Row Operation to Clear First Column of Matrix/Examples/Arbitrary Matrix 1

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Example of Use of Row Operation to Clear First Column of Matrix

Let $\mathbf A$ be the matrix:

$\mathbf A = \begin {pmatrix} 0 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end {pmatrix}$

The matrix $\mathbf R$ corresponding to the row operation to clear the first column is:

$\mathbf R = \begin {pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -1 & 1 \end {pmatrix}$

and the matrix $\mathbf B$ which results from the row operation $\mathbf R$ is:

$\mathbf B = \begin {pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & -1 & 1 \end {pmatrix}$


Proof

We use Row Operation to Clear First Column of Matrix as follows:


$(1): \quad$ Apply the elementary row operation $r_1 \leftrightarrow r_2$ to exchange row $1$ with row $2$.

From Elementary Matrix corresponding to Elementary Row Operation: Exchange Rows, this is accomplished by pre-multiplying by the matrix:

$\mathbf E_1 := \begin {pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end {pmatrix}$

which converts $\mathbf A$ to $\mathbf A_1$ as follows:

\(\ds \mathbf A_1\) \(=\) \(\ds \mathbf E_1 \mathbf A\)
\(\ds \) \(=\) \(\ds \begin {pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end {pmatrix} \begin {pmatrix} 0 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end {pmatrix}\)
\(\ds \) \(=\) \(\ds \begin {pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end {pmatrix}\)


$(2): \quad$ Apply the elementary row operation $r_3 \to r_3 + \paren {-1} r_1$ 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:

\(\ds \mathbf A_2\) \(=\) \(\ds \mathbf E_2 \mathbf A_1\)
\(\ds \) \(=\) \(\ds \begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end {pmatrix} \begin {pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end {pmatrix}\)
\(\ds \) \(=\) \(\ds \begin {pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & -1 & 1 \end {pmatrix}\)

The matrix is in the correct form, and so:

$\mathbf B = \mathbf A_2$

$\Box$


Hence we can calculate $\mathbf R$:

\(\ds \mathbf R\) \(=\) \(\ds \mathbf E_2 \mathbf E_1\)
\(\ds \) \(=\) \(\ds \begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end {pmatrix} \begin {pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end {pmatrix}\)
\(\ds \) \(=\) \(\ds \begin {pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -1 & 1 \end {pmatrix}\)

$\blacksquare$


Sources

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