Row Operation to Clear First Column of Matrix/Examples/Arbitrary Matrix 2
Example of Use of Row Operation to Clear First Column of Matrix
Let $\mathbf A$ be the matrix:
- $\mathbf A = \begin {pmatrix} 1 & 0 & 1 & 1 \\ -1 & 0 & 2 & 1 \\ -1 & 1 & 5 & 9 \end {pmatrix}$
The matrix $\mathbf R$ corresponding to the row operation to clear the first column is:
- $\mathbf R = \begin {pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end {pmatrix}$
and the matrix $\mathbf B$ which results from the row operation $\mathbf R$ is:
- $\mathbf B = \begin {pmatrix} 1 & 0 & 1 & 1 \\ 0 & 0 & 3 & 2 \\ 0 & 1 & 6 & 10 \end {pmatrix}$
Proof
We use Row Operation to Clear First Column of Matrix as follows:
- $(1): \quad$ Apply the elementary row operation $r_2 \to r_2 + r_1$ to to add $1$ times row $1$ to row $2$.
From Elementary Matrix corresponding to Elementary Row Operation: Scale Row and Add, this is accomplished by pre-multiplying by the matrix:
- $\mathbf E_1 := \begin {pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end {pmatrix}$
which converts $\mathbf A$ to $\mathbf A_1$ as follows:
\(\ds \mathbf A_1\) | \(=\) | \(\ds \mathbf E_1 \mathbf A\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin {pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end {pmatrix} \begin {pmatrix} 1 & 0 & 1 & 1 \\ -1 & 0 & 2 & 1 \\ -1 & 1 & 5 & 9 \end {pmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin {pmatrix} 1 & 0 & 1 & 1 \\ 0 & 0 & 3 & 2 \\ -1 & 1 & 5 & 9 \end {pmatrix}\) |
- $(2): \quad$ Apply the elementary row operation $r_3 \to r_3 + r_1$ to add $1$ times row $1$ to row $3$.
From Elementary Matrix corresponding to Elementary Row Operation: Scale Row and Add, this is accomplished by pre-multiplying by the matrix:
- $\mathbf E_2 := \begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end {pmatrix}$
which converts $\mathbf A_1$ to $\mathbf A_2$ as follows:
\(\ds \mathbf A_2\) | \(=\) | \(\ds \mathbf E_2 \mathbf A_1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end {pmatrix} \begin {pmatrix} 1 & 0 & 1 & 1 \\ 0 & 0 & 3 & 2 \\ -1 & 1 & 5 & 9 \end {pmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin {pmatrix} 1 & 0 & 1 & 1 \\ 0 & 0 & 3 & 2 \\ 0 & 1 & 6 & 10 \end {pmatrix}\) |
The matrix is in the correct form, and so:
- $\mathbf B = \mathbf A_2$
$\Box$
Hence we can calculate $\mathbf R$:
\(\ds \mathbf R\) | \(=\) | \(\ds \mathbf E_2 \mathbf E_1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end {pmatrix} \begin {pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end {pmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin {pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end {pmatrix}\) |
$\blacksquare$
Sources
- 1998: Richard Kaye and Robert Wilson: Linear Algebra ... (previous) ... (next): Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.2 \ \text {(b)}$