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Let $A$ and $B$ be real matrices of the form $\left[ {\begin{array}{*{20}{c}}
\alpha &0\\
0&\beta
\end{array}} \right]$ and $\left[ {\begin{array}{*{20}{c}}
0&\gamma \\
\delta &0
\end{array}} \right]$, respectively
Statement $1$ : $AB - BA$ is always an invertible matrix
Statement $2$ : $AB -BA$ is never an identity matrix
Statement $1$ is true, Statement $2$ is false
Statement $1$ is false, Statement $2$ is true
Statement $1$ is true, Statement $2$ is true;Statement $2$ is a correct explanation of Statement $1$.
Statement $1$ 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
$A = \left[ {\begin{array}{*{20}{c}}
\alpha &0\\
0&\beta
\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}
0&\lambda \\
\delta &0
\end{array}} \right]$
Now, $AB = \left[ {\begin{array}{*{20}{c}}
0&{\alpha \gamma }\\
{\beta \delta }&0
\end{array}} \right]$
and $BA = \left[ {\begin{array}{*{20}{c}}
0&{\gamma \beta }\\
{\delta \alpha }&0
\end{array}} \right]$
Statement – $1$:
$AB – BA = \left[ {\begin{array}{*{20}{c}}
0&{\gamma \left( {\alpha – \beta } \right)}\\
{\delta \left( {\beta – \alpha } \right)}&0
\end{array}} \right]$
$\left| {AB – BA} \right| = {\left( {\alpha – \beta } \right)^2}\delta \ne 0$
$\therefore AB – BA$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.
