Gaussian Elimination/Examples/Arbitrary Matrix 1

Example of Use of Matrix is Row Equivalent to Reduced Echelon Matrix
Let $\mathbf A$ denote the matrix:


 * $\mathbf A = \begin {bmatrix}

0 & 0 & 5 & 35 & -24 & 1 \\ 0 & 2 & 1 & -1 &  1 & 0 \\ 0 & 3 & 2 &  2 &  -1 & 1 \\ 0 & 0 & 0 &  0 &   0 & 0 \\ 0 & 5 & 3 &  1 &   0 & 1 \end {bmatrix}$

The reduced echelon form of $\mathbf A$ is:


 * $\mathbf E = \begin {bmatrix}

0 & 1 & 0 & -4 & 0 & 26 \\ 0 & 0 & 1 &  7 & 0 & -43 \\ 0 & 0 & 0 &  0 & 1 &  -9 \\ 0 & 0 & 0 &  0 & 0 &   0 \\ 0 & 0 & 0 &  0 & 0 &   0 \end {bmatrix}$

Proof
$(1): \quad$ Use the elementary row operation $r_4 \leftrightarrow r_5$ to move the zero row to the bottom

$(2): \quad$ Use the elementary row operation $r_2 \to \dfrac {r_2} 2$ to put a leading $1$ in row $2$

$(3): \quad$ Use the elementary row operation $r_2 \leftrightarrow r_1$ to put the row with the leftmost leading $1$ at the top.

Hence we get matrix $\mathbf A_1$:


 * $\mathbf A_1 = \begin {bmatrix}

0 & 1 & \dfrac 1 2 & -\dfrac 1 2 & \dfrac 1 2 & 0 \\ 0 & 0 & 5 & 35 & -24 & 1 \\ 0 & 3 & 2 & 2 &  -1 & 1 \\ 0 & 5 & 3 &  1 &   0 & 1 \\ 0 & 0 & 0 &  0 &   0 & 0 \end {bmatrix}$

$(4): \quad$ Use the following elementary row operations:
 * $r_3 \leftrightarrow r_3 - 3 r_1$
 * $r_4 \leftrightarrow r_4 - 5 r_1$

to clear column $2$.

Hence we get matrix $\mathbf A_2$:


 * $\mathbf A_2 = \begin {bmatrix}

0 & 1 & \dfrac 1 2 & -\dfrac 1 2 & \dfrac 1 2 & 0 \\ 0 & 0 & 5 & 35 & -24 & 1 \\ 0 & 0 & \dfrac 1 2 & \dfrac 7 2 & -\dfrac 5 2 & 1 \\ 0 & 0 & \dfrac 1 2 & \dfrac 7 2 & -\dfrac 5 2 & 1 \\ 0 & 0 & 0 & 0 &   0 & 0 \end {bmatrix}$

$(5): \quad$ Use the following elementary row operations:
 * $r_2 \leftrightarrow r_3$
 * $r_1 \to r_1 - r_2$
 * $r_4 \to r_4 - r_2$
 * $r_3 \to r_3 - 10 r_1$
 * $r_2 \to 2 r_2$

Hence we get matrix $\mathbf A_3$:


 * $\mathbf A_3 = \begin {bmatrix}

0 & 1 & 0 & -4 & 3 &  1 \\ 0 & 0 & 1 &  7 & -5 &  2 \\ 0 & 0 & 0 &  0 &  1 & -9 \\ 0 & 0 & 0 &  0 &  0 &  0 \\ 0 & 0 & 0 &  0 &  0 &  0 \end {bmatrix}$

$(6): \quad$ Use the following elementary row operations:
 * $r_1 \to r_1 - 3 r_3$
 * $r_2 \to r_2 + 5 r_3$

Hence we get matrix $\mathbf A_3$:


 * $\mathbf A_3 = \begin {bmatrix}

0 & 1 & 0 & -4 & 0 & 26 \\ 0 & 0 & 1 &  7 &  0 & 43 \\ 0 & 0 & 0 &  0 &  1 & -9 \\ 0 & 0 & 0 &  0 &  0 &  0 \\ 0 & 0 & 0 &  0 &  0 &  0 \end {bmatrix}$

Different sequences of elementary row operations can be used, but it is necessary to be precise about the order in which they are applied.