Question

Given matrix $\left[A\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 2& 1\\ 0& 0& 4\end{array}\right]$ if $\lambda$ is selected from the values satisfying the equation $|A-{\lambda }^{2}I|=0,$ then the probability that the roots of the equation $x^2+\lambda x+1=0$ will have real roots is

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. None of these

Answer 1

The correct option is C $\frac{1}{3}$

$A-{\lambda }^{2}I=\left|\begin{array}{ccc}1& 0& 0\\ 0& 2& 1\\ 0& 0& 4\end{array}\right|-{\lambda }^2\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|$
$\Rightarrow A-{\lambda }^{2}I=\left|\begin{array}{ccc}1-{\lambda }^2& 0& 0\\ 0& 2-{\lambda }^2& 1\\ 0& 0& 4-{\lambda }^2\end{array}\right|$
$\Rightarrow |A-{\lambda }^{2}I|=(1-{\lambda }^2)(2-{\lambda }^2)(4-{\lambda }^2)=0$
$\Rightarrow \lambda =\pm 1,\pm \sqrt{2},\pm 2$ $(i.e.6$ values of $\lambda )$
$\Rightarrow {\lambda }^2+\lambda x+1=0\Rightarrow {\lambda }^2-4\ge 0$
$\Rightarrow \lambda \ge 2$ or $\lambda \le -2$
two values of $\lambda i.e.-2$ and $2$
Required probability $=\frac{2}{6}=\frac{1}{3}$