If $A = [a\,\,b],B = [ – b – a]$ and $C = \left[ \begin{array}{l}\,\,\,\,a\\ – a\end{array} \right]$, then the correct statement is

Show that

$\left[ {\begin{array}{*{20}{c}}
  5&{ – 1} \\ 
  6&7 
\end{array}} \right]\left[ {\begin{array}{*{20}{l}}
  2&1 \\ 
  3&4 
\end{array}} \right]$ $ \ne \left[ {\begin{array}{*{20}{l}}
  2&1 \\ 
  3&4 
\end{array}} \right]\left[ {\begin{array}{*{20}{l}}
  5&{ – 1} \\ 
  6&7 
\end{array}} \right]$

Find $x$ and $y,$ if $2\left[\begin{array}{ll}1 & 3 \\ 0 & x\end{array}\right]+\left[\begin{array}{ll}y & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right]$

Find $\mathrm{X}$ and $\mathrm{Y}$, if $\mathrm{X}+\mathrm{Y}=\left[\begin{array}{ll}5 & 2 \\ 0 & 9\end{array}\right]$ and $\mathrm{X}-\mathrm{Y}=\left[\begin{array}{cc}3 & 6 \\ 0 & -1\end{array}\right]$.

$A,B$ are n-rowed square matrices such that $AB = O$ and $B$ is non-singular. Then