A) \[\left[ \begin{matrix} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 1 & -1 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} -9 & 8 & 5 \\ -8 & 7 & -4 \\ 2 & 2 & -1 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\]
Correct Answer: A
Solution :
Given that,\[A=\left[ \begin{matrix} 1 & -2 & 3 \\ 0 & -1 & 4 \\ -2 & 2 & 1 \\ \end{matrix} \right]\] \[\therefore \] \[A'=\left[ \begin{matrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \\ \end{matrix} \right]\] \[\Rightarrow \] \[|A'|=1(-1-8)+0-2(-8+3)\] \[=-9+10=1\] Cofactors of \[A'\] are \[{{C}_{11}}=-9,\,\,{{C}_{12}}=8,\,\,{{C}_{13}}=-5\] \[{{C}_{21}}=-8,\,\,{{C}_{22}}=7,\,\,{{C}_{23}}=-4\] \[{{C}_{31}}=-2,\,\,{{C}_{32}}=2,\,\,{{C}_{33}}=-1\] \[\therefore \] \[adj\,\,(A')=\left[ \begin{matrix} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & 1 \\ \end{matrix} \right]\] \[\therefore \] \[{{(A')}^{-1}}=\frac{adj\,\,(A')}{|A'|}\] \[=\frac{1}{1}\left[ \begin{matrix} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \\ \end{matrix} \right]\] Note: For the existence of an inverse, the value of that determinant is not equal to zero.You need to login to perform this action.
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