Q178 / 270 · NDA 2016 (Phase 2) · Mathematics
$A=\left[\begin{array}{rr}1 & -1 \\ 2 & 3\end{array}\right]$ and $B=\left[\begin{array}{rr}2 & 3 \\ -1 & -2\end{array}\right]$, then which of the following is/are correct?
1. $\quad \mathrm{AB}\left(\mathrm{A}^{-1} \mathrm{~B}^{-1}\right)$ is a unit matrix.
2. $(\mathrm{AB})^{-1}=\mathrm{A}^{-1} \mathrm{~B}^{-1}$
Select the correct answer using the code given below:
- A1 only
- B2 only
- CBoth 1 only 2
- Correct answerDNeither 1 nor 2
Solution
Here, $A=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}2 & 3 \\ -1 & -2\end{array}\right]$
$|A|=3-(-2)=5$ and $|B|=-4-(-3)=-1$ $\Rightarrow A^{-1}=\frac{1}{5}\left[\begin{array}{cc}3 & 1 \\ -2 & 1\end{array}\right]$ and $B^{-1}=-1\left[\begin{array}{cc}-2 & -3 \\ 1 & 2\end{array}\right]$ $A B=\left[\begin{array}{ll}3 & 5 \\ 1 & 0\end{array}\right]$ and $A^{-1} B^{-1}=\frac{1}{5}\left[\begin{array}{cc}5 & 7 \\ -5 & 8\end{array}\right]$ $\Rightarrow A B\left(A^{-1} B^{-1}\right)=\frac{1}{5}\left[\begin{array}{cc}-10 & -61 \\ 5 & 7\end{array}\right] \neq 1$. $|A B|=0-5=-5$
$\therefore(A B)^{-1}=\frac{-1}{5}\left[\begin{array}{cc}
0 & -5 \\
-1 & 3
\end{array}\right] \neq A^{-1} B^{-1}$
Hence, the correct answer is option: 'D'.