Row Equivalence/Examples/Arbitrary Example 3

Example of Row Equivalence
Let $\mathbf A$ be the matrix defined as:


 * $\mathbf A := \begin {bmatrix}

1 & -1 & 1 & 2 \\ -2 &  3 &  0 & 1 \\ 1 &  0 & -1 & 3 \\ \end {bmatrix}$

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


 * $\mathbf B := \begin {bmatrix}

0 & -1 & 2 & 3 \\ 1 &  2 & -1 & 0 \\ -2 & -5 &  4 & 3 \\ \end {bmatrix}$

Then $\mathbf A$ and $\mathbf B$ are not 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 $\mathbf A := \begin {bmatrix} 1 & -1 & 1 & 2 \\ -2 &  3 &  0 & 1 \\ 1 &  0 & -1 & 3 \\ \end {bmatrix}$.

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

$e_2 := r_3 \to r_3 - r_1$


 * $\mathbf A_2 := \begin {bmatrix}

1 & -1 & 1 & 2 \\ 0 &  1 &  2 & 5 \\ 0 &  1 & -2 & 1 \\ \end {bmatrix}$

$e_3 := r_3 \to r_3 - r_2$

$e_4 := r_1 \to r_1 + r_2$


 * $\mathbf A_4 := \begin {bmatrix}

1 & 0 & 3 &  7 \\ 0 & 1 &  2 &  5 \\ 0 & 0 & -4 & -4 \\ \end {bmatrix}$

$e_5 := r_3 \to -\dfrac {r_3} 4$


 * $\mathbf A_5 := \begin {bmatrix}

1 & 0 & 3 & 7 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & 1 & 1 \\ \end {bmatrix}$

$e_6 := r_1 \to r_1 - 3 r 3$

$e_7 := r_2 \to r_1 - 2 r 3$


 * $\mathbf A_7 := \begin {bmatrix}

1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 1 \\ \end {bmatrix}$

Then we take $\mathbf B := \begin {bmatrix} 0 & -1 & 2 & 3 \\ 1 &  2 & -1 & 0 \\ -2 & -5 &  4 & 3 \\ \end {bmatrix}$.

$e_1 := r_1 \leftrightarrow r_2$


 * $\mathbf B_1 := \begin {bmatrix}

1 & 2 & -1 & 0 \\ 0 & -1 &  2 & 3 \\ -2 & -5 &  4 & 3 \\ \end {bmatrix}$

$e_2 := r_3 \to r_3 + 2 r_1$


 * $\mathbf B_2 := \begin {bmatrix}

1 & 2 & -1 & 0 \\ 0 & -1 &  2 & 3 \\ 0 & -1 &  2 & 3 \\ \end {bmatrix}$

$e_3 := r_3 \to r_3 - r_2$


 * $\mathbf B_3 := \begin {bmatrix}

1 & 2 & -1 & 0 \\ 0 & -1 &  2 & 3 \\ 0 &  0 &  0 & 0 \\ \end {bmatrix}$

A zero row has appeared.

There is no need to proceed further.

The reduced echelon matrices for $\mathbf A$ and $\mathbf B$ are different.

The result follows by Row Equivalence is Equivalence Relation.