Question

If $B,C$ are square matrices of same order such that $C^{2}=BC-CB$ and $B^{2}=-I,$ where $I$ is an identity matrix, then the inverse of matrix $\left(C - B\right)$ is

• A. $C$
• B. $C+B$
• C. $C-B$
• D. $I$

Answer 1

Answer: (B)

Given, $C^{2}-BC+CB=0$
$\Rightarrow C^{2}-BC+CB-B^{2}+B^{2}=0$
$\Rightarrow \left(C - B\right)\left(C + B\right)+B^{2}=0$
$\Rightarrow \left(C - B\right)\left(C + B\right)=I$ (as $B^{2}=-I$ )
Hence, $\left(C - B\right)^{- 1}=C+B$