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

Marks: 6M

Year: May 2015

The characteristic Equation:

$\begin{bmatrix} 2 - \lambda & 2 & 1 \\ 1 & 3 - \lambda & 1 \\ 1 & 1 & 2 - \lambda \end{bmatrix} = 0$

Simplifying we get

$\lambda^3 - 7\lambda^2 + 11\lambda - 5 = 0\\ (\lambda - 1)( \lambda - 1) (\lambda + 5) = 0 \\ \therefore \lambda = 1, 1, 5$

1) Now for $\lambda = 1 \hspace{0.25cm} (A - \lambda I)X = 0 \\ \begin{bmatrix} 1 & 2 & 1 \\ 1 & 2 & 1 \\ 1 & 2 & 1 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ \therefore R_{2} - R_{1}, R_{3} - R_{1} \\ \begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ \therefore X_{1} + X_{2} + X_{3} = 0 $

Rank of matrix = 1

And No. of variables = 3

$\therefore$ 3 - 2 = 1 linearly independent solution.

Putting $X_{3} = 0, X_{2} = -1$

We get $X_{1} = 2$

Putting $X_{2} = 0, X_{3} = -1$

We get $X_{1} = 1$

Hence corresponding to repeated eigen value $\lambda$ = 1

We get

$X_{1} = [2, -1, 0] \\ X_{2} = [1, 0 , -1]$

2) For $\lambda = 5 \hspace{0.25cm} [A - \lambda_{2} I]X = 0$

Given

$\begin{bmatrix} -3 & 2 & 1 \\ 1 & -2 & 1 \\ 1 & 2 & -3 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ R_{1 \lt - \gt 3} \begin{bmatrix} 1 & 2 & -3 \\ 1 & -2 & 1 \\ -3 & 2 & 1 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ R_{2} - R_{1} \begin{bmatrix} 1 & 2 & -3 \\ 0 & -4 & 4 \\ -3 & 2 & 1 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ R_{3} + 3R_{1} \begin{bmatrix} 1 & 2 & -3 \\ 0 & -4 & 4 \\ -3 & 2 & 1 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ R_{3} + 2R_{2} \begin{bmatrix} 1 & 2 & -3 \\ 0 & -4 & -4 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0\\ 0 \end{bmatrix} \\ \therefore X_{1} + 2X_{2} - 3X_{3} = 0, -4X_{2} + 4X_{3} = 0$

Putting $X_{3} = 1$ we get $X_{2} = 1$and then $X_{1} = 1$ Hence for Eigen value 5 Eigen vector is X[1, 1, 1]