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

## 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*}

## 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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