Find the matrix $A$ such that $\left[ {\begin{array}{cccccccccccccccccccc}2&{ - 1}\\1&0\\{ - 3}&4\end{array}} \right]A = \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&{ - 8}&{ - 10}\\1&{ - 2}&{ - 5}\\9&{22}&{15}\end{array}} \right]$.

We have, ${\left[ {\begin{array}{cccccccccccccccccccc}2&{ - 1}\\1&0\\{ - 3}&4\end{array}} \right]_{3 \times 2}}A$

$= {\left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&{ - 8}&{ - 10}\\1&{ - 2}&{ - 5}\\9&{22}&{15}\end{array}} \right]_{3 \times 3}}$

From the given equation, it is clear that order of $A$ should be $2 \times 3$.

Let $A = \left[ {\begin{array}{llllllllllllllllllll}a&b&c\\d&e&f\end{array}} \right]$

$\therefore$ $\left[ {\begin{array}{cccccccccccccccccccc}2&{ - 1}\\1&0\\{ - 3}&4\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}a&b&c\\d&e&f\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&{ - 8}&{ - 10}\\1&{ - 2}&{ - 5}\\9&{22}&{15}\end{array}} \right]$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}{2a - d}&{2b - e}&{2c - f}\\{a + 0d}&{b + 0 \cdot e}&{c + 0 \cdot f}\\{ - 3a + 4d}&{ - 3b + 4e}&{ - 3c + 4f}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&{ - 8}&{ - 10}\\1&{ - 2}&{ - 5}\\9&{22}&{15}\end{array}} \right]$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}{2a - d}&{2b - e}&{2c - f}\\a&b&c\\{ - 3a + 4d}&{ - 3b + 4e}&{ - 3c + 4f}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&{ - 8}&{ - 10}\\1&{ - 2}&{ - 5}\\9&{22}&{15}\end{array}} \right]$

By equality of matrices,

we get
$a = 1,b = - 2,c = - 5$
and $2a - d = - 1 \Rightarrow d = 2a + 1 = 3$;

$\Rightarrow$ $2b - e = - 8 \Rightarrow e = 2( - 2) + 8 = 4$
$2c - f = - 10 \Rightarrow f = 2c + 10 = 0$

$\therefore$ $A = \left[ {\begin{array}{cccccccccccccccccccc}1&{ - 2}&{ - 5}\\3&4&0\end{array}} \right]$