Problems and solutions in Linear Algebra" width="720" height="340" />Problems and solutions in Linear Algebra" width="720" height="340" />
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
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
The augmented matrix of the given system is
\begin
\left[\begin
1 & 0 & -1 & 0 &-3 & 1 \\
3 & 1 & -1 & 1 & -9 & 3 \\
1 & 0 & -1 & 1 & -2 & 1 \\
\end \right].
\end
We apply the elementary row operations as follows.
We have
\begin
\left[\begin
1 & 0 & -1 & 0 &-3 & 1 \\
3 & 1 & -1 & 1 & -9 & 3 \\
1 & 0 & -1 & 1 & -2 & 1 \\
\end \right] \xrightarrow>
\left[\begin
1 & 0 & -1 & 0 &-3 & 1 \\
0 & 1 & 2 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 1 & 0 \\
\end \right]\\[10pt] \xrightarrow
\left[\begin
1 & 0 & -1 & 0 &-3 & 1 \\
0 & 1 & 2 & 0 & -1 & 0 \\
0 & 0 & 0 & 1 & 1 & 0 \\
\end \right].
\end
The last matrix is in reduced row echelon form.
From this reduction, we see that the general solution is
\begin
x_1&=x_3+3x_5+1\\
x_2&=-2x_3+x_5\\
x_4&=-x_5.
\end
Here $x_3, x_5$ are free (independent) variables and $x_1, x_2, x_4$ are dependent variables.
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