Question

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

Answer 1

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