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Let $\mathrm{A}=\left(\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ c & \mathrm{~d}\end{array}\right)$ be a $2 \times 2$ real matrix with $\operatorname{det} \mathrm{A}=1$. If the equation $\operatorname{det}\left(\mathrm{A}-\lambda \mathrm{I}{2}\right)=0$ has imaginary roots $\left(\mathrm{I}{2}\right.$ be the
Identity matrix of order 2), then
Options:
Identity matrix of order 2), then
Solution:
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Verified Answer
The correct answer is:
$(a+d)^{2} < 4$
Hint:
$|\mathrm{A}|=0 \therefore \mathrm{ad}-\mathrm{bc}=1$
$\left|\mathrm{~A}-\lambda \mathrm{I}_{2}\right|=0$
$\left|\begin{array}{cc}\mathrm{a}-\lambda & \mathrm{b} \\ \mathrm{c} & \mathrm{d}-\lambda\end{array}\right|=0$
$\therefore \mathrm{ad}-(\mathrm{a}+\mathrm{d}) \lambda+\lambda^{2}-\mathrm{bc}=0$
$\lambda^{2}-(\mathrm{a}+\mathrm{d}) \lambda+1=0$
$\therefore(\mathrm{a}+\mathrm{d})^{2} < 4$
$|\mathrm{A}|=0 \therefore \mathrm{ad}-\mathrm{bc}=1$
$\left|\mathrm{~A}-\lambda \mathrm{I}_{2}\right|=0$
$\left|\begin{array}{cc}\mathrm{a}-\lambda & \mathrm{b} \\ \mathrm{c} & \mathrm{d}-\lambda\end{array}\right|=0$
$\therefore \mathrm{ad}-(\mathrm{a}+\mathrm{d}) \lambda+\lambda^{2}-\mathrm{bc}=0$
$\lambda^{2}-(\mathrm{a}+\mathrm{d}) \lambda+1=0$
$\therefore(\mathrm{a}+\mathrm{d})^{2} < 4$
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