# Row Equivalence/Examples/Arbitrary Example 2

## Example of Row Equivalence

Let $\mathbf A$ be the matrix defined as:

$\mathbf A := \begin {bmatrix} 1 & 0 & -1 \\ 2 & 1 & 0 \\ 1 & -1 & 1 \end {bmatrix}$

Let $\mathbf B$ be the matrix defined as:

$\mathbf B := \begin {bmatrix} 3 & -1 & 1 \\ 0 & 2 & 1 \\ 1 & -1 & 1 \end {bmatrix}$

Then $\mathbf A$ and $\mathbf B$ are row equivalent.

## Proof

We use elementary row operations to convert both $\mathbf A$ and $\mathbf B$ to reduced echelon form.

In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to either $\mathbf A$ or $\mathbf B$ accordingly.

The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\mathbf A_k$ or $\mathbf B_k$.

First we take $\mathbf A := \begin {bmatrix} 1 & 0 & -1 \\ 2 & 1 & 0 \\ 1 & -1 & 1 \end {bmatrix}$.

$e_1 := r_2 \to r_2 - 2 r_1$

$e_2 := r_3 \to r_3 - r_1$

$\mathbf A_2 := \begin {bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & -1 & 2 \end {bmatrix}$

$e_3 := r_3 \to r_3 + r_2$

$\mathbf A_3 := \begin {bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end {bmatrix}$

$e_4 := r_3 \to \dfrac {r_3} 4$

$\mathbf A_4 := \begin {bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end {bmatrix}$

$e_5 := r_1 \to r_1 + r_3$

$e_6 := r_2 \to r_2 - 2 r 3$

$\mathbf A_6 := \begin {bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {bmatrix}$

$\Box$

Then we take $\mathbf B := \begin {bmatrix} 3 & -1 & 1 \\ 0 & 2 & 1 \\ 1 & -1 & 1 \end {bmatrix}$.

$e_1 := r_1 \leftrightarrow r_3$

$\mathbf B_1 := \begin {bmatrix} 1 & -1 & 1 \\ 0 & 2 & 1 \\ 3 & -1 & 1 \\ \end {bmatrix}$

$e_2 := r_3 \to r_3 - 3 r_1$

$\mathbf B_2 := \begin {bmatrix} 1 & -1 & 1 \\ 0 & 2 & 1 \\ 0 & 2 & -2 \\ \end {bmatrix}$

$e_3 := r_3 \to \dfrac {r_3} 2$

$\mathbf B_3 := \begin {bmatrix} 1 & -1 & 1 \\ 0 & 2 & 1 \\ 0 & 1 & -1 \\ \end {bmatrix}$

$e_4 := r_2 \leftrightarrow r_3$

$\mathbf B_4 := \begin {bmatrix} 1 & -1 & 1 \\ 0 & 1 & -1 \\ 0 & 2 & 1 \\ \end {bmatrix}$

$e_5 := r_1 \to r_1 + r_2$

$e_6 := r_1 \to r_1 - 2 r 1$

$\mathbf B_6 := \begin {bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 3 \\ \end {bmatrix}$

$e_7 := r_3 \to \dfrac {r_3} 3$

$\mathbf B_7 := \begin {bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \\ \end {bmatrix}$

$e_8 := r_2 \to r_2 + r 3$

$\mathbf B_8 := \begin {bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end {bmatrix}$

$\Box$

It is seen that both $\mathbf A$ and $\mathbf B$ are row equivalent to the same reduced echelon matrix.

The result follows by Row Equivalence is Equivalence Relation.

$\blacksquare$