Subject: Applied Mathematics 2

Topic: Matrices

Difficulty: Medium

The Characteristic Equation is

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

$\therefore (4-\lambda)[(3-\lambda)(-2-\lambda)+10]-6[-2-\lambda+2]+6[-5+3-\lambda]=0$

$\therefore (4-\lambda)(-6-3\lambda+2\lambda+\lambda^2+10)+6\lambda-12-6\lambda=0\\ \therefore (4-\lambda)(\lambda^2-\lambda+4)-12=0\\ \therefore \lambda^3-5\lambda^2+8\lambda-4=0\\ \therefore \text { On simplifying we get } \\ \lambda=1,2,2\\ \therefore \text { the eigen values are 1 & 2} \\ (1) for \space \lambda=1[A-\lambda,I]X=0\space \space given$

$$\begin{bmatrix} 3 & 6 & 6 \\ 1 & 2 & 2 \\ -1 & -5 & -3 \\ \end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end {bmatrix}\begin{bmatrix}0\\0\\0\end {bmatrix}$$

Expanding $R_1$ and $R_3$

$\therefore x_1 + 2x_2 + 2x_3 = 0$

$\therefore -x_1 - 5x_2 - 3x_3 = 0$

By cramer's rule

$\frac{x_1} {\begin{vmatrix} 2 & 2 \\ -5 & -3 \\ \end{vmatrix}} = \frac{-x_2}{\begin{vmatrix} 1 & 2 \\ -1 & -3 \\ \end{vmatrix}} = \frac{-x_2}{\begin{vmatrix} 1 & 2 \\ -1 & -5 \\ \end{vmatrix}} $

$\frac{x_1} {4} = \frac{-x_2}{-1} = \frac{-x_2}{-3} $

$\therefore $ For Eigen value 1 Eigen vector is $\text {[ 4, 1, -3]'}$

.

$(2) for \space \lambda=2[A-\lambda,I]X=0 \space \space given$

$$\begin{bmatrix} 2 & 6 & 6 \\ 1 & 1 & 2 \\ -1 & -5 & -4 \\\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end {bmatrix}\begin{bmatrix}0\\0\\0\end {bmatrix}$$

Expanding $R_1$ and $R_2$

$x_1 + 3x_2 + 3x_3 = 0$

$x_1 + x_2 + 2x_3 = 0$

By Cramer's rule

$\frac{x_1} {\begin{vmatrix} 3 & 3 \\ 1 & 2 \\ \end{vmatrix}} = \frac{-x_2}{\begin{vmatrix} 1 & 3 \\ 1 & 2 \\ \end{vmatrix}} = \frac{-x_2}{\begin{vmatrix} 1 & 3 \\ 1 & 1 \\ \end{vmatrix}} $

$\frac{x_1} {3} = \frac{-x_2}{-1} = \frac{-x_2}{-2} $

For Eigen Value 2 we get Eigen Vector as $\text { [3, 1,-2]'}$