If $[m\ n]\left[ {\begin{array}{*{20}{c}}m\\n\end{array}} \right] = [25]$ and $m< n$, then $(m, n) =$
$(2, 3)$
$(3, 4)$
$(4, 3)$
None of these
(b) It is obvious that $(m,n) =(3, 4).$
Find $\mathrm{X}$ and $\mathrm{Y}$, if $\mathrm{X}+\mathrm{Y}=\left[\begin{array}{ll}5 & 2 \\ 0 & 9\end{array}\right]$ and $\mathrm{X}-\mathrm{Y}=\left[\begin{array}{cc}3 & 6 \\ 0 & -1\end{array}\right]$.
If $A=\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right],$ find $k$ so that $A^{2}=k A-2 I$
If $A = \left[ {\begin{array}{*{20}{c}}\lambda &1\\{ – 1}&{ – \lambda }\end{array}} \right]$, then for what value of $\lambda ,\,{A^2} = O$
If $A = \left[ {\begin{array}{*{20}{c}}{ – 1}&0&0\\0&{ – 1}&0\\0&0&{ – 1}\end{array}} \right]$, then ${A^2}$is
If $A = \left[ {\begin{array}{*{20}{c}}4&2\\{ – 1}&1\end{array}} \right]$and $I$ is the identity matrix of order $2$, then $(A – 2I)(A – 3I) = $
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