Worked Example 2
Basis & Dimension of a
Homogeneous System with 6 Variables
[A/M-23-R-21]
[A/M-26-R-25]
Tutorial Video: Basis & Dimension of Homogeneous System
📌 Problem Statement
Find a basis and the dimension of the solution space of the homogeneous system:
\[
\begin{aligned}
x_1 + 3x_2 - 2x_3 + 2x_5 &= 0 \\
2x_1 + 6x_2 - 5x_3 - 2x_4 + 4x_5 - 3x_6 &= 0 \\
5x_3 + 10x_4 + 15x_6 &= 0 \\
2x_1 + 6x_2 + 8x_4 + 4x_5 + 18x_6 &= 0
\end{aligned}
\]
[A/M-23-R-21] · [A/M-26-R-25]
1
Set Up the Augmented Matrix
We write the system as $AX = 0$. Since all right-hand sides are zero, we work directly with the coefficient matrix $A$:
\[
A = \begin{bmatrix}
1 & 3 & -2 & 0 & 2 & 0 \\
2 & 6 & -5 & -2 & 4 & -3 \\
0 & 0 & 5 & 10 & 0 & 15 \\
2 & 6 & 0 & 8 & 4 & 18
\end{bmatrix}
\]
2
Row Reduce to Reduced Row Echelon Form
We apply a sequence of elementary row operations to reach RREF.
Operations: $R_2 \leftarrow R_2 - 2R_1$, $R_4 \leftarrow R_4 - 2R_1$
\[
\begin{bmatrix}
1 & 3 & -2 & 0 & 2 & 0 \\
0 & 0 & -1 & -2 & 0 & -3 \\
0 & 0 & 5 & 10 & 0 & 15 \\
0 & 0 & 4 & 8 & 0 & 18
\end{bmatrix}
\]
Operation: $R_2 \leftarrow -R_2$ (make leading entry positive)
\[
\begin{bmatrix}
1 & 3 & -2 & 0 & 2 & 0 \\
0 & 0 & 1 & 2 & 0 & 3 \\
0 & 0 & 5 & 10 & 0 & 15 \\
0 & 0 & 4 & 8 & 0 & 18
\end{bmatrix}
\]
Operations: $R_1 \leftarrow R_1 + 2R_2$, $R_3 \leftarrow R_3 - 5R_2$, $R_4 \leftarrow R_4 - 4R_2$
\[
\begin{bmatrix}
1 & 3 & 0 & 4 & 2 & 6 \\
0 & 0 & 1 & 2 & 0 & 3 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 6
\end{bmatrix}
\]
Operation: $R_3 \leftrightarrow R_4$ (move non-zero row up)
\[
\begin{bmatrix}
1 & 3 & 0 & 4 & 2 & 6 \\
0 & 0 & 1 & 2 & 0 & 3 \\
0 & 0 & 0 & 0 & 0 & 6 \\
0 & 0 & 0 & 0 & 0 & 0
\end{bmatrix}
\]
Operation: $R_3 \leftarrow R_3 \div 6$ (normalize pivot to 1)
\[
\begin{bmatrix}
1 & 3 & 0 & 4 & 2 & 6 \\
0 & 0 & 1 & 2 & 0 & 3 \\
0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 0
\end{bmatrix}
\]
Operations: $R_1 \leftarrow R_1 - 6R_3$, $R_2 \leftarrow R_2 - 3R_3$
\[
A \approx \begin{bmatrix}
1 & 3 & 0 & 4 & 2 & 0 \\
0 & 0 & 1 & 2 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 0
\end{bmatrix}
\quad \textbf{(RREF)}
\]
3
Identify Pivot and Free Variables
Pivot columns (columns with leading 1s) give the pivot variables; the remaining columns give free variables:
$x_1$ — pivot
$x_2$ — free ($= s$)
$x_3$ — pivot
$x_4$ — free ($= t$)
$x_5$ — free ($= u$)
$x_6$ — pivot
Rank–Nullity Theorem check: rank$(A) = 3$ (three non-zero rows) + nullity $= 3$ (three free variables) $= 6$ (total variables) ✓
4
Express Pivot Variables via Free Variables
Reading from the RREF rows, with $x_2 = s,\ x_4 = t,\ x_5 = u$:
\[
\begin{aligned}
x_1 + 3x_2 + 4x_4 + 2x_5 &= 0 &\implies x_1 &= -3s - 4t - 2u \\
x_3 + 2x_4 &= 0 &\implies x_3 &= -2t \\
x_6 &= 0 &\implies x_6 &= 0
\end{aligned}
\]
5
Write the General Solution
Collecting all components into vector form and separating by parameters $s$, $t$, and $u$:
\[
\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{bmatrix}
=
\begin{bmatrix} -3s - 4t - 2u \\ s \\ -2t \\ t \\ u \\ 0 \end{bmatrix}
=\;
s\begin{bmatrix} -3 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}
+\;
t\begin{bmatrix} -4 \\ 0 \\ -2 \\ 1 \\ 0 \\ 0 \end{bmatrix}
+\;
u\begin{bmatrix} -2 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix},
\quad s,t,u \in \mathbb{R}
\]
✅ Final Answer
The basis of the null space (solution space) is:
$\mathbf{v}_1$
\[\begin{bmatrix}-3\\1\\0\\0\\0\\0\end{bmatrix}\]
,
$\mathbf{v}_2$
\[\begin{bmatrix}-4\\0\\-2\\1\\0\\0\end{bmatrix}\]
,
$\mathbf{v}_3$
\[\begin{bmatrix}-2\\0\\0\\0\\1\\0\end{bmatrix}\]
The dimension of the solution space is:
dim = 3
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