A) \[\left[ \begin{matrix} 4 & 8 & 3 \\ 2 & 1 & 6 \\ 0 & 2 & 1 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 11 & 9 & 3 \\ 1 & 2 & 8 \\ 6 & 9 & 1 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 1 & -2 & 1 \\ -1 & 3 & 3 \\ -2 & 3 & -3 \\ \end{matrix} \right]\]
Correct Answer: B
Solution :
Let A = \[\left[ \begin{matrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \\ \end{matrix} \right]\] Then, \[{{A}_{11}}=1,\,{{A}_{12}}=-2,\,{{A}_{13}}=-2\] \[{{A}_{21}}=-1\] ,\[{{A}_{22}}=3\],\[{{A}_{23}}=3\] \[{{A}_{31}}=0\], \[{{A}_{32}}=-4\],\[{{A}_{33}}=-3\] \[adj\,(A)=\left[ \begin{matrix} {{A}_{11}} & {{A}_{21}} & {{A}_{31}} \\ {{A}_{12}} & {{A}_{22}} & {{A}_{32}} \\ {{A}_{13}} & {{A}_{23}} & {{A}_{33}} \\ \end{matrix} \right]\]\[=\left[ \begin{matrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \\ \end{matrix} \right]\].
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