Question

If $A=\left[\begin{array}{cc}a& b\\ 0& c\end{array}\right],$ then $A^{-1}+(A-aI)(A-cI)=$

• A. $\frac{1}{ac}\left[\begin{array}{cc}a& b\\ 0& -c\end{array}\right]$
• B. $\frac{1}{ac}\left[\begin{array}{cc}-a& b\\ 0& c\end{array}\right]$
• C. $\frac{1}{ac}\left[\begin{array}{cc}c& -b\\ 0& a\end{array}\right]$
• D. $\frac{1}{ac}\left[\begin{array}{cc}c& b\\ 0& a\end{array}\right]$

Answer 1

The correct option is C $\frac{1}{ac}\left[\begin{array}{cc}c& -b\\ 0& a\end{array}\right]$

The characteristic equation of A :
$|A-\lambda I|=0\Rightarrow (\lambda -a)(\lambda -c)=0\text{From Caley-Hamilton therom,}\Rightarrow (A-aI)(A-cI)=0\Rightarrow A^{-1}+(A-aI)(A-cI)=A^{-1}\Rightarrow A^{-1}=\frac{\text{adj}A}{\text{det}A}$
$\therefore A^{-1}+(A-aI)(A-cI)=\frac{1}{ac}\left[\begin{array}{cc}c& -b\\ 0& a\end{array}\right]$