Question

If $A$ and $B$ are non-singular matrix of order $3$ and $(I-AB)$ is invertible, then which of the following statements is/are correct:

• A. $I-BA$ is not invertible
• B. $I-BA$ is invertible
• C. inverse of $(I-BA)$ is $I+B(I-AB{)}^{-1}A$
• D. inverse of $(I-BA)$ is $I+A(I-BA{)}^{-1}B$

Answer 1

Answer: (B), (C)

inverse of $(I-BA)$ is $I+B(I-AB{)}^{-1}A$

$I-BA=BI{B}^{-1}-BAB{B}^{-1}$ (multiplying by unit matrix $i.e.B{B}^{-1}$)
$\Rightarrow I-BA=B(I-AB)B^{-1}\cdots \left(i\right)$
$\Rightarrow |I-BA|=|B||I-AB||B^{-1}|$
$\Rightarrow |I-BA|=|I-AB|$
Given: $I-AB$ is invertible $\Rightarrow |I-AB|\ne 0\Rightarrow |I-BA|\ne 0$
$\therefore I-BA$ is also invertible.

Now consider, $X=(I-BA)[I+B(I-AB{)}^{-1}A]$
$\Rightarrow X=I-BA+(I-BA)B(I-AB{)}^{-1}A$
Using equation $\left(i\right)$, we have:
$X=I-BA+B(I-AB)B^{-1}B(I-AB{)}^{-1}A$
$\Rightarrow X=I-BA+BA=I$
$\therefore X=(I-BA)[I+B(I-AB{)}^{-1}A]=I$
$\Rightarrow (I-BA{)}^{-1}=I+B(I-AB{)}^{-1}A$