Gauss-Jordan Elimination

We now start with Step 5 of Example 2.2.11 and apply the elementary operations once again. But this time, we start with the $3^{\mbox{rd}}$ row.

1. Add $-1$ times the third equation to the second equation (or $R_{23}(-1)$ ).
   $$\begin{array}{cr} x + \frac{3}{2} z &= \frac{5}{2} \\ y &= 2 \\ ... ... 0 & \frac{3}{2} & \frac{5}{2} \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{bmatrix}.$$

2. Add $\frac{-3}{2}$ times the third equation to the first equation (or $R_{13}(-\frac{3}{2})$ ).
   $$\begin{array}{cr} x &= 1 \\ y &= 1 \\ z &= 1 \end{array} \hspace... ... \begin{bmatrix}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{bmatrix}.$$

3. From the above matrix, we directly have the set of solution as $(x,y,z)^t = (1,1,1)^t.$

DEFINITION 2.3.6 (Row Reduced Echelon Form of a Matrix)   A matrix $C$ is said to be in the row reduced echelon form if

1. $C$ is already in the row reduced form;
2. The rows consisting of all zeros comes below all non-zero rows; and
3. the leading terms appear from left to right in successive rows. That is, for $1 \leq \ell \leq k,$ let $i_{\ell}$ be the leading column of the $\ell^{\mbox{th}}$ row. Then $i_1 < i_2 < \cdots < i_k.$

EXAMPLE 2.3.7   Suppose $A = \begin{bmatrix}0 & 1 & 0 & 2\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix}0 & 0 & 0 & 1 & 0 \\ 1 & 1 & 0& 0 & 0\\ 0 & 0 & 0 & 0 & 1\end{bmatrix}$ are in row reduced form. Then the corresponding matrices in the row reduced echelon form are respectively, $\begin{bmatrix}0 & 1 & 0 & 2\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0 \end{bmatrix}$ and $\begin{bmatrix}1 & 1 & 0& 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{bmatrix}.$

DEFINITION 2.3.8 (Row Reduced Echelon Matrix)   A matrix which is in the row reduced echelon form is also called a row reduced echelon matrix.

DEFINITION 2.3.9 (Back Substitution/Gauss-Jordan Method)   The procedure to get to Step II of Example 2.2.11 from Step 5 of Example 2.2.11 is called the back substitution.

The elimination process applied to obtain the row reduced echelon form of the augmented matrix is called the Gauss-Jordan elimination.

That is, the Gauss-Jordan elimination method consists of both the forward elimination and the backward substitution.

Method to get the row-reduced echelon form of a given matrix $A$
Let $A$ be an $m \times n$ matrix. Then the following method is used to obtain the row-reduced echelon form the matrix $A.$

Step 1: Consider the first column of the matrix $A.$

If all the entries in the first column are zero, move to the second column.

Else, find a row, say $i^{\mbox{th}}$ row, which contains a non-zero entry in the first column. Now, interchange the first row with the $i^{\mbox{th}}$ row. Suppose the non-zero entry in the $(1,1)$ -position is ${\alpha}\neq 0.$ Divide the whole row by ${\alpha}$ so that the $(1,1)$ -entry of the new matrix is $1.$ Now, use the $1$ to make all the entries below this $1$ equal to $0.$

Step 2: If all entries in the first column after the first step are zero, consider the right $m \times (n-1)$ submatrix of the matrix obtained in step 1 and proceed as in step 1.

Else, forget the first row and first column. Start with the lower $(m-1) \times (n-1)$ submatrix of the matrix obtained in the first step and proceed as in step 1.

Step 3: Keep repeating this process till we reach a stage where all the entries below a particular row, say $r$ , are zero. Suppose at this stage we have obtained a matrix $C.$ Then $C$ has the following form:

1. THE FIRST NON-ZERO ENTRY IN EACH ROW of $C$ is $1.$ These $1$ 's are the leading terms of $C$ and the columns containing these leading terms are the leading columns.
2. THE ENTRIES OF MATHEND000# BELOW THE LEADING TERM ARE ALL ZERO.

Step 4: Now use the leading term in the $r^{\mbox{th}}$ row to make all entries in the $r^{\mbox{th}}$ leading column equal to zero.

Step 5: Next, use the leading term in the $(r-1)^{\mbox{th}}$ row to make all entries in the $(r-1)^{\mbox{th}}$ leading column equal to zero and continue till we come to the first leading term or column.

The final matrix is the row-reduced echelon form of the matrix $A.$

Remark 2.3.10   Note that the row reduction involves only row operations and proceeds from LEFT TO RIGHT. Hence, if $A$ is a matrix consisting of first $s$ columns of a matrix $C,$ then the row reduced form of $A$ will be the first $s$ columns of the row reduced form of $C.$

The proof of the following theorem is beyond the scope of this book and is omitted.

THEOREM 2.3.11   The row reduced echelon form of a matrix is unique.

EXERCISE 2.3.12  

1. Solve the following linear system.
   1. $x + y + z + w = 0, \; x - y + z + w = 0$ and $-x + y + 3 z + 3 w = 0.$
   2. $x + 2 y + 3 z = 1$ and $x + 3 y + 2 z = 1.$
   3. $x + y + z = 3, \; x + y - z = 1$ and $x + y + 7 z = 6.$
   4. $x + y + z = 3, \; x + y - z = 1$ and $x + y + 4 z = 6.$
   5. $x + y + z = 3, \; x + y - z = 1, \; x + y + 4 z = 6$ and $x+y-4z= -1.$
2. Find the row-reduced echelon form of the following matrices.
   $$1. \; \begin{bmatrix}-1 & 1 & 3 &5\\ 1& 3 & 5 &7 \\ 9 & 11 & 13 &... ...4 \\ 2 & 0 & -2 & -4 \\ -6 & -8 & -10 & -12 \\ -2 & -4 & -6 & -8 \end{bmatrix}$$