### Author Topic: section 7.3 question 24  (Read 1138 times)

#### youjianz

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• Posts: 2
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##### section 7.3 question 24
« on: October 31, 2018, 08:26:06 PM »
I have no idea how to find the root for this one

can some post the solution (including the progress work) to get the root?

thanks

#### Tianfangtong Zhang

• Full Member
• Posts: 16
• Karma: 15
##### Re: section 7.3 question 24
« Reply #1 on: November 01, 2018, 01:18:46 PM »
\begin{equation*}
A=
\begin{bmatrix}
11 & -2 & 8\\ -2 & 2 & 10 \\ 8 & 10 & 5
\end{bmatrix}
\end{equation*}
then
\begin{equation*}
(A -\lambda I)x = 0
\end{equation*}
then
\begin{equation*}
\begin{bmatrix}
11-\lambda & -2 & 8\\ -2 & 2-\lambda & 10 \\ 8 & 10 & 5-\lambda
\end{bmatrix}
\begin{bmatrix}
x_1 \\ x_2 \\x_3
\end{bmatrix} =
\begin{bmatrix}
0 \\ 0 \\ 0
\end{bmatrix}
\end{equation*}
then the eigenvalues are the roots of equation det(A-$\lambda$ I) = 0
\begin{equation*}
\left|\begin{matrix}
11-\lambda & -2 & 8\\ -2 & 2-\lambda & 10 \\ 8 & 10 & 5-\lambda
\end{equation*}
$C_2 -> 4C_2 + C_3$
\begin{equation*}
\begin{bmatrix}
11-\lambda & 4(-2)+8 & 8\\ -2 & 4(2-\lambda)+10 & 10 \\ 8 & 4(10)+(5-\lambda) & 5-\lambda
\end{bmatrix} = 0
\end{equation*}
\begin{equation*}
\begin{bmatrix}
11-\lambda &0 & 8\\ -2 & 18-4\lambda & 10 \\ 8 & 45-\lambda & 5-\lambda
\end{bmatrix} = 0
\end{equation*}
\begin{equation*}
(11-\lambda)
\begin{bmatrix}
18-4\lambda & 10 \\ 45-\lambda & 5-\lambda
\end{bmatrix} - 0
\begin{bmatrix}
-2 & 10 \\ 8 & 5-\lambda
\end{bmatrix} + 8
\begin{bmatrix}
-2 & 18-4\lambda \\ 8 & 45 -\lambda
\end{bmatrix} = 0
\end{equation*}

(11-$\lambda$)(18-4$\lambda$)(5-$\lambda$)(10)(45-$\lambda$)+8(-2)(45-$\lambda$)8(18-4$\lambda$)=0
then -4$\lambda^3$ + 8 $\lambda^2$ + 4$\lambda$ - 8 = 0
($\lambda$ -1 )($\lambda$ - 2) ($\lambda$ +1) = 0
then $\lambda$ = 1,2,-1
then the eigenvalues are $\lambda_1$ = 1, $\lambda_2$ = 2, $\lambda_3$ = -1
« Last Edit: November 01, 2018, 05:01:52 PM by Victor Ivrii »