2.5 Elementary Row Operations on Matrices
The following are called elementary row operations on a matrix.
- 1.
- Interchange the \(i^{\text {th}}\) and \(j^{\text {th}}\) rows; \(r_i\longleftrightarrow r_j\). For instance \[ \begin {pmatrix} 1&2\\3&4\\ \end {pmatrix} \hspace {0.5cm} \begin {matrix} r_1 \longleftrightarrow r_2\\ \longrightarrow \\ \end {matrix} \hspace {0.5cm} \begin {pmatrix} 3&4\\1&2\\ \end {pmatrix} \]
- 2.
- Multiply the \(i^{\text {th}}\) row by \(\alpha \), where \(\alpha \) is a non-zero scalar; \(r_i\longrightarrow \alpha r_j\). For example \[ \begin {pmatrix} 1&2\\3&4\\ \end {pmatrix} \hspace {0.5cm} \begin {matrix} r_1\longrightarrow 2r_1\\ \end {matrix} \hspace {0.5cm} \begin {pmatrix} 2&4\\3&4\\ \end {pmatrix} \]
- 3.
- Add the \(i^{\text {th}}\) row to \(\alpha \) times the \(j^{\text {th}}\) row; \(r_i\longrightarrow r_i+\alpha r_j\). For example \[ \begin {pmatrix} 1&2\\3&4\\ \end {pmatrix} \hspace {0.5cm} \begin {matrix} r_1\longrightarrow r_1+2r_2\\ \end {matrix} \hspace {0.5cm} \begin {pmatrix} 7&10\\3&4\\ \end {pmatrix} \]
Definition 2.5.1. Two \(m\) by \(n\) matrices \(A\) and \(B\) are said to be row equivalent if \(B\) can be obtained from \(A\) by a finite sequence of elementary row operations.
Example 2.5.2. \( \begin {pmatrix} 1&-3&1&0\\-1&2&4&1\\1&0&-2&1\\ \end {pmatrix} \) is row equivalent to \( \begin {pmatrix} 1&-3&1&0\\0&-1&5&1\\0&1&1&1\\ \end {pmatrix} \) since \begin {align*} \begin {pmatrix} 1&-3&1&0\\-1&2&4&1\\1&0&-2&1\\ \end {pmatrix} &\hspace {0.5cm} \begin {matrix} r_3\longrightarrow r_3+r_2\\ \end {matrix} \hspace {0.5cm} \begin {pmatrix} 1&-3&1&0\\-1&2&4&1\\0&2&2&2\\ \end {pmatrix}\\\\ &\hspace {0.5cm} \begin {matrix} r_2\longrightarrow r_2+r_1\\ r_3\longrightarrow \frac {1}{2}r_3\\ \end {matrix} \hspace {0.5cm} \begin {pmatrix} 1&-3&1&0\\0&-1&5&1\\0&1&1&1\\ \end {pmatrix} \end {align*}
Definition 2.5.3. An \(m\) by \(n\) matrix \(A\) is called an echelon matrix if it satisfies the following:
- i.
- the leading non-zero element in every row is 1.
- ii.
- the leading 1 in each row occurs to the right of the leading 1 in any preceding row.
- iii.
- non-zero rows appear before zero rows.
Definition 2.5.4. A reduced echelon matrix is an echelon matrix with the property that the leading 1 in every row is the only non-zero element in the column in which it occurs.
Example 2.5.5. \( \begin {pmatrix} 1&0&3&4\\0&1&-1&0\\0&0&1&5\\ \end {pmatrix} \) is an echelon matrix but not a reduced echelon matrix.
\( \begin {pmatrix} 1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1 \end {pmatrix} \) is a reduced echelon matrix.
Example 2.5.6. Find the reduced echelon of (a) \( \begin {pmatrix} 1&1&0\\1&-3&0\\ \end {pmatrix} \hspace {0.5cm}\) (b) \( \begin {pmatrix} 3&1&-2\\-4&2&1\\-1&0&1\\ \end {pmatrix} \).
Solution. \begin {align*} \text {(a)}\hspace {0.4cm} \begin {pmatrix} 1&1&0\\1&-3&0\\ \end {pmatrix} & \hspace {0.3cm} \begin {matrix} r_2\longrightarrow r_2-r_1\\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&1&0\\0&-4&0\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_2\longrightarrow -\frac {1}{2}r_2\\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&1&0\\0&1&0\\ \end {pmatrix}\\\\ &\hspace {0.3cm} \begin {matrix} r_1\longrightarrow r_1-r_2\\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&0&0\\0&1&0\\ \end {pmatrix}\\ \end {align*}
\begin {align*} \text {(b)}\hspace {0.4cm} \begin {pmatrix} 3&1&-2\\ -4&2&1\\-1&0&1\\ \end {pmatrix} &\hspace {0.3cm} \begin {matrix} r_1\rightarrow r_1+3r_2\\ \longrightarrow \\ r_2\rightarrow r_2-4r_3\\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 0&1&1\\0&2&-3\\-1&0&1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_1\leftrightarrow r_3\\ \longrightarrow \\ \end {matrix} \begin {pmatrix} -1&0&1\\0&2&-3\\0&1&1\\ \end {pmatrix}\\\\ &\hspace {0.3cm} \begin {matrix} r_2\rightarrow r_2-2r_3\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} -1&0&1\\0&0&-5\\0&1&1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_2\leftrightarrow r_3\\ \longrightarrow \\ r_1\rightarrow -r_1\\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&0&-1\\0&1&1\\0&0&-5\\ \end {pmatrix}\\\\ &\hspace {0.3cm} \begin {matrix} r_3\rightarrow -\frac {1}{5}r_3\\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&0&-1\\0&1&1\\0&0&1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_1\rightarrow r_1+r_3\\ \longrightarrow \\ r_2\rightarrow r_2-r_3\\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&0&0\\ 0&1&0\\0&0&1\\ \end {pmatrix}\\ \end {align*}
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