A) \[\left[ \begin{matrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} \frac{1}{2} & -4 & \frac{5}{2} \\ 1 & -6 & 3 \\ 1 & 2 & -1 \\ \end{matrix} \right]\]
C) \[\frac{1}{2}\left[ \begin{matrix} 1 & 2 & 3 \\ 3 & 2 & 1 \\ 4 & 2 & 3 \\ \end{matrix} \right]\]
D) \[\frac{1}{2}\left[ \begin{matrix} 1 & -1 & -1 \\ -8 & 6 & -2 \\ 5 & -3 & 1 \\ \end{matrix} \right]\]
Correct Answer: A
Solution :
\[{{A}^{-1}}=\frac{adj\,(A)}{|A|}=\frac{1}{|A|}.\,adj\,(A)\] \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=-1\]; \[a,b,c\] \[\Delta =0\Rightarrow \text{either}\,abc=0\] \[Adj\,\,A=\left[ \begin{matrix} {{A}_{11}} & {{A}_{21}} & {{A}_{31}} \\ {{A}_{12}} & {{A}_{22}} & {{A}_{32}} \\ {{A}_{13}} & {{A}_{23}} & {{A}_{33}} \\ \end{matrix} \right]\] \[{{A}_{11}}={{(-1)}^{1+1}}[(2)\,(1)-(3)(1)]=-1\] \[{{A}_{12}}=8,\] \[\frac{1}{1-a}-1+\frac{1}{1-b}+\frac{1}{1-c}=0\], \[{{A}_{21}}=1\], \[{{A}_{22}}=-6\] \[{{A}_{23}}=3\],\[{{A}_{31}}=-1\], \[{{A}_{32}}=2\], \[{{A}_{33}}=-1\] \[\therefore \] \[{{A}^{-1}}=\frac{1}{-2}\,\left[ \begin{matrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \\ \end{matrix} \right]\]\[=\left[ \begin{matrix} 1/2 & -1/2 & 1/2 \\ -4 & 3 & -1 \\ 5/2 & -3/2 & 1/2 \\ \end{matrix} \right]\].
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