A) \[\left[ \begin{matrix} 1 & -2 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 1 & -2 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{matrix} \right]\]
D) None of these
Correct Answer: B
Solution :
Let \[A=\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{matrix} \right]\]\[\Rightarrow \] \[|A|=1\,(1+0)=1\] \[Adj\,(A)=\left[ \begin{matrix} {{A}_{11}} & {{A}_{21}} & {{A}_{31}} \\ {{A}_{12}} & {{A}_{22}} & {{A}_{32}} \\ {{A}_{13}} & {{A}_{23}} & {{A}_{33}} \\ \end{matrix} \right]\] \[Adj\,(A)=\left[ \begin{matrix} 1 & -2 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \\ \end{matrix} \right]\] \[\Rightarrow \]\[{{A}^{-1}}=\frac{Adj\,(A)}{|A|}=\left[ \begin{matrix} 1 & -2 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \\ \end{matrix} \right]\].
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