Mumbai University > Electronics and Telecommunication > Sem 4 > Applied Maths 4

Marks: 6M

Year: May 2014

Given A = $\begin{bmatrix} -2 & 5 & 4 \\ 5 & 7 & 5 \\ 4 & 5 & -2 \end{bmatrix}$

The characteristic equation of matrix A is

$$|A - \lambda I| = 0 \\ (-1)^3 \lambda^3 + (-1)^2 s_{1} \lambda^2 + (-1) s_{2} \lambda + |A| = 0.......(1)$$

Where $$s_{1} = trace(A) = (-2 + 7 - 2) = 3 \\ \therefore s_{1} = 3 \\ s_{2} = \begin{vmatrix} 7 & 5 \\ 5 & -2 \end{vmatrix} + \begin{vmatrix} -2 & 4 \\ 4 & -2 \end{vmatrix} + \begin{vmatrix} -2 & 5 \\ 5 & 7 \end{vmatrix} \\ = -39 - 12 - 39 = -90 \\ \therefore s_{2} = -90 \\ |A| = 78 + 150 - 12 \\ \therefore |A| = 216$$

Equation (1) becomes,

$$-\lambda^3 + 3\lambda^2 + 90\lambda + 216 = 0$$

Hence eigen values of A are $$\lambda = 12, -3, -6$$

To find eigen vectors corresponding to these eigen values, we use the system of homogeneous equations $(A - \lambda I)X = 0$

Using Cramer's rule

When $\lambda = 12$, corresponding systen of homogeneous equations is

$$\begin{bmatrix} -14 & 5 & 4 \\ 5 & -5 & 5 \\ 4 & 5 & -14 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

The linear equations are:

$$-14x + 5y + 4z = 0 \\ 5x - 5y + 5z = 0 \\ 4x - 5y - 14z = 0$$

Considering any two linear equations and solving them by Cramer's rule, we get

$$-14x + 5y + 4z = 0 \\ 5x - 5y + 5z = 0$$

$$\frac{x}{\begin{vmatrix} 5 & 4 \\ -5 & 5 \end{vmatrix}} = \frac{-y}{\begin{vmatrix} -14 & 4 \\ 5 & 5 \end{vmatrix}} = \frac{z}{\begin{vmatrix} -14 & 5 \\ 5 & -5 \end{vmatrix}} \\ \frac{x}{45} = -\frac{y}{-90} = \frac{z}{45} \\ \frac{x}{1} = \frac{y}{2} = \frac{z}{1}$$

Hence, the eigen vector corresponding to the eigen vlaue $\lambda = 12$ is

$$X_{1} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$$

When $\lambda = -3$, corresponding systen of homogeneous equations is

$$\begin{bmatrix} 1 & 5 & 4 \\ 5 & 10 & 5 \\ 4 & 5 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

$$x + 5y + 4z = 0 \\ 5x + 10y + 5z = 0 \\ 4x + 5y + z = 0$$

Considering any two linear equations and solving them by Cramer's rule, we get

$$x + 5y + 4z = 0 \\ 5x + 10y + 5z = 0$$

$$\frac{x}{\begin{vmatrix} 5 & 4 \\ 10 & 5 \end{vmatrix}} = \frac{-y}{\begin{vmatrix} 1 & 4 \\ 5 & 5 \end{vmatrix}} = \frac{z}{\begin{vmatrix} 1 & 5 \\ 5 & 10 \end{vmatrix}} \\ \frac{x}{-15} = -\frac{y}{-15} = \frac{z}{-15} \\ \frac{x}{-1} = \frac{y}{1} = \frac{z}{-1}$$

Hence, the eigen vector corresponding to the eigen value $\lambda$ = -3 is

$$X_{2} = \begin{bmatrix} -1 \\ 1 \\ -1 \end{bmatrix}$$

When $\lambda$ = -6, corresponding system of homogeneous equations is

$$\begin{bmatrix} 4 & 5 & 4 \\ 5 & 13 & 5 \\ 4 & 5 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

$$4x + 5y + 4z = 0 \\ 5x + 13y + 5z = 0 \\ 4x + 5y + 4z = 0$$

Considering any two linear equations and solving them by Cramer's rule, we get

$$5x + 13y + 5z = 0 \\ 4x + 5y + 4z = 0$$

$$\frac{x}{\begin{vmatrix} 13 & 5 \\ 5 & 4 \end{vmatrix}} = \frac{-y}{\begin{vmatrix} 5 & 5 \\ 4 & 4 \end{vmatrix}} = \frac{z}{\begin{vmatrix} 5 & 13 \\ 4 & 5 \end{vmatrix}} \\ \frac{x}{27} = -\frac{y}{0} = \frac{z}{-1} \\ \frac{x}{1} = \frac{y}{0} = \frac{z}{-1}$$

Hence, the eigen vector corresponding to the eigen value $\lambda$ = -6 is

$$X_{3} = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$$

Hence, the eigen values are $\lambda$ = 12, -3, -6 and the corresponding eigen vectors are $$X_{1} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} \hspace{1cm} X_{2} = \begin{bmatrix} -1 \\ 1 \\ -1 \end{bmatrix} \hspace{1cm} X_{3} = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$$