Solved Exercise Using the Gauss Jordan Algorithm
I have a 4x4 matrix M, and I need to compute its row echelon form using the Gauss Jordan method.
$$ M_{4,4} = \begin{bmatrix} 1 & 1 & 1 & 2 \\ 0 & 3 & -3 & 3 \\ 1 & 0 & 2 & 1 \\ 2 & 0 & 4 & 2 \\ \end{bmatrix} $$
The first column already has a pivot at the top left.
So, I'll nullify the non-zero elements below the pivot, using Gauss' rule:
$$ R_i - ( q_j / p_k ) \cdot R_k $$
First, I nullify the element qj=1
$$ M_{4,4} = \begin{bmatrix} 1 & 1 & 1 & 2 \\ 0 & 3 & -3 & 3 \\ [1] & 0 & 2 & 1 \\ 2 & 0 & 4 & 2 \\ \end{bmatrix} $$
$$ M_{4,4} = \begin{bmatrix} 1 & 1 & 1 & 2 \\ 0 & 3 & -3 & 3 \\ 1-(1)·1 & 0-(1)·1 & 2-(1)·1 & 1-(1)·2 \\ 2 & 0 & 4 & 2 \\ \end{bmatrix} $$
$$ M_{4,4} = \begin{bmatrix} 1 & 1 & 1 & 2 \\ 0 & 3 & -3 & 3 \\ 0 & -1 & 1 & -1 \\ 2 & 0 & 4 & 2 \\ \end{bmatrix} $$
Then, I nullify the element qj=2
$$ M_{4,4} = \begin{bmatrix} 1 & 1 & 1 & 2 \\ 0 & 3 & -3 & 3 \\ 0 & -1 & 1 & -1 \\ [2] & 0 & 4 & 2 \\ \end{bmatrix} $$
$$ M_{4,4} = \begin{bmatrix} 1 & 1 & 1 & 2 \\ 0 & 3 & -3 & 3 \\ 0 & -1 & 1 & -1 \\ 2-(2)·1 & 0-(2)·1 & 4-(2)·1 & 2-(2)·2 \\ \end{bmatrix} $$
$$ M_{4,4} = \begin{bmatrix} 1 & 1 & 1 & 2 \\ 0 & 3 & -3 & 3 \\ 0 & -1 & 1 & -1 \\ 0 & -2 & 2 & -2 \\ \end{bmatrix} $$
I swap rows R2 and R3.
$$ M_{4,4} = \begin{bmatrix} 1 & 1 & 1 & 2 \\ 0 & -1 & 1 & -1 \\ 0 & 3 & -3 & 3 \\ 0 & -2 & 2 & -2 \\ \end{bmatrix} $$
Then, I multiply R2 → R2·(-1)
$$ M_{4,4} = \begin{bmatrix} 1 & 1 & 1 & 2 \\ 0 & -1·(-1) & 1·(-1) & -1·(-1) \\ 0 & 3 & -3 & 3 \\ 0 & -2 & 2 & -2 \\ \end{bmatrix} $$
$$ M_{4,4} = \begin{bmatrix} 1 & 1 & 1 & 2 \\ 0 & 1 & -1 & 1 \\ 0 & 3 & -3 & 3 \\ 0 & -2 & 2 & -2 \\ \end{bmatrix} $$
Thus, I've also obtained a pivot in the second column.
Now, I nullify the elements below qj=3 and qj=-2.
$$ M_{4,4} = \begin{bmatrix} 1 & 1 & 1 & 2 \\ 0 & 1 & -1 & 1 \\ 0 & 3-(3/1)·1 & -3-(3/1)·(-1) & 3-(3/1)·1 \\ 0 & -2-(-2/1)·1 & 2-(-2/1)·(-1) & -2-(-2/1)·1 \\ \end{bmatrix} $$
$$ M_{4,4} = \begin{bmatrix} 1 & 1 & 1 & 2 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} $$
The matrix now has only two pivots.
Note: The number of pivots in the row echelon form corresponds to the rank of the original matrix M4,4. Therefore, the matrix M4,4 has a rank of 2.
