Question

The maximum value of 'a' such that the matrix $\left[\begin{array}{ccc}-3& 0& -2\\ 1& -1& 0\\ 0& a& -2\end{array}\right]$ has three linearly independent real eigen vectors is

• A. $\frac{2}{3\sqrt{3}}$
• B. $\frac{1}{3\sqrt{3}}$
• C. $\frac{1+2\sqrt{3}}{3\sqrt{3}}$
• D. $\frac{1+\sqrt{3}}{3\sqrt{3}}$

Answer 1

The correct option is B $\frac{1}{3\sqrt{3}}$

The characteristic equation is $|A-\lambda I|=0$

$\Rightarrow \left[\begin{array}{ccc}-3-\lambda & 0& -2\\ 1& -1-\lambda & 0\\ 0& a& -2-\lambda \end{array}\right]=0$
$\Rightarrow (-3-\lambda )(-1-\lambda )(-2-\lambda )=0$
$\Rightarrow a=\frac{-1}{2}(3+\lambda )(1+\lambda )(2+\lambda )$ .....(i)
Now using maxima-minima concept for 'a'

$\frac{da}{d\lambda }=0$
$\Rightarrow \frac{-1}{2}(3{\lambda }^2+12\lambda +11)=0$
$\Rightarrow$ $\lambda =-2\pm \frac{1}{\sqrt{3}}$
and $\frac{d{a}^2}{d{\lambda }^2}=(-6\lambda -12)$
At $\lambda =(-2+\frac{1}{\sqrt{3}})$:
$\frac{d{a}^2}{d{\lambda }^2}=\frac{-6}{\sqrt{3}}<0$

At $\lambda =(-2-\frac{1}{\sqrt{3}})$:
$\frac{d{a}^2}{d{\lambda }^2}=\frac{6}{\sqrt{3}}>0$

$\therefore$ We get maximum value at $\lambda =(-2+\frac{1}{\sqrt{3}})$
So $\lambda =(-2+\frac{1}{\sqrt{3}})$ is the point maxima and max value of 'a' $=[-\frac{1}{2}(3+\lambda \left)\right(1+\lambda \left)\right(2+\lambda ){]}_{\lambda =-2+\frac{1}{\sqrt{3}}}$ $=\frac{1}{3\sqrt{3}}$