Question

$A=\left[\begin{array}{cc}a& b\\ 0& c\end{array}\right]$ 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 $\frac{1}{ac}\left[\begin{array}{cc}c& -b\\ 0& a\end{array}\right]$

$A=\left[\begin{array}{cc}a& b\\ 0& c\end{array}\right]$

$A^{-1}=\frac{1}{ac-0\left(b\right)}\left[\begin{array}{cc}c& -b\\ 0& a\end{array}\right]=\frac{1}{ac}\left[\begin{array}{cc}c& -b\\ 0& a\end{array}\right]$

$(A-aI)(A-cI)=A^2-(a+c)A+acI=\left[\begin{array}{cc}a& b\\ 0& c\end{array}\right]\left[\begin{array}{cc}a& b\\ 0& c\end{array}\right]-(a+c)\left[\begin{array}{cc}a& b\\ 0& c\end{array}\right]+acI=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

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