Linear Algebra (Pt. 4): Gauss-Jordan Elimination

We've previously solved a system of linear equation previously. It looked something like

System of Linear Equation	Augmented Matrix
$x+2y+z=2\\ 3x+8y+z=12\\ 4y+z=2$	$\left[\begin{array}{cc} 1 & 2 & 1 && 2\\ 3 & 8 & 1 && 12\\ 0 & 4 & 1 && 2 \end{array}\right]$

However, what if the question begs to solve 2 systems of linear equation at once 🤠?

Given a new example of:

System of Linear Equation

where

• The first row is$a+3b=1$, while

• The second row is $2a+7b=0$

What if we were to arbitrarily tweak "b" and also state that

• $a+3b=0$, and

• $2a+7b=1$

As you might have remembered under the "ways to think about MM", for:

$$A\times B = C$$

• The entire Matrix A multiplies with

• Columns of Matrix B, will

• Return columns of Matrix C

Hence, it is legal to thus propose the change of Matrix C from $\left[\begin{array}{cc} 1\\0 \end{array}\right]$to $\left[\begin{array}{cc} 1 & 0\\0 & 1\end{array}\right]$.

Gauss-Jordan Elimination

The only difference between Gauss, and Gauss-Jordan is that Gauss would stop REF, while Gauss-Jordan Elimination would continue to RREF (Reduced Row Echelon Form).

Reduced Row Echelon Form

Gauss-Jordan, compared to Gauss, is capable of solving multiple systems of linear equation as mentioned. However, the other differences is also to reach RREF (Reduced Row Echelon Form).

To recap, the REF is an upper-triangular form (u) in which only the elements below the pivoted element has the value 0. For example,

REF

The above is in REF as all elements below a possible pivot element is 0.

However, RREF requires a matrix to be firstly in REF form, and

1. Have all pivots as the value 1, and

2. elements above pivot to be of value 0

Thankfully, all possible pivots of (1, 1) and (2, 2) are the value 1. If not, for example if (1, 1) has the value of 3, we can easily divide the entire row by 3 easily (may result in a few fractions in other elements in the row 💀).

We would subsequently carry on to eliminate (1, 2) as it is above the pivot (2, 2), and is not of the value 0.

RREF

Since all pivots are 1, and elements both below and above the pivots are 0, this satisfies the RREF condition.

Inverse Matrix

If you are observant, you might have realized that we have started off with a Matrix A (LHS) and an identity matrix (RHS), to reach RREF of an identity matrix (LHS) and final Matrix (RHS).

RREF

This final matrix is essentially $A^{\mathrm{-1}}$, the inverse matrix of A. However, we have also found the inverse matrix along the journey to RREF.

Looking only at LHS, we have transformed A into I through a series of elimination (2 steps only actually). Hence, to re-use the notation as used before,

$$E_{\mathrm{21}}E_{\mathrm{12}}A=I$$

We divide over A by both side, we can essentially get

$$E_{\mathrm{21}}E_{\mathrm{12}}=A^{\mathrm{-1}}$$

Thus, the product of all elimination matrices are similarly equal to the inverse of Matrix A 🙉.