Question
\[A = \left[ \begin{array}{l}
1\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,0\\
0\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,1\\
0\,\,\,\,\, - 2\,\,\,\,\,\,\,\,4
\end{array} \right]\,{\rm{and }}\,\,I = \left[ \begin{array}
\,\,1\,\,\,\,\,\,\,0\,\,\,\,\,\,\,0\\
0\,\,\,\,\,\,\,1\,\,\,\,\,\,\,0\\
0\,\,\,\,\,\,\,0\,\,\,\,\,\,\,1
\end{array} \right]\,{\rm{and}}\] $${A^{ - 1}} = \left[ {\frac{1}{6}\left( {{A^2} + cA + dI} \right)} \right],$$ then the value of $$c$$ and $$d$$ are
A.
$$(- 6, - 11)$$
B.
$$(6, 11)$$
C.
$$(- 6, 11)$$
D.
$$(6, - 11)$$
Answer :
$$(- 6, 11)$$
Solution :
\[{\rm{Given }}\,\,A = \left[ \begin{array}{l}
0\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,0\\
0\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,1\\
0\,\,\,\,\, - 2\,\,\,\,\,\,\,\,4
\end{array} \right]\]
∴ Characteristic eqn of above matrix $$A$$ is given by
\[\left| {A - \lambda I} \right| = 0 \Rightarrow \,\left[ \begin{array}{l}
1 - \lambda \,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\\
\,\,\,\,0\,\,\,\,\,\,\,\,\,\,1 - \lambda \,\,\,\,\,\,\,\,\,\,\,\,1\\
\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\, - 2\,\,\,\,\,\,\,\,4 - \lambda
\end{array} \right] = 0\]
$$\eqalign{
& \Rightarrow \,\,\left( {1 - \lambda } \right)\left( {4 - 5\lambda + {\lambda ^2} + 2} \right) = 0 \cr
& \Rightarrow \,\,{\lambda ^3} - 6{\lambda ^2} + 11\lambda - 6 = 0 \cr} $$
Also by Cayley Hamilton thm (every square matrix satisfies its characteristic equation) we obtain
$${A^3} - 6{A^2} + 11A - 6I = 0$$
Multiplying by $${A^{ - 1}},$$ we get
$$\eqalign{
& {A^2} - 6A + 11I - 6{A^{ - 1}} = 0 \cr
& \Rightarrow \,\,{A^{ - 1}} = \frac{1}{6}\left( {{A^2} - 6A + 11I} \right) \cr} $$
Comparing it with given relation,
$${A^{ - 1}} = \frac{1}{6}\left( {{A^2} - cA + dI} \right)$$
$${\text{we get }}c = - 6\,\,{\text{and}}\,{\text{ }}d = 11$$