Vector Form for the General Solution of a System of Linear Equations

Problems and solutions in Linear Algebra

Problem 267

Solve the following system of linear equations by transforming its augmented matrix to reduced echelon form (Gauss-Jordan elimination).

Find the vector form for the general solution.
\begin{align*}
x_1-x_3-3x_5&=1\\
3x_1+x_2-x_3+x_4-9x_5&=3\\
x_1-x_3+x_4-2x_5&=1.
\end{align*}

 
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Solution.

The augmented matrix of the given system is
\begin{align*}
\left[\begin{array}{rrrrr|r}
1 & 0 & -1 & 0 &-3 & 1 \\
3 & 1 & -1 & 1 & -9 & 3 \\
1 & 0 & -1 & 1 & -2 & 1 \\
\end{array} \right].
\end{align*}
We apply the elementary row operations as follows.
We have
\begin{align*}
\left[\begin{array}{rrrrr|r}
1 & 0 & -1 & 0 &-3 & 1 \\
3 & 1 & -1 & 1 & -9 & 3 \\
1 & 0 & -1 & 1 & -2 & 1 \\
\end{array} \right] \xrightarrow{\substack{R_2-3R_1\\R_3-R_1}}
\left[\begin{array}{rrrrr|r}
1 & 0 & -1 & 0 &-3 & 1 \\
0 & 1 & 2 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 1 & 0 \\
\end{array} \right]\\[10pt] \xrightarrow{R_2-R_3}
\left[\begin{array}{rrrrr|r}
1 & 0 & -1 & 0 &-3 & 1 \\
0 & 1 & 2 & 0 & -1 & 0 \\
0 & 0 & 0 & 1 & 1 & 0 \\
\end{array} \right].
\end{align*}
The last matrix is in reduced row echelon form.
From this reduction, we see that the general solution is
\begin{align*}
x_1&=x_3+3x_5+1\\
x_2&=-2x_3+x_5\\
x_4&=-x_5.
\end{align*}
Here $x_3, x_5$ are free (independent) variables and $x_1, x_2, x_4$ are dependent variables.

To find the vector form for the general solution, we substitute these equations into the vector $\mathbf{x}$ as follows.
We have
\begin{align*}
\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5
\end{bmatrix}&=
\begin{bmatrix}
x_3+3x_5+1 \\
-2x_3+x_5 \\
x_3 \\
-x_5 \\
x_5
\end{bmatrix}\\[10pt] &=
\begin{bmatrix}
x_3 \\
-2x_3 \\
x_3 \\
0 \\
0
\end{bmatrix}
+
\begin{bmatrix}
3x_5 \\
x_5 \\
0 \\
-x_5 \\
x_5
\end{bmatrix}
+
\begin{bmatrix}
1 \\
0 \\
0 \\
0 \\
0
\end{bmatrix}\\[10pt] &=x_3\begin{bmatrix}
1 \\
-2 \\
1 \\
0 \\
0
\end{bmatrix}+x_5 \begin{bmatrix}
3 \\
1 \\
0 \\
-1 \\
1
\end{bmatrix}+\begin{bmatrix}
1 \\
0 \\
0 \\
0 \\
0
\end{bmatrix}.
\end{align*}
Therefore the vector form for the general solution is given by
\[\mathbf{x}=x_3\begin{bmatrix}
1 \\
-2 \\
1 \\
0 \\
0
\end{bmatrix}+x_5 \begin{bmatrix}
3 \\
1 \\
0 \\
-1 \\
1
\end{bmatrix}+\begin{bmatrix}
1 \\
0 \\
0 \\
0 \\
0
\end{bmatrix},\] where $x_3, x_5$ are free variables.

Related Question.

For a similar question, check out the post ↴
Solve the System of Linear Equations and Give the Vector Form for the General Solution.


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