Toronto Math Forum
MAT244--2018F => MAT244--Tests => Quiz-6 => Topic started by: Victor Ivrii on November 17, 2018, 03:52:31 PM
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Find the general solution of the given system of equations:
$$\mathbf{x}'=
\begin{pmatrix}
1 &1 &2\\
1 &2 &1\\
2 &1 &1
\end{pmatrix}\mathbf{x}.$$
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This is my answer!
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\begin{equation*}
det
\begin{pmatrix}
1-\lambda &1 &2 \\
1 & 2-\lambda &1 \\
2 & 1 & 1-\lambda
\end{pmatrix}
=-\lambda^3+4\lambda^2+\lambda-4=-(\lambda-1)(\lambda-4)(\lambda+1)=0
\end{equation*}
$$
\lambda=1,\lambda=4,\lambda=-1
$$
when $\lambda$=1
\begin{equation*}
\begin{pmatrix}
0 &1 &2 \\
1 & 1 &1 \\
2 & 1 & 0
\end{pmatrix}
\sim
\begin{pmatrix}
2 &1 &0 \\
0 & 1 &2 \\
1 & 1 & 1
\end{pmatrix}
\sim
\begin{pmatrix}
2 &0 &-2 \\
-1 & 0 &1 \\
1 & 1 & 1
\end{pmatrix}
\sim
\begin{pmatrix}
2 &0 &-2 \\
0 & 0 &0 \\
1 & 1 & 1
\end{pmatrix}
\sim
\begin{pmatrix}
2 &0 &-2 \\
1 & 1 &1 \\
0 & 0 & 0
\end{pmatrix}
\sim
\begin{pmatrix}
x_1 \\ x_2 \\ x_3
\end{pmatrix}=0
\end{equation*}
$$
\text{let } x_3=t,x_1=t,x_2=-2t
x=
\begin{pmatrix}
1 \\ -2 \\ 1
\end{pmatrix}t
$$
when $\lambda$=4
\begin{equation*}
\begin{pmatrix}
-3 &1 &2 \\
1 & -2 &1 \\
2 & 1 & -3
\end{pmatrix}
\sim
\begin{pmatrix}
-3 &1 &2 \\
0 & -5 &5 \\
2 & 1 & -3
\end{pmatrix}
\begin{pmatrix}
x_1 \\ x_2 \\ x_3
\end{pmatrix}
=0
\end{equation*}
$$
x=\begin{pmatrix} 1 \\1 \\1 \end{pmatrix}t
$$
when $\lambda$=-1
\begin{equation*}
\begin{pmatrix}
2 &1 &2 \\
1 & 3 &1 \\
2 & 1 & 2
\end{pmatrix}
\sim
\begin{pmatrix}
2 &1 &2 \\
0 & 5 &0 \\
0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
x_1 \\ x_2 \\ x_3
\end{pmatrix}
=0
\end{equation*}
$$
x=\begin{pmatrix}-1\\0\\1\end{pmatrix}
$$
$$
x(t)=c_1e^4t\begin{pmatrix}1\\1\\1\end{pmatrix}+c_2e^{-t}\begin{pmatrix}-1\\0\\1\end{pmatrix}+c_3e^t\begin{pmatrix}1\\-2\\1\end{pmatrix}
$$
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Sometimes we do not have to compute all the ref or rref. For example, when λ= 4, it is easy to observe that the sum of three columns of the matrix is 0. Sometimes it saves your time during the quiz if you can observe it directly.
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Let's solve this question step by step!!!