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 $A^{-1}=\left[\frac{1}{6}\right(A^2+cA+dI\left)\right]$, 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 $-6,11$

$A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& -2& 4\end{array}\right]\left|A-\lambda I\right|=0\left|\begin{array}{ccc}1-\lambda & 0& 0\\ 0& 1-\lambda & 1\\ 0& -2& 4-\lambda \end{array}\right|=0\left(1-\lambda \right)\left[\left(1-\lambda \right)\left(4-\lambda \right)+2\right]=0\left(1-\lambda \right)\left(6-5\lambda +{\lambda }^2\right)=0\left(\lambda -1\right)\left({\lambda }^2-5\lambda +6\right)=0{\lambda }^3-5{\lambda }^2+6\lambda -{\lambda }^2+5\lambda -6=0{\lambda }^3-6{\lambda }^2+11\lambda -6=0A^3-A^2+11A-6I=0A^2-6A+11I-6A^{-1}=0$

$Now,A^{-1}=\frac{1}{2}\left[A^2-6A+11I\right]C=-6d=11$

$Hence,optionCisthecorrectanswer.$