# 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*}

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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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