Find the eigen value and the eigen vectors of the following matrix:

$A=\begin{bmatrix} 4 & 6 & 6 \\ 1 & 3 & 2 \\ -1 & -4 & -3 \end{bmatrix}$

Solution:

$A=\begin{bmatrix} 4 & 6 & 6 \\ 1 & 3 & 2 \\ -1 & -4 & -3 \end{bmatrix}$

$|A|=-4$

Characteristic equation is given by: $|A-\lambda I|=0$

$\begin{vmatrix} 4-\lambda & 6 & 6 \\ 1 & 3-\lambda & 2 \\ -1 & -4 & -3-\lambda \end{vmatrix}=0$

$\lambda^{3}-(4) \lambda^{2} + (12-9-12-6+6+8)\lambda-(-4)=0$

$\lambda^{3}-4\lambda^{2}-\lambda+4=0$

$\therefore \ \lambda=4, \ 1, \ -1$

$\therefore \ $Eigen values are = 4, 1, -1

Now to find the eigen vectors,

$[A-\lambda I]X=0$

$\begin{bmatrix} 4-\lambda & 6 & 6 \\ 1 & 3-\lambda & 2 \\ -1 & -4 & -3-\lambda \end{bmatrix} \begin{bmatrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ ------------(1)

Put $\lambda=4$ in equation (1),

$\begin{bmatrix} 0 & 6 & 6 \\ 1 & -1 & 2 \\ -1 & -4 & -7 \end{bmatrix} \begin{bmatrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

$0x_{1}+6x_{2}+6x_{3}=0$

$x_{1}-x_{2}+2x_{3}=0$

$-x_{1}-4x_{2}-7x_{3}=0$

By Cramer's rule,

$\cfrac{x_{1}}{\begin{vmatrix} 6 & 6 \\ -1 & 2 \end{vmatrix}}=\cfrac{-x_{2}}{\begin{vmatrix} 0 & 6 \\ 1 & 2 \end{vmatrix}}=\cfrac{x_{3}}{\begin{vmatrix} 0 & 6 \\ 1 & -1 \end{vmatrix}}$

$\cfrac{x_{1}}{12-(-6)}=\cfrac{-x_{2}}{0-6}=\cfrac{x_{3}}{0-6}$

$\cfrac{x_{1}}{18}=\cfrac{-x_{2}}{-6}=\cfrac{x_{3}}{-6}$

$\therefore X_{1}= \begin{bmatrix} 18 \\ 6 \\ -6 \end{bmatrix}= \begin{bmatrix} 3 \\ 1 \\ -1 \end{bmatrix}$ --------------(2)

Put $\lambda=1$ in equation (1),

$\begin{bmatrix} 3 & 6 & 6 \\ 1 & 2 & 2 \\ -1 & -4 & -4 \end{bmatrix} \begin{bmatrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

$3x_{1}+6x_{2}+6x_{3}=0$

$x_{1}+2x_{2}+2x_{3}=0$

$-x_{1}-4x_{2}-4x_{3}=0$

By Cramer's rule,

$\cfrac{x_{1}}{\begin{vmatrix} 2 & 2 \\ -4 & -4 \end{vmatrix}}=\cfrac{-x_{2}}{\begin{vmatrix} 1 & 2 \\ -1 & -4 \end{vmatrix}}=\cfrac{x_{3}}{\begin{vmatrix} 1 & 2 \\ -1 & -4 \end{vmatrix}}$

$\cfrac{x_{1}}{-8-(-6)}=\cfrac{-x_{2}}{-4-(-2)}=\cfrac{x_{3}}{-4-(-2)}$

$\cfrac{x_{1}}{0}=\cfrac{-x_{2}}{-2}=\cfrac{x_{3}}{-2}$

$\therefore X_{2}= \begin{bmatrix} 0 \\ 2 \\ -2 \end{bmatrix}= \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}$ --------------(3)

Put $\lambda=-1$ in equation (1),

$\begin{bmatrix} 5 & 6 & 6 \\ 1 & 4 & 2 \\ -1 & -4 & -2 \end{bmatrix} \begin{bmatrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

$5x_{1}+6x_{2}+6x_{3}=0$

$x_{1}+4x_{2}+2x_{3}=0$

$-x_{1}-4x_{2}-2x_{3}=0$

By Cramer's rule,

$\cfrac{x_{1}}{\begin{vmatrix} 6 & 6 \\ 4 & 2 \end{vmatrix}}=\cfrac{-x_{2}}{\begin{vmatrix} 5 & 6 \\ 1 & 2 \end{vmatrix}}=\cfrac{x_{3}}{\begin{vmatrix} 5 & 6 \\ 1 & 4 \end{vmatrix}}$

$\cfrac{x_{1}}{12-24}=\cfrac{-x_{2}}{10-6}=\cfrac{x_{3}}{20-6}$

$\cfrac{x_{1}}{-12}=\cfrac{-x_{2}}{4}=\cfrac{x_{3}}{14}$

$\therefore X_{3}= \begin{bmatrix} -12 \\ -4 \\ 14 \end{bmatrix}= \begin{bmatrix} -6 \\ -2 \\ 7 \end{bmatrix}$ --------------(4)

Eigen values are:

$X_{1}= \begin{bmatrix} 3 \\ 1 \\ -1 \end{bmatrix}, \ X_{2}= \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}, \ X_{3}= \begin{bmatrix} -6 \\ -2 \\ 7 \end{bmatrix}$