A) \[\left[ \begin{matrix} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right]\]
Correct Answer: C
Solution :
Given, \[A=\left[ \begin{matrix} 1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \\ \end{matrix} \right]\] \[{{C}_{11}}=8-6=2\] \[{{C}_{12}}=-(0+9)=-9,\]\[{{C}_{31}}=+6-4=2\] \[{{C}_{13}}=0-6=-6,\] \[{{C}_{32}}=-(-3-0)=3\] \[{{C}_{21}}=-(-8+4)=4,\] \[{{C}_{33}}=2-0=2\] \[{{C}_{22}}=4-6=-2,\] \[{{C}_{23}}=-(-2+6)=-4\] \[adj\,(A)={{\left[ \begin{matrix} 2 & -9 & -6 \\ 4 & -2 & -4 \\ 2 & 3 & 2 \\ \end{matrix} \right]}^{T}}=\left[ \begin{matrix} 2 & 4 & 2 \\ -9 & -2 & 3 \\ -6 & -4 & 2 \\ \end{matrix} \right]\] Now, A. \[adj\,(A)=\left[ \begin{matrix} 1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & 4 & 2 \\ -9 & -2 & 3 \\ -6 & -4 & 2 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \\ \end{matrix} \right]\] Alternative : \[{{A}^{-1}}=\frac{adj\,\,(A)}{|A|}\] \[\Rightarrow \] \[A\,{{A}^{-1}}=\frac{A.[adj\,(A)]}{|A|}\] \[\Rightarrow \] \[I.|A|=A.[adj\,(A)]\] \[|A|=8\] \[\therefore \] A \[[adj\,(A)]=8I=\left[ \begin{matrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \\ \end{matrix} \right]\]
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