Subject: Applied Mathematics 2

Topic: Matrices

Difficulty: High

The Characteristic Equation is

$$\begin{bmatrix} -9-\lambda & 4 & 4 \\ -8 & 3-\lambda & 4 \\ -16 & 8 & 7-\lambda \\ \end{bmatrix}=0$$

$\therefore(-9-\lambda)[(3-\lambda)(7-\lambda)-32]-4[-8(7-\lambda)+64]+4[-64+48+-16\lambda]=0$

$\therefore(-9-\lambda)[21-3\lambda-7\lambda+\lambda^2-32]-4[-56+8\lambda+64]+4[-16-16\lambda]=0\\ \therefore -9\lambda^2+90\lambda+99-\lambda^3+10\lambda^2+11\lambda-32-32\lambda-64-64\lambda=0\\ \therefore -\lambda^3+\lambda^2+5\lambda+3=0\\ \therefore \lambda=-1,-1,3\\ (1)\ for \space \lambda=-1[A-\lambda_1 I]X=0 \space \space Given $

$$\begin{bmatrix} -8 & 4 & 4 \\ -8 & 4 & 4 \\ -16 & 8 & 8 \\ \end{bmatrix} .\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$

Expanding $R_1$

$ -8x_1 + 4x_2 + 4 x_3 = 0$

$OR$

$-2x_1 + x_2 + x_3 = 0$

Put $ x_2 = 0, x_3 = 1 $ in above we get $ x_1 = \frac{1}{2} $

$\therefore $ for $ \lambda = -1$, eigen vector

$X_1 = \begin{bmatrix} 1/2 \\ 0 \\ 1 \end{bmatrix} $ or $ \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} $

for repeated eigen value $\lambda = -1$ again,

we get $-8x_1 + 4x_2 + 4x_3 = 0$

Now put $ x_2 = 1 $ and $x_3 = 0$

we get $x = \frac{1}{2}$

i.e. $X_2 = \begin{bmatrix} 1/2 \\ 1 \\ 0 \end{bmatrix} $ or $ \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} $

$$\therefore X_1=[1,0,2] \space \space \space X_2=[1,2,0]$$

.

$(2)for\space \lambda=3[A-\lambda_2I]X=0 \space \space given$

$$\begin{bmatrix} -12 & 4 & 4 \\ -8 & 0 & 4 \\ -16 & 8 & 4 \\ \end{bmatrix} .\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$

Expanding $R_1$ and $R_2$

$ -12x_1 + 4x_2 + 4 x_3 = 0$

$-8x_1 + 0x_2 + 4x_3 = 0$

By Cramer's rule

$ \frac{x_1}{16} = \frac{-x_2}{-16} = \frac{x_3}{32}$

$OR$

$ \frac{x_1}{1} = \frac{-x_2}{1} = \frac{x_3}{2}$

$\therefore $ Corresponding to Eigen Value 3

We get Eigen vector

$$\therefore X_3=[1,1,2]$$

Although Eigen Value of A are not distinct the geometric multiplicity of each Eigen value Is equal to its algebraic multiplicity, As is diagonalizable

Since $M^{-1} AM=D$

The matrix $$A=\begin{bmatrix} -9 & 4 & 4 \\ -8 & 3 & 4 \\ -16 & 8 & 7 \\ \end{bmatrix}$$ will be digitalized to $$D=\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 3 \\ \end{bmatrix}$$ by the transforming $$M=\begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 2 & 0 & 2 \\ \end{bmatrix}$$