A) \[\left[ \begin{matrix} 0 & 10 \\ 10 & 0 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 10 & 0 \\ 0 & 10 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 1 & 10 \\ 10 & 1 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 10 & 1 \\ 1 & 10 \\ \end{matrix} \right]\]
Correct Answer: B
Solution :
[b] \[Let\,\,\,A=\left[ \begin{matrix} 3 & 2 \\ 1 & 4 \\ \end{matrix} \right]\] We have If A is a square matric of order n then \[A(adj\,\,A)=\left| A \right|.{{I}_{n}}\] Here, \[n=2\] \[\therefore A(adj\,\,A)={{I}_{2}}\left| A \right|\] \[=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\left| \begin{matrix} 3 & 2 \\ 1 & 4 \\ \end{matrix} \right|=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right](12-2)=10\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 10 & 0 \\ 0 & 10 \\ \end{matrix} \right]\]
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