A) \[\left[ \begin{matrix} 1 & 2 & 3 \\ 3 & 3 & 7 \\ -2 & -4 & -5 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 1 & -3 & 5 \\ 7 & 4 & 6 \\ 4 & 2 & 7 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 1 & -3 & 5 \\ 7 & 4 & 6 \\ 4 & 2 & -7 \\ \end{matrix} \right]\]
Correct Answer: C
Solution :
Let \[A=\left[ \begin{matrix} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1 \\ \end{matrix} \right]\], then \[|A|=\left| \,\begin{matrix} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1 \\ \end{matrix}\, \right|=1\] The matrix of cofactors of A = \[\left[ \begin{matrix} {{c}_{11}} & {{c}_{12}} & {{c}_{13}} \\ {{c}_{21}} & {{c}_{22}} & {{c}_{23}} \\ {{c}_{31}} & {{c}_{32}} & {{c}_{33}} \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 2 & -2 \\ 2 & 5 & -4 \\ 3 & 7 & -5 \\ \end{matrix} \right]\] Therefore, \[adj(A)=\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \\ \end{matrix} \right]\] \[\therefore \] \[{{A}^{-1}}=\frac{1}{|A|}\,.\,adjA=\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \\ \end{matrix} \right]\] , (\[\because \,|A|=1\]).
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