Question

If ${\lambda }_1,{\lambda }_2$ and ${\lambda }_3$ are the eigen values (i.e. characteristic values) of the matrix $\left|\begin{array}{ccc}1& 2& 15\\ 3& 4& 11\\ 5& 6& 7\end{array}\right|$ then $\left(1-{\lambda }_1\right)\left(1+{\lambda }_2\right)\left(1+{\lambda }_3\right)$ equals

• A. 0
• B. -108
• C. -95
• D. 95

Answer 1

Answer: (C)

Let $A=\left|\begin{array}{ccc}1& 2& 15\\ 3& 4& 11\\ 5& 6& 7\end{array}\right|$ then $|A-\lambda I|=0$ is characteristic equation.
$i.e.,\left|\begin{array}{ccc}1-\lambda & 2& 15\\ 3& 4-\lambda & 11\\ 5& 6& 7-\lambda \end{array}\right|=0$
Expanding along $R_1$, we get
$\Rightarrow \left(1-\lambda \right)\left[\left(4-\lambda \right)\left(7-\lambda \right)-66\right]+2\left[55-3\left(7-\lambda \right)\right]+15\left[18-5\left(4-\lambda \right)\right]=0$
$\Rightarrow \left(1-\lambda \right)\left(28-11\lambda +{\lambda }^2-66\right)+2\left(34+3\lambda \right)+15\left(-2+5\lambda \right)=0$
$\Rightarrow \left(1-\lambda \right)\left({\lambda }^2-11\lambda -38\right)+68+6\lambda -30+75\lambda =0$
$\Rightarrow {\lambda }^2-11\lambda -38-{\lambda }^2+38\lambda +38+81\lambda =0$
$\Rightarrow -{\lambda }^3+12{\lambda }^2+108\lambda =0$
$\Rightarrow -\lambda \left({\lambda }^2-12\lambda -108\right)=0$
$\Rightarrow \lambda =0or{\lambda }^2-12\lambda -108=0$
$\Rightarrow \lambda =0or\left(\lambda -18\right)\left(\lambda +6\right)=0$
$\Rightarrow \lambda =0,18,-6$
Let ${\lambda }_1=-6,{\lambda }_2=0,{\lambda }_3=18$
$\therefore \left(1+{\lambda }_1\right)\left(1+{\lambda }_2\right)\left(1+{\lambda }_3\right)=\left(-5\right)\left(1\right)\left(19\right)=-95$