Q197 / 270 · NDA 2015 (Phase 1) · Mathematics
Consider the following in respect of two non-singular matrices $A$ and $B$ of same order:
1. $\operatorname{det}(A+B)=\operatorname{det} A+\operatorname{det} B$
2. $\quad(A+B)^{-1}=A^{-1}+B^{-1}$
Which of the above is/are correct?
- A1 only
- B2 only
- CBoth 1 and 2
- Correct answerDNeither 1 nor 2
Solution
Non-singular matrix is a matrix whose determinate Value is non-zero.
Let $A=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right]$ and $B=\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]$
Here, $A$ and $B$ are non-singular matrix
Now from Statement 1:
$\begin{aligned}
& A+B=\left[\begin{array}{ll}
1 & 1 \\
1 & 2
\end{array}\right]+\left[\begin{array}{ll}
2 & 1 \\
1 & 2
\end{array}\right]=\left[\begin{array}{ll}
3 & 2 \\
2 & 4
\end{array}\right] \\
& \operatorname{det}(A+B)=12-4=8 \\
& \text { Now det }(A)=\left|\begin{array}{ll}
1 & 1 \\
1 & 2
\end{array}\right|=1 \\
& \text { and } \operatorname{det}(B)=\left|\begin{array}{ll}
2 & 1 \\
1 & 2
\end{array}\right|=3
\end{aligned}$
Now, $\operatorname{det}(A+B)=4 \neq \operatorname{det}(A+B)$
$\therefore \quad$ Statement 1 is wrong.
Now from Statement 2:
$\begin{aligned}
& (A+B)^{-1}=\frac{1}{|A+B|} \operatorname{adj}(A+B) \\
& =\frac{1}{8}\left[\begin{array}{cc}
4 & -2 \\
-2 & 3
\end{array}\right] \\
& \text { and } \mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj}(\mathrm{A})=\frac{1}{1}\left[\begin{array}{cc}
2 & -1 \\
-1 & 1
\end{array}\right] \\
& \text { and } B^{-1}=\frac{1}{|B|} \operatorname{adj}(B)=\frac{1}{3}\left[\begin{array}{cc}
2 & -1 \\
-1 & 2
\end{array}\right] \\
& \text { Now, } A^{-1}+B^{-1}=\left[\begin{array}{cc}
2 & -1 \\
-1 & 1
\end{array}\right]+\left[\begin{array}{cc}
\frac{2}{3} & \frac{-1}{3} \\
\frac{-1}{3} & \frac{2}{3}
\end{array}\right]
\end{aligned}$
$=\left[\begin{array}{cc}
\frac{8}{3} & \frac{-4}{3} \\
\frac{-4}{3} & \frac{5}{3}
\end{array}\right] \neq(A+B)^{-1}$
$\therefore \quad$ Statement 2 is wrong
$\therefore \quad$ Option (D) is correct.
Hence, the correct answer is option: 'D'.