Q110 / 180 · COMEDK 2017 · Mathematics
If $A=\left[\begin{array}{cc}2 & -3 \\ 5 & -7\end{array}\right]$, then $A+A^{-1}=$
- A$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
- Correct answerB$\left[\begin{array}{cc}-5 & 0 \\ 0 & -5\end{array}\right]$
- C$\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]$
- D$\left[\begin{array}{cc}4 & 0 \\ 0 & -5\end{array}\right]$
Solution
$\quad A=\left[\begin{array}{cc}2 & -3 \\ 5 & -7\end{array}\right]$
$\therefore \quad|A|=-14+15=1 \neq 0$
So, $A^{-1}$ exists.
$\begin{aligned}
&\therefore \quad \operatorname{adj} A=\left[\begin{array}{ll}
-7 & 3 \\
-5 & 2
\end{array}\right] \\
&\Rightarrow \quad A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\left[\begin{array}{cc}
-7 & 3 \\
-5 & 2
\end{array}\right]
\end{aligned}$
$\begin{aligned}
\Rightarrow A+A^{-1} &=\left[\begin{array}{cc}
2 & -3 \\
5 & -7
\end{array}\right]+\left[\begin{array}{ll}
-7 & 3 \\
-5 & 2
\end{array}\right] \\
&=\left[\begin{array}{cc}
-5 & 0 \\
0 & -5
\end{array}\right]
\end{aligned}$
Hence, the correct answer is option: 'B'.