Search any question & find its solution
Question:
Answered & Verified by Expert
If $A=\left[\begin{array}{cc}2 & -3 \\ 5 & -7\end{array}\right]$, then $A+A^{-1}=$
Options:
Solution:
1761 Upvotes
Verified Answer
The correct answer is:
$\left[\begin{array}{cc}-5 & 0 \\ 0 & -5\end{array}\right]$
$\quad A=\left[\begin{array}{cc}2 & -3 \\ 5 & -7\end{array}\right]$
$\therefore \quad|A|=-14+15=1 \neq 0$
So, $A^{-1}$ exists.
$\begin{aligned}
&\therefore \quad \operatorname{adj} A=\left[\begin{array}{ll}
-7 & 3 \\
-5 & 2
\end{array}\right] \\
&\Rightarrow \quad A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\left[\begin{array}{cc}
-7 & 3 \\
-5 & 2
\end{array}\right]
\end{aligned}$
$\begin{aligned}
\Rightarrow A+A^{-1} &=\left[\begin{array}{cc}
2 & -3 \\
5 & -7
\end{array}\right]+\left[\begin{array}{ll}
-7 & 3 \\
-5 & 2
\end{array}\right] \\
&=\left[\begin{array}{cc}
-5 & 0 \\
0 & -5
\end{array}\right]
\end{aligned}$
$\therefore \quad|A|=-14+15=1 \neq 0$
So, $A^{-1}$ exists.
$\begin{aligned}
&\therefore \quad \operatorname{adj} A=\left[\begin{array}{ll}
-7 & 3 \\
-5 & 2
\end{array}\right] \\
&\Rightarrow \quad A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\left[\begin{array}{cc}
-7 & 3 \\
-5 & 2
\end{array}\right]
\end{aligned}$
$\begin{aligned}
\Rightarrow A+A^{-1} &=\left[\begin{array}{cc}
2 & -3 \\
5 & -7
\end{array}\right]+\left[\begin{array}{ll}
-7 & 3 \\
-5 & 2
\end{array}\right] \\
&=\left[\begin{array}{cc}
-5 & 0 \\
0 & -5
\end{array}\right]
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.