Question

Let $A$ and $B$ be two square matrices of order $n$ with real entries and $ABA-BAB=I$, Also $A^{2}B+B^{2}A=0$ (where $I$ is the identity matrix), then

• A. $A$ and $B$ are invertible
• B. $A$ and $B$ can not be invertible
• C. either one of $A$ or $B$ is invertible
• D. $|A{|}^2|B|=\frac{1}{2^n}$

Answer 1

Answer: (A), (D)

The correct options are
A $A$ and $B$ are invertible
D $|A{|}^2|B|=\frac{1}{2^n}$

Now $(A-iB)(BA-iAB)=ABA-BAB-i{B}^{2}A-i{A}^{2}B=I$
$\Rightarrow A-iB\text{and}BA-iAB$ are inverse of each other
$\Rightarrow (BA-iAB)(A-iB)=I$
$\Rightarrow B{A}^2-iBAB-iABA-A{B}^2=I$
$\Rightarrow B{A}^2-A{B}^2=I,BAB+ABA=0$
Given that $ABA-BAB=I$
$\Rightarrow$ 2ABA = I
$\Rightarrow ABA=\frac{1}{2}I$
$\Rightarrow |A{|}^2|B|=\frac{1}{2^n}$
$\Rightarrow |A{|}^2|B|=\frac{1}{2^n}$