Question: Show that the matrix $$A= \begin{bmatrix} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \\ \end{bmatrix}$$ is diagonalizable. Find the transforming matrix and the diagonal matrix
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Subject: Applied Mathematics 2

Topic: Matrices

Difficulty: High

m4c(14) • 739 views
 modified 17 months ago by Juilee • 2.3k written 3.0 years ago by sonipathak13 • 10
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Let $λ$ be Eigen value & X be corresponding Eigen Vector of Matrix A. $$\text {Characteristic equation is } |A-λI|=0$$

$$\therefore \begin{bmatrix} 8-λ & -6 & 2 \\ -6 & 7-λ & -4 \\ 2 & -4 & 3-λ \\ \end{bmatrix}=0$$

On solving ,we get

$$λ^3(8+7+3)λ^2+(5+20+20)λ -0=0$$

$\therefore λ^3-18λ^2+45λ-0=0$

$\therefore$ Eigen value $(λ)$ are $0,3,15$

### case 1: $λ=0$

$$\\ \therefore [A-λI]X=0$$

$$\therefore \begin{bmatrix} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \\ \end{bmatrix}.\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$

$\therefore 8x_1-6x_2+2x_3=0 \space \space \&\space \space -6x_1+7x_2-4x_3=0$

Using Cramer’s rules,

$$\dfrac {x_1}{\begin{vmatrix}-6&2\\7&-4\end{vmatrix}}=\dfrac {-x_2}{\begin{vmatrix}8&2\\-6&-4\end{vmatrix}}=\dfrac {x_3}{\begin{vmatrix}8&-6\\-6&7\end{vmatrix}}$$

$\therefore \dfrac {x_1}{10}=\dfrac {-x_2}{-20}=\dfrac {x_3}{20}\\ \therefore \dfrac {x_1}{1}=\dfrac {x_2}{2}=\dfrac {x_3}{2}$

$\therefore$ Eigen vector $x_1=[1\space\space2\space\space2]'$

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### Case 2: $λ=3$

$$\therefore [A-λI]=0$$

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

$\therefore 5x_1-6x_2+2x_3=0 \space \space \&\space \space -6x_1+4x_2-4x_3=0$

Using Cramer’s rules;

$$\dfrac {x_1}{\begin{vmatrix}-6&2\\4&-4\end{vmatrix}}=\dfrac {-x_2}{\begin{vmatrix}5&2\\-6&-4\end{vmatrix}}=\dfrac {x_3}{\begin{vmatrix}5&-6\\-6&4\end{vmatrix}}$$

$\therefore \dfrac {x_1}{16}=\dfrac {-x_2}{-8}=\dfrac {x_3}{16}\\ \therefore \dfrac {x_1}{2}=\dfrac {x_2}{1}=\dfrac {x_3}{-2}$

$\therefore$ Eigen vector $x_1=[2\space\space1\space\space -2]'$

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### Case 3: $λ=15$

$$\therefore [A-λI]=0$$

$$\therefore \begin{bmatrix} -7 & -6 & 2 \\ -6 & -8 & -4 \\ 2 & -4 & -12 \\ \end{bmatrix}.\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$

$\therefore -7x_1-6x_2+2x_3=0 \space \space \&\space \space -6x_1-8x_2-4x_3=0$

Using Gummer's rules;

$$\dfrac {x_1}{\begin{vmatrix}-6&2\\-8&-4\end{vmatrix}}=\dfrac {-x_2}{\begin{vmatrix}-7&2\\-6&-4\end{vmatrix}}=\dfrac {x_3}{\begin{vmatrix}5&-6\\-6&-8\end{vmatrix}}$$

$\therefore \dfrac {x_1}{40}=\dfrac {-x_2}{40}=\dfrac {x_3}{40}\\ \therefore \dfrac {x_1}{2}=\dfrac {x_2}{-2}=\dfrac {x_3}{1}$

$\therefore$ Eigen vector $x_1=[2\space\space-2\space\space 1]'$

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Since, all Eigen values are distinct, matrix A is diagonalized.

$$\therefore M^{-1}AM=D$$

So the given Matrix A is diagonalized to diagonal Matrix D by the transforming Matrix M;

Where $$D=\begin {bmatrix}0&0&0\\0&3&0\\0&0&15\end {bmatrix}$$ and $$M=\begin {bmatrix}1&2&2\\2&1&-2\\2&-2&1\end {bmatrix}$$