Question

Let $A=\left(\begin{array}{cc}m& n\\ p& q\end{array}\right),d=|A|\ne 0$ and $|A-d(\mathrm{Adj}A)|=0$. 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)

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