In each of 5-8, solve the given initial-value problem

Step 1 of 2

Given: The matrices \(\bar{x}=\left(\begin{array}{lll}
-1 & 1 & 2 \\
-1 & 1 & 1 \\
-2 & 1 & 3
\end{array}\right) x, x(0)=\left(\begin{array}{l}
1 \\
0 \\
1
\end{array}\right)\)

Consider \(A=\left(\begin{array}{lll}
-1 & 1 & 2 \\
-1 & 1 & 1 \\
-2 & 1 & 3
\end{array}\right)\) :

Find Eigen values.

\(\begin{aligned}
|A-\lambda| \mid & =0 \\
\left(\begin{array}{ccc}
-1-a & 1 & 2 \\
-1 & 1-a & 1 \\
-2 & 1 & 3-a
\end{array}\right) & =0 \\
-(1+a)\{(1-a)(3-a)-1\}-1\{-(3-a)+2\}-2\{-1+2(1-a)\} & =0 \\
-a^{3}+3 a^{2}-3 a+1 & =0 \\
(a-1)^{3} & =0 \\
a & =1,1,1
\end{aligned}\)

Now calculate Eigen vector for Eigen value a = 1

\(\left(\begin{array}{lll}
-2 & 1 & 2 \\
-1 & 0 & 1 \\
-2 & 1 & 2
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=0\)

Therefore, Eigen vector \(\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{l}
1 \\
0 \\
1
\end{array}\right)\)

Now find other two vectors as shown below:

\(\begin{aligned}
\left(\begin{array}{rrr}
-2 & 1 & 2 \\
-1 & 0 & 1 \\
-2 & 1 & 2
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right) & =\left(\begin{array}{l}
1 \\
0 \\
1
\end{array}\right) \\
-2 x+y+2 z & =1 \\
-x+z & =0
\end{aligned}\)

Therefore, Eigen vector \(\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{l}
1 \\
1 \\
1
\end{array}\right)\)

\(\begin{aligned}
\left(\begin{array}{rrr}
-2 & 1 & 2 \\
-1 & 0 & 1 \\
-2 & 1 & 2
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right) & =\left(\begin{array}{l}
1 \\
1 \\
1
\end{array}\right) \\
-2 x+y+2 z & =1 \\
-x+z & =1
\end{aligned}\)

Therefore, Eigen vector \(\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{c}
1 \\
-1 \\
2
\end{array}\right)\)