Question

$A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& -2& 4\end{array}\right],I=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$ and $A^{-1}=\left[\frac{1}{6}\right(A^2+cA+dI\left)\right]$

The value of $(c,d)$ is

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

Answer 1

The correct option is C $(-6,11)$

Given $A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& -2& 4\end{array}\right]$
The characteristic equation of $A$ is given by
$|A-\lambda I|=0$
$\left|\begin{array}{ccc}1-\lambda & 0& 0\\ 0& 1-\lambda & 1\\ 0& -2& 4-\lambda \end{array}\right|=0$

$\Rightarrow {\lambda }^3-6{\lambda }^2+11\lambda -6=0$
$\Rightarrow A^3-6A^2+11A-6=0$ ($\because$ Every square matrix satisfies its characteristic equation )
$\Rightarrow A^2-6A+11I-6A^{-1}=0$
$\Rightarrow A^{-1}=\frac{1}{6}(A^2-6A+11I)$

Comparing this with $A^{-1}=\left[\frac{1}{6}\right(A^2+cA+dI\left)\right]$, we get $c=-6,d=11$

$\therefore (c,d)=(-6,11)$

Hence, option C.