b = - 10$

$56 + 2c = 0 \Rightarrow c = - 28$

and $76 + 2d = 0 \Rightarrow d = - 38$

$\therefore$ $C = \left[ {\begin{array}{llllllllllllllllllll}{ - 24}&{ - 10}\\{ - 28}&{ - 38}\end{array}} \right]$