Find A, if $\left[ {\begin{array}{llllllllllllllllllll}4\\1\\3\end{array}} \right]A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 4}&8&4\\{ - 1}&2&1\\{ - 3}&6&3\end{array}} \right]$.

We have ${\left[ {\begin{array}{llllllllllllllllllll}4\\1\\3\end{array}} \right]_{3 \times 1}}A = {\left[ {\begin{array}{llllllllllllllllllll}{ - 4}&8&4\\{ - 1}&2&1\\{ - 3}&6&3\end{array}} \right]_{3 \times 3}}$

Let $A = [xyz]$
$\therefore$ ${\left[ {\begin{array}{llllllllllllllllllll}4\\1\\3\end{array}} \right]_{3 \times 1}}{[xyz]_{1 \times 3}} = {\left[ {\begin{array}{llllllllllllllllllll}{ - 4}&8&4\\{ - 1}&2&1\\{ - 3}&6&3\end{array}} \right]_{3 \times 3}}$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}{4x}&{4y}&{4z}\\x&y&z\\{3x}&{3y}&{3z}\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}{ - 4}&8&4\\{ - 1}&2&1\\{ - 3}&6&3\end{array}} \right]$
$\Rightarrow$ $4x = - 4 \Rightarrow x = - 1,4y = 8$
$\Rightarrow$ $y = 2$

and $4z = 4$
$\Rightarrow$ $z = 1$
$\therefore$ $A = [ - 121]$