A) \[\frac{-1}{8}\,\left[ \begin{matrix} 2 & 3 \\ 4 & 2 \\ \end{matrix} \right]\]
B) \[\frac{-1}{8}\,\left[ \begin{matrix} 3 & 2 \\ 2 & 4 \\ \end{matrix} \right]\]
C) \[\frac{1}{8}\,\left[ \begin{matrix} 2 & 3 \\ 4 & 2 \\ \end{matrix} \right]\]
D) \[\frac{1}{8}\,\left[ \begin{matrix} 3 & 2 \\ 2 & 4 \\ \end{matrix} \right]\]
Correct Answer: A
Solution :
Let \[A=\left[ \begin{matrix} 2 & -3 \\ -4 & 2 \\ \end{matrix} \right]\,,\text{ }\therefore \,|A|\,=\,\left| \,\begin{matrix} 2 & -3 \\ -4 & 2 \\ \end{matrix}\, \right|=4-12=-8\] The matrix of cofactors of the elements of A viz. \[\left[ \begin{matrix} {{c}_{11}} & {{c}_{12}} \\ {{c}_{21}} & {{c}_{22}} \\ \end{matrix} \right]=\left[ \begin{matrix} 2 & -(-4) \\ -(-3) & 2 \\ \end{matrix} \right]=\left[ \,\begin{matrix} 2 & 4 \\ 3 & 2 \\ \end{matrix}\, \right]\] \[\therefore \] \[adjA=\] transpose of the matrix of cofactors of elements of \[A=\left[ \begin{matrix} 2 & 3 \\ 4 & 2 \\ \end{matrix} \right]\] \[\therefore \] \[A(adj\,A)=|A|\,I\].
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