Question

Let $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ be a $2\times 2$ real matrix with $detA=1$. If the equation det $(A-\lambda I_2)=0$ has imaginary roots $(I_2$ be the Identify matrix of order $2)$, then

• A. $(a+d{)}^2<4$
• B. $(a+d{)}^2=4$
• C. $(a+d{)}^2>4$
• D. $(a+d{)}^2=16$

Answer 1

The correct option is A $(a+d{)}^2<4$

$detA=1ad-cb=1$
Now
$A-\lambda I_2=\left(\begin{array}{cc}a-\lambda & b\\ c& d-\lambda \end{array}\right)\Rightarrow det(A-\lambda I_2)=0{\lambda }^2-\lambda (a+d)+ad-cb=0$
$\because$ roots are imaginary
$\therefore D<0(a+d{)}^2-4(ad-cb)<0(a+d{)}^2<4$