Question

If $\lambda_1 , \lambda_2$ and $\lambda_3$ are the eigen values (i.e. characteristic values) of the matrix $\begin{vmatrix}1&2&15\\ 3&4&11\\ 5&6&7\end{vmatrix}$ 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 = \begin{vmatrix}1&2&15\\ 3&4&11\\ 5&6&7\end{vmatrix} $ then $|A - \lambda I| = 0$ is characteristic equation.
$i.e., \ \ \begin{vmatrix}1 -\lambda &2&15\\ 3&4-\lambda &11\\ 5&6&7-\lambda \end{vmatrix} =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 = 0 or \lambda^{2} - 12\lambda - 108 = 0 $
$\Rightarrow \lambda = 0 or \left(\lambda -18\right)\left(\lambda +6\right)= 0$
$ \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 $