If $\left[\begin{array}{cc}3& -2\\ 3& 0\\ 2& 4\end{array}\right]\left[\begin{array}{l}\mathrm{y}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{y}\\ \mathrm{x}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}\end{array}\right]=\left[\begin{array}{cc}3& 3\\ 3\mathrm{y}& 3\mathrm{y}\\ 10& 10\end{array}\right]$ then the value of  10xy is

$\begin{array}{l}\left[\begin{array}{cc}3& -2\\ 3& 0\\ 2& 4\end{array}\right]\left[\begin{array}{cc}\mathrm{y}& \mathrm{y}\\ \mathrm{x}& \mathrm{x}\end{array}\right]\\ =\left[\begin{array}{lc}3\mathrm{y}-2\mathrm{x}& 3\mathrm{y}-2\mathrm{x}\\ 3\mathrm{y}& 3\mathrm{y}\\ 2\mathrm{y}+4\mathrm{x}& 2\mathrm{y}+4\mathrm{x}\end{array}\right]\\ =\left[\begin{array}{cc}3& 3\\ 3\mathrm{y}& 3\mathrm{y}\\ 10& 10\end{array}\right]\text{ (Given) }\end{array}$

$\begin{array}{l}\Rightarrow 3\mathrm{y}-2\mathrm{x}=3----\left(1\right)\\ \text{ and }2\mathrm{y}+4\mathrm{x}=10----\left(2\right)\end{array}$

solving we get $\mathrm{x}=\frac{3}{2},\mathrm{y}=2$