Question

Let A and B be real matrices of the form $\begin{bmatrix}\alpha&0\\\ 0&\beta\end{bmatrix}$ and $\begin{bmatrix}0&\gamma\\\ \delta&0\end{bmatrix}$, respectively. AB - BA is always an invertible matrix. AB-BA is never an identity matrix.

• A. Statement 1 is true. Statement 2 is false
• B. Statement 1 is folse. Statement 2 is true
• C. Statement 1 is true. Statement 2 is true; Statement 2 is a correct explanation of Statement 1
• D. Statement 1 is true. Statement 2 is true, Statement 2 is not a correct explanation of Statement 1

Answer 1

Answer: (A)

Statement 1 is true. Statement 2 is false

Answer 2

Let A and B be real matrices such that $A =\begin{bmatrix}\alpha&0\\\ 0&\beta\end{bmatrix}$ and $B = \begin{bmatrix}0&\gamma\\\ \delta&0\end{bmatrix}$ Now, $AB = \begin{bmatrix}0&\alpha\gamma \\\ \beta\delta &0\end{bmatrix}$ and $BA =\begin{bmatrix}0&\gamma \beta\\\ \delta\alpha &0\end{bmatrix}$ $AB-BA = \begin{bmatrix}0&\gamma \left(\alpha-\beta\right) \\\ \delta \left(\beta-\alpha\right) &0\end{bmatrix}$ $\left|AB-BA\right| = \left(\alpha-\beta^{2}\right)\delta \ne 0$ $\therefore AB - BA$ is always an invertible matrix. Hence, statement -1 is true. But $AB - BA$ can be identity matrix if $\gamma=- \delta$ or $\delta = -\gamma$ So, statement - 2 is false.