Question

If $A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& -2& 4\end{array}\right],I=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$ and $A^{-1}=\frac{1}{6}\left(A^2+cA+dI\right)$ then the sum of values of $c$ and $d$ is

Answer 1

Answer: 5

Method $I$
We evaluate $A^2$ and $A^3$ and write the given equation as $A{A}^{-1}=I=\frac{1}{6}\left[A^3+c{A}^2+dA\right]$ .
Comparing the corresponding elements on both the sides, we get $c=-6,d=11$ .
Method $II$
Characteristic equation for $A$ is $\left|A-xI\right|=0=\left|\begin{array}{ccc}1-x& 0& 0\\ 0& 1-x& 1\\ 0& -2& 4-x\end{array}\right|=0$
$\Rightarrow \left(1-x\right)\left[\left(1-x\right)\left(4-x\right)+2\right]=0$
Hence characteristic equation is $x^3-6x^2+11x-6=0$
Then by Caley Hamilton theorem
$A^3-6A^2+11A-6I=0$
multiply by $A^{-1}$ both the sides,
we get $\frac{1}{6}\left(A^2-6A+11I\right)=A^{-1}$ ...(i)
given $A^{-1}=\frac{1}{6}\left(A^2+cA+d\right)$ ...(ii)
then from equation (i) and (ii)
we get $c=-6$ & $d=11$
then $c+d=5$