If B,C are square matrices of same order such that C2=BC-CB and B2=-I, where I is an identity matrix, then the inverse of matrix (C - B) is

Question

If B,C are square matrices of same order such that C2=BC-CB and B2=-I, where I is an identity matrix, then the inverse of matrix (C - B) is (A) C (B) C+B (C) C-B (D) I

Tardigrade Admin · Accepted Answer

Given, C2-BC+CB=0 ⇒ C2-BC+CB-B2+B2=0 ⇒ (C - B)(C + B)+B2=0 ⇒ (C - B)(C + B)=I (as B2=-I ) Hence, (C - B)- 1=C+B

Q. 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 $(C - B)$ is

$C$

$C + B$

$C - B$

$I$

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

Q. If B,C are square matrices of same order such that C2=BC−CB and B2=−I, where I is an identity matrix, then the inverse of matrix (C−B) is

C

C+B

C−B

I

Given, C2−BC+CB=0 ⇒C2−BC+CB−B2+B2=0 ⇒(C−B)(C+B)+B2=0 ⇒(C−B)(C+B)=I (as B2=−I ) Hence, (C−B)−1=C+B

Q. 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 $(C - B)$ is

$C$

$C + B$

$C - B$

$I$

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