Question

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

• A. $(–6,–11)$
• B. $(6,11)$
• C. $(–6,11)$
• D. $(6,–11)$

Answer 1

The correct option is C

$(–6,11)$

Explanation for the correct option

Every square matrix A satisfies its characteristic equation i.e $\left|A-\lambda I\right|=0$

$\Rightarrow \left|\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& -2& 4\end{array}\right]-\left[\begin{array}{ccc}\lambda & 0& 0\\ 0& \lambda & 0\\ 0& 0& \lambda \end{array}\right]\right|=0\begin{array}{}\\ \\ \end{array}$

$\Rightarrow \left[\begin{array}{ccc}1-\lambda & 0& 0\\ 0& 1-\lambda & 1\\ 0& -2& 4-\lambda \end{array}\right]=0\begin{array}{}\\ \\ \end{array}$
$$\begin{gathered} \Rightarrow {\lambda }^3-6{\lambda }^2+11\lambda -6=0 \\ \Rightarrow A^3-6A^2+11A-6I=0 \end{gathered}$$

$\Rightarrow 6I=A^3-6A^2+11A$

Now, Multiplying both sides by $A^{-1}$
$\Rightarrow 6A^{-1}=A^2-6A+11I$

Comparing it with the given equation,

$\Rightarrow 6A^{-1}=A^2+cA+dI$

We get, $c=-6;d=11$

Hence, $\mathrm{Option}\left(\mathrm{C}\right)$ is the correct answer.