Solve for $x$ and $y$,$x\left[ {\begin{array}{llllllllllllllllllll}2\\1\end{array}} \right] + y\left[ {\begin{array}{llllllllllllllllllll}3\\5\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}{ - 8}\\{ - 11}\end{array}} \right] = 0$.

We have, $x\left[ {\begin{array}{llllllllllllllllllll}2\\1\end{array}} \right] + y\left[ {\begin{array}{llllllllllllllllllll}3\\5\end{array}} \right] + \left[ {\begin{array}{llllllllllllllllllll}{ - 8}\\{ - 11}\end{array}} \right] = 0$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}{2x}\\x\end{array}} \right] + \left[ {\begin{array}{llllllllllllllllllll}{3 \cdot y}\\{5 \cdot y}\end{array}} \right] + \left[ {\begin{array}{llllllllllllllllllll}{ - 8}\\{ - 11}\end{array}} \right] = 0$

$\Rightarrow$ $\left[ {\begin{array}{llllllllllllllllllll}{2x}&{3y}&{ - 8}\\x&{5y}&{ - 11}\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}0\\0\end{array}} \right]$
$\therefore$ $2x + 3y - 8 = 0$

$\Rightarrow$ $4x + 6y = 16$

……(i)
and $x + 5y - 11 = 0$
$\Rightarrow$ $4x + 20y = 44$

…….(ii)
On subtracting Eq. (i) from Eq. (ii),

we get
$14y = 28 \Rightarrow y = 2$
$\therefore$ $2x + 3 \times 2 - 8 = 0$

$\Rightarrow$ $2x = 2 \Rightarrow x = 1$
$\therefore$ $x = 1$

and $y = 2$