Question

Let $A$ and $B$ be two square matrices satisfying $AB-BA=A$. Then which of the following is (are) TRUE?

• A. If $A$ is non-singular, then $B=O$, where $O$ is a null matrix.
• B. If $A$ is non-singular, then $\text{det}(B-I)=\text{det}(B+I)$, where $I$ is an identity matrix.
• C. If $A$ and $B$ are symmetric matrices of order $3$, then $A$ is singular.
• D. If $A$ and $B$ are skew-symmetric matrices of order $3$, then $A$ is non-singular.

Answer 1

Answer: (B), (C)

The correct options are
B If $A$ is non-singular, then $\text{det}(B-I)=\text{det}(B+I)$, where $I$ is an identity matrix.
C If $A$ and $B$ are symmetric matrices of order $3$, then $A$ is singular.

If $A$ is non-singular, then $|A|\ne 0$
Given, $AB-BA=A$
$\Rightarrow |AB-BA|=|A|\cdots \left(1\right)$
Then $B$ can't be a null matrix as if $B$ is a null matrix, then LHS of eqn$\left(1\right)$ is zero which contradicts $|A|\ne 0$.

$AB-BA=A$
$\Rightarrow AB-A=BA$
$\Rightarrow A(B-I)=BA$
$\Rightarrow |A||B-I|=|B||A|$
$\Rightarrow |B-I|=|B|\cdots \left(2\right)(\because |A|\ne 0)$

Also, $AB=BA+A$
$\Rightarrow AB=(B+I)A$
$\Rightarrow |A||B|=|B+I||A|$
$\Rightarrow |B|=|B+I|\cdots \left(3\right)$

From $\left(2\right)$ and $\left(3\right)$,
$|B-I|=|B+I|$

If $A$ and $B$ are symmetric matrices of order $3$, then $A^T=A$ and $B^T=B$
Now, $(AB-BA{)}^T$
$=(AB{)}^T-(BA{)}^T$
$=B^T{A}^T-A^T{B}^T$
$=BA-AB$
$=-(AB-BA)$
Hence, $AB-BA$ is a skew-symmetric matrix of order $3$ (odd).
$\therefore |AB-BA|=0$
$\Rightarrow |A|=0\left[\text{From (1)}\right]$
$\Rightarrow A$ is singular.

If $A$ and $B$ are skew-symmetric matrices of order $3$, then $A^T=-A$ and $B^T=-B$
Again, $(AB-BA{)}^T=-(AB-BA)$
Hence, $AB-BA$ is a skew-symmetric matrix of order $3$ (odd).
$\therefore |AB-BA|=0$
$\Rightarrow |A|=0$
$\Rightarrow A$ is singular.