A) A
B) \[{{A}^{T}}\]
C) \[\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 1 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{matrix} \right]\]
Correct Answer: A
Solution :
\[A=\left[ \begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\] \[\Rightarrow \] \[|A|=-1(1+0)=-1\] \[adj(A)=\,\left[ \,\begin{matrix} {{A}_{11}} & {{A}_{21}} & {{A}_{31}} \\ {{A}_{12}} & {{A}_{22}} & {{A}_{32}} \\ {{A}_{13}} & {{A}_{23}} & {{A}_{33}} \\ \end{matrix}\, \right]\] \[\frac{{{(-1)}^{3+2}}{{M}_{32}}}{-2},\,\text{ }=\frac{-(-2)}{-2}=-1\] \[{{A}_{11}}=0,\,{{A}_{12}}=-1,\,{{A}_{13}}=0\] \[{{A}_{21}}=-1,\,{{A}_{22}}=0,\,{{A}_{23}}=0\] \[{{A}_{31}}=0\], \[{{A}_{32}}=0,{{A}_{33}}=-1\] \[\frac{{{(-1)}^{3+2}}{{M}_{32}}}{-2},\,\text{ }=\frac{-(-2)}{-2}=-1\] \[{{A}^{-1}}=\frac{adj\,(A)}{|A|}=\left[ \,\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{matrix}\, \right]=A\].
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