Question

If $A$ is a $n\times n$ non-singular matrix such that $3AB{A}^{-1}+A=2A^{-1}BA,$ then which of the following options is (are) CORRECT?

• A. $|A+B|=1$
• B. $|A+B|=0$
• C. $|AB{A}^{-1}-A^{-1}BA|=0$
• D. $|AB{A}^{-1}-A^{-1}BA|=1$

Answer 1

Answer: (B), (C)

The correct options are
B $|A+B|=0$
C $|AB{A}^{-1}-A^{-1}BA|=0$

$3AB{A}^{-1}+A=2A^{-1}BA$
Adding $2A$ both sides,
$3AB{A}^{-1}+3A=2A^{-1}BA+2A$
$\Rightarrow 3A(B{A}^{-1}+I)=(A^{-1}B+I)2A$
$\Rightarrow 3A(B{A}^{-1}+A{A}^{-1})=(A^{-1}B+A^{-1}A)2A$
$\Rightarrow 3A(B+A)A^{-1}=A^{-1}(B+A)2A$
Taking determinant both sides,
$|3A(B+A)A^{-1}|=|A^{-1}(B+A)2A|$
$\Rightarrow 3^n|A|\cdot |B+A|\cdot |A^{-1}|=|A^{-1}|\cdot |B+A|\cdot 2^n|A|$
$\Rightarrow 3^n|B+A|=2^n|B+A|$
$\Rightarrow |B+A|=0$


$3AB{A}^{-1}+A=2A^{-1}BA$
$\Rightarrow 3AB{A}^{-1}-3A^{-1}BA=-A-A^{-1}BA$
$\Rightarrow 3(AB{A}^{-1}-A^{-1}BA)=-(A+A^{-1}BA)$
$\Rightarrow 3(AB{A}^{-1}-A^{-1}BA)=-(I+A^{-1}B)A$
$\Rightarrow 3(AB{A}^{-1}-A^{-1}BA)=-A^{-1}(A+B)A$
Taking determinant both sides,
$3^n|AB{A}^{-1}-A^{-1}BA|=(-1{)}^n|A^{-1}||A+B||A|$
So, $|AB{A}^{-1}-A^{-1}BA|=0$