Gaussian Elimination/Examples/Arbitrary Matrix 5
Example of Use of Gaussian Elimination
Let $\mathbf A$ denote the matrix:
- $\mathbf A = \begin {bmatrix}
0 & 1 & 1 & 1 & 2 & 2 \\
-1 & 4 & 3 & 3 & 4 & 7 \\
2 & 1 & 3 & 2 & 8 & 3 \\ 3 & 1 & 4 & -1 & 4 & 0 \\ 5 & 2 & 7 & 0 & 10 & 2 \\
\end {bmatrix}$
The reduced echelon form of $\mathbf A$ is:
- $\mathbf E = \begin {bmatrix}
1 & 0 & 1 & 0 & 2 & 0 \\ 0 & 1 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\
\end {bmatrix}$
Proof
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\mathbf A$.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\mathbf A_k$.
$e_1 := r_2 \leftrightarrow r_1$
Hence:
- $\mathbf A_1 = \begin {bmatrix}
-1 & 4 & 3 & 3 & 4 & 7 \\
0 & 1 & 1 & 1 & 2 & 2 \\ 2 & 1 & 3 & 2 & 8 & 3 \\ 3 & 1 & 4 & -1 & 4 & 0 \\ 5 & 2 & 7 & 0 & 10 & 2 \\
\end {bmatrix}$
$e_2 := r_1 \to -r_1$
- $\mathbf A_2 = \begin {bmatrix}
1 & -4 & -3 & -3 & -4 & -7 \\ 0 & 1 & 1 & 1 & 2 & 2 \\ 2 & 1 & 3 & 2 & 8 & 3 \\ 3 & 1 & 4 & -1 & 4 & 0 \\ 5 & 2 & 7 & 0 & 10 & 2 \\
\end {bmatrix}$
$e_3 := r_1 \to r_1 + 4 r_2$
- $\mathbf A_3 = \begin {bmatrix}
1 & 0 & 1 & 1 & 4 & 1 \\ 0 & 1 & 1 & 1 & 2 & 2 \\ 2 & 1 & 3 & 2 & 8 & 3 \\ 3 & 1 & 4 & -1 & 4 & 0 \\ 5 & 2 & 7 & 0 & 10 & 2 \\
\end {bmatrix}$
$e_4 := r_3 \to r_3 - 2 r_1$
$e_5 := r_4 \to r_4 - 3 r_1$
$e_6 := r_5 \to r_5 - 5 r_1$
Hence:
- $\mathbf A_6 = \begin {bmatrix}
1 & 0 & 1 & 1 & 4 & 1 \\ 0 & 1 & 1 & 1 & 2 & 2 \\ 0 & 1 & 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & -4 & -8 & -3 \\ 0 & 2 & 2 & -5 & -10 & -3 \\
\end {bmatrix}$
$e_7 := r_3 \to r_3 - r_2$
$e_8 := r_4 \to r_4 - r_2$
- $\mathbf A_8 = \begin {bmatrix}
1 & 0 & 1 & 1 & 4 & 1 \\ 0 & 1 & 1 & 1 & 2 & 2 \\ 0 & 0 & 0 & -1 & -2 & -1 \\ 0 & 0 & 0 & -5 & -10 & -5 \\ 0 & 2 & 2 & -5 & -10 & -3 \\
\end {bmatrix}$
$e_9 := r_3 \to -r_3$
$e_{10} := r_5 \to r_5 - r_4$
- $\mathbf A_{10} = \begin {bmatrix}
1 & 0 & 1 & 1 & 4 & 1 \\ 0 & 1 & 1 & 1 & 2 & 2 \\ 0 & 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & -5 & -10 & -5 \\ 0 & 2 & 2 & 0 & 0 & 2 \\
\end {bmatrix}$
$e_{12} := r_4 \to r_4 + 5 r_3$
$e_{13} := r_2 \to r_2 - r_3$
$e_{14} := r_1 \to r_1 - r_3$
- $\mathbf A_{14} = \begin {bmatrix}
1 & 0 & 1 & 0 & 2 & 0 \\ 0 & 1 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 2 & 0 & 0 & 2 \\
\end {bmatrix}$
$e_{15} := r_5 \to r_5 - 2 r_2$
- $\mathbf A_{15} = \begin {bmatrix}
1 & 0 & 1 & 0 & 2 & 0 \\ 0 & 1 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\
\end {bmatrix}$
and it is seen that $\mathbf A_{16}$ is the required reduced echelon form.
$\blacksquare$
Sources
- 1982: A.O. Morris: Linear Algebra: An Introduction (2nd ed.) ... (previous) ... (next): Chapter $1$: Linear Equations and Matrices: $1.2$ Elementary Row Operations on Matrices: Exercises $1.2$: $1 \ \text {(iv)}$