Question

Let $A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& -2& 4\end{array}\right]$ and $I$ be an identity matrix of order $3.$ If $A^{-1}=\frac{1}{6}(A^2+cA+dI)$ where $c$ and $d$ are real numbers, then the value of $c+d$ is

• A. $11$
• B. $5$
• C. $0$
• D. $1$

Answer 1

The correct option is B $5$

$A^{-1}=\frac{1}{6}(A^2+cA+dI)$
Multiplying with matrix $A$ on both sides, we get
$6I=A^3+c{A}^2+dA$
$\Rightarrow A^3+c{A}^2+dA-6I=0\cdots \left(1\right)$

Also, $A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& -2& 4\end{array}\right]$
Characteristic equation is given by
$|A-\lambda I|=0$
$\Rightarrow \left|\begin{array}{ccc}1-\lambda & 0& 0\\ 0& 1-\lambda & 1\\ 0& -2& 4-\lambda \end{array}\right|=0$
$\Rightarrow (1-\lambda )\left[\right(1-\lambda \left)\right(4-\lambda )+2]=0$
$\Rightarrow {\lambda }^3-6{\lambda }^2+11\lambda -6=0$

We know that every matrix satisfies its characteristic equation.
Therefore, $A^3-6A^2+11A-6I=0\cdots \left(2\right)$
Comparing $\left(1\right)$ and $\left(2\right),$ we have
$c=-6$ and $d=11$
$\therefore c+d=5$