Q33 / 90 · JEE Main 2012 (12 May Online) · Mathematics
Let $A$ and $B$ be real matrices of the form $\left[\begin{array}{ll}\alpha & 0 \\ 0 & \beta\end{array}\right]$ and $\left[\begin{array}{ll}0 & \gamma \\ \delta & 0\end{array}\right]$, respectively.
Statement 1: $A B-B A$ is always an invertible matrix.
Statement $2: A B-B A$ is never an identity matrix.
- Correct answerAStatement 1 is true, Statement 2 is false.
- BStatement 1 is false, Statement 2 is true.
- CStatement 1 is true, Statement 2 is true; Statement 2 is a correct explanation of Statement 1 .
- DStatement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.
Solution
Let $A$ and $B$ be real matrices such that $A=\left[\begin{array}{ll}\alpha & 0 \\ 0 & \beta\end{array}\right]$ and $B=\left[\begin{array}{ll}0 & \gamma \\ \delta & 0\end{array}\right]$ Now, $A B=\left[\begin{array}{cc}0 & \alpha \gamma \\ \beta \delta & 0\end{array}\right]$ and $B A=\left[\begin{array}{cc}0 & \gamma \beta \\ \delta \alpha & 0\end{array}\right]$ Statement-1:
$$
\begin{aligned}
& \left.A B-B A=\left[\begin{array}{cc}
0 & \gamma(\alpha-\beta \\
\delta(\beta-\alpha) & 0
\end{array}\right]\right) \\
& |A B-B A|=\left(\alpha-\beta^2\right\rangle \delta \neq 0
\end{aligned}
$$
$\therefore A B-B A$ is always an invertible matrix. Hence, statement $-1$ is true. But $A B-B A$ can be identity matrix if $\gamma=-\delta$ or $\delta=-\gamma$ So, statement $-2$ is false.
Hence, the correct answer is option: 'A'.