Question

If $A=\left[\begin{array}{ccc}a& b& c\\ x& y& z\\ p& q& r\end{array}\right],B=\left[\begin{array}{ccc}q& -b& y\\ -p& a& -x\\ r& -c& z\end{array}\right]$ and if $A$ is invertible, then which of the following is/are true:

• A. $|adjA|=|adjB|$
• B. $|A|=|B|$
• C. $|A|=-|B|$
• D. $B$ is invertible iff $A$ is invertible

Answer 1

Answer: (A), (C), (D)

$B$ is invertible iff $A$ is invertible

$|B|=\left|\begin{array}{ccc}q& -b& y\\ -p& a& -x\\ r& -c& z\end{array}\right|$
Multiplying $R_2\text{by}-1$
$|B|=-\left|\begin{array}{ccc}q& -b& y\\ p& -a& x\\ r& -c& z\end{array}\right|$
Multiplying $C_2\text{by}-1$
$|B|=\left|\begin{array}{ccc}q& b& y\\ p& a& x\\ r& c& z\end{array}\right|$
Changing $R_1\text{with}{R}_2$
$|B|=-\left|\begin{array}{ccc}p& a& x\\ q& b& y\\ r& c& z\end{array}\right|$
Changing $C_1\text{with}{C}_2$ followed by $C_2\text{with}{C}_3$
$|B|=-\left|\begin{array}{ccc}a& x& p\\ b& y& q\\ c& z& r\end{array}\right|$

Hence, $|A^T|=|A|=-|B|$. Clearly when $|A|\ne 0$ $\Rightarrow |B|\ne 0.$
$\therefore B$ is also invertible if $A$ is invertible.
Also, $|adjB|=|B{|}^2=(-|A|{)}^2=|A{|}^2$