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QUESTION

Find the eigenvalues of the matrix

$$\left(\begin{array}{ccc}2&-1&2\\0&3&3\\0&2&4\end{array}\right),$$

and determine the corresponding eigenvectors.

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ANSWER

$$\left(\begin{array}{ccc}2&-1&2\\0&3&3\\0&2&4\end{array}\right)$$
Eigenvalues satisfy $|A-\lambda I|=0$
\begin{eqnarray*}
\left|\begin{array}{ccc}2-\lambda&-1&2\\0&3-\lambda&3\\0&2&4-\lambda\end{array}\right|&=&(2-\lambda)\{(3-\lambda)(4-\lambda)-6\}\\
&=&(2-\lambda)\{12-7\lambda+\lambda^2-6\}\\
&=&(2-\lambda)(6-7\lambda+\lambda^2)\\
&=&(2-\lambda)(\lambda-6)(\lambda-1)
\end{eqnarray*}
This is zero if $\lambda=1,2,6$ so these are the eigenvalues.

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${\lambda=1}:\ $ Eigenvector satisfies $(A-1I)\textbf{x}=0\\
\left(\begin{array}{ccc}1&-1&2\\0&2&3\\0&2&3\end{array}\right)
\left(\begin{array}{c}x_1\\x_2\\x_3\end{array}\right)=
\left(\begin{array}{c}0\\0\\0\end{array}\right)$

\begin{eqnarray*}
x_1-x_2+2x_3&=&0\\2x_2+3x_3&=&0\\ 2x_2+3x_3&=&0
\end{eqnarray*}

Choose $\ds x_3=c,\Rightarrow x_2=-\frac{3c}{2},\Rightarrow
x_1=x_2-2x_3=-\frac{3c}{2}-2c=-\frac{7c}{2}$\\ Therefore the
eigenvector is
$\left(\begin{array}{c}x_1\\x_2\\x_3\end{array}\right)
=\left(\begin{array}{c}-\frac{7c}{2}\\-\frac{3c}{2}\\c\end{array}\right),$
where $c$ is any constant.

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${\lambda=6:\ }$ Eigenvector satisfies $(A-6I)\textbf{x}=0\\
\left(\begin{array}{ccc}-4&-1&2\\0&-3&3\\0&2&-2\end{array}\right)
\left(\begin{array}{c}x_1\\x_2\\x_3\end{array}\right)=
\left(\begin{array}{c}0\\0\\0\end{array}\right)$

\begin{eqnarray*}
-4x_1-x_2+2x_3&=&0\\-3x_2+3x_3&=&0\\ 2x_2-2x_3&=&0
\end{eqnarray*}

Choose $\ds x_2=d, \Rightarrow x_3=d,\Rightarrow
4x_1=2x_3-x_2=2d-d=d,\ x_1=\frac{d}{4}$\\ Therefore the
eigenvector is
$\left(\begin{array}{c}x_1\\x_2\\x_3\end{array}\right)
=\left(\begin{array}{c}\frac{d}{4}\\d\\d\end{array}\right),$ where
$d$ is any constant.

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${\lambda=2:\ }$ Eigenvector satisfies $(A-2I)\textbf{x}=0\\
\left(\begin{array}{ccc}0&-1&2\\0&1&3\\0&2&2\end{array}\right)
\left(\begin{array}{c}x_1\\x_2\\x_3\end{array}\right)=
\left(\begin{array}{c}0\\0\\0\end{array}\right)$

\begin{eqnarray*}
-x_2+2x_3&=&0\\x_2+3x_3&=&0\\ 2x_2+2x_3&=&0
\end{eqnarray*}

Hence $x_3=0,\ x_2=0$, choose $x_1=e.$\\ Therefore the eigenvector
is $\left(\begin{array}{c}x_1\\x_2\\x_3\end{array}\right)
=\left(\begin{array}{c}e\\0\\0\end{array}\right)$ where $e$ is any
constant.




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