Find the value of $x$, if $[1\,x\,1]\left[ {\begin{array}{cccccccccccccccccccc}1&3&2\\2&5&1\\{15}&3&2\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}1\\2\\x\end{array}} \right] = 0$.

We have, ${[1\,x\,1]_{1 \times 3}}{\left[ {\begin{array}{cccccccccccccccccccc}1&3&2\\2&5&1\\{15}&3&2\end{array}} \right]_{3 \times 3}}{\left[ {\begin{array}{cccccccccccccccccccc}1\\2\\x\end{array}} \right]_{3 \times 1}} = 0$
$\Rightarrow$ ${[1 + 2x + 15\,\,\,\,\,\,\,\,3 + 5x + 3\,\,\,\,\,\,\,\,2 + x + 2]_{1 \times 3}}{\left[ {\begin{array}{llllllllllllllllllll}1\\2\\x\end{array}} \right]_{3 \times 1}} = 0$

$\Rightarrow$ ${[16 + 2x\,\,\,\,\,5x + 6\,\,\,\,\,x + 4]_{1 \times 3}}{\left[ {\begin{array}{cccccccccccccccccccc}\begin{array}{l}1\\2\end{array}\\x\end{array}} \right]_{3 \times 1}} = 0$

$\Rightarrow$ ${[16 + 2x + (5x + 6) \cdot 2 + (x + 4) \cdot x]_{1 \times 1}} = 0$
$\Rightarrow$ $\left[ {16 + 2x + 10x + 12 + {x^2} + 4x} \right] = 0$

$\Rightarrow$ $\left[ {{x^2} + 16x + 28} \right] = 0$
$\Rightarrow$ $\left[ {{x^2} + 2x + 14x + 28} \right] = 0$

$\Rightarrow$ $(x + 2)(x + 14) = 0$
$\therefore$ $x = - 2, - 14$