Question

Let A, B are two non-singular matrices of order 3 with real entries such that $\text{adj}\left(A\right)=3B$ and $\text{adj}\left(B\right)=2A$ then -

• A. $\text{adj}\left(A^{2}B\right)+\text{adj}\left(A{B}^2\right)=36(2A+3B)$
• B. $B=6A^{-1}$
• C. $AB=12I$
• D. $\text{adj}\left(A{B}^{-1}\right)=9A^{-2}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $\text{adj}\left(A^{2}B\right)+\text{adj}\left(A{B}^2\right)=36(2A+3B)$
B $B=6A^{-1}$
D $\text{adj}\left(A{B}^{-1}\right)=9A^{-2}$

$\begin{array}{cc}|adjA|=|3B|& |adjB|=|2A|\\ |A{|}^2=27|B|& |B{|}^2=8|A|\end{array}$

$|A{|}^4=27\times 27\times |B{|}^2=27\times 27\times 8|A|\Rightarrow |A{|}^3=3^6\times 2^3\Rightarrow |A|=18$
$|B|=12$
$AB=\frac{adj\left(B\right).B}{2}=\frac{|B|.I}{2}=6I$
$\text{adj}\left(A^{2}B\right)+\text{adj}\left(A{B}^2\right)=\text{adj}\left(6A\right)+\text{adj}\left(6B\right)$
$=36(\text{adj}A+\text{adj}B)$
$=36(3B+2A)$
$\text{adj}\left(A{B}^{-1}\right)=\text{adj}\left(6B^{-2}\right)=36\times \left(\text{adj}\right(B){)}^{-2}$
$=36\times (2A{)}^{-2}$
$=36\times \frac{1}{4}\times A^{-2}=9A^{-2}$