A) (6, 11)
B) (6, -11)
C) (-6, 11)
D) (-6, -11)
Correct Answer: C
Solution :
Given \[A=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \\ \end{matrix} \right],\,\,{{A}^{-1}}=\frac{1}{6}\left[ \begin{matrix} 6 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 2 & 1 \\ \end{matrix} \right]\] \[{{A}^{2}}=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \\ \end{matrix} \right]\,\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \\ \end{matrix} \right]\,=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & -1 & 5 \\ 0 & -10 & 14 \\ \end{matrix} \right]\] \[cA=\left[ \begin{matrix} c & 0 & 0 \\ 0 & c & c \\ 0 & -2c & 4c \\ \end{matrix} \right]\,\]; \[dI=\left[ \begin{matrix} d & 0 & 0 \\ 0 & d & 0 \\ 0 & 0 & d \\ \end{matrix} \right]\] \[\therefore \]By \[{{A}^{-1}}=\frac{1}{6}[{{A}^{2}}+cA+dI]\] \[\Rightarrow 6=1+c+d\],(By equality of matrices) \ (-6, 11) satisfy the relation.
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