Question

Let $A= \begin{pmatrix} m & n \\ p & q\end{pmatrix}, d=|A| \neq 0$ and $|A-d(\operatorname{Adj} A)|=0$. Then

• A. $1+d^2=m^2+q^2$
• B. $1+d^2=(m+q)^2$
• C. $(1+d)^2=m^2+q^2$
• D. $(1+d)^2=(m+q)^2$

Answer 1

Answer: (D)

$ A =\begin{bmatrix} m & n \\ p & q \end{bmatrix}, | A - d (\operatorname{adj} A )|=0 $
$ \Rightarrow | A - d (\operatorname{adj} A )|=\left|\begin{bmatrix} m & n \\ p & q \end{bmatrix}- d \begin{bmatrix} q & - n \\ - p & m \end{bmatrix}\right| $
$ =\begin{vmatrix} m - qd & n (1+ d ) \\ p (1+ d ) & q - md \end{vmatrix}=0 $
$\Rightarrow ( m - qd )( q - md )- np (1+ d )^2=0 $
$ \Rightarrow mq - m ^2 d - q ^2 d + mqd d ^2- np (1+ d )^2=0$
$\Rightarrow ( mq - np )+ d ^2( mq - np )- d \left( m ^2+ q ^2+2 np \right)=0 $
$ \Rightarrow d + d ^3- d \left(( m + q )^2-2 d \right)=0$
$ \Rightarrow 1+ d ^2=( m + q )^2-2 d$
$\Rightarrow (1+ d )^2=( m + q )^2$
$ \therefore Option (1) $ is correct.