Programming + Errors: Rank of a Matrix Continued.....

In previous sections, we solved linear systems using Gauss elimination method or the Gauss-Jordan method. In the examples considered, we have encountered three possibilities, namely

existence of a unique solution,
existence of an infinite number of solutions, and
no solution.

Based on the above possibilities, we have the following definition.

DEFINITION 2.4.1 (Consistent, Inconsistent) A linear system is called CONSISTENT if it admits a solution and is called INCONSISTENT if it admits no solution.

The question arises, as to whether there are conditions under which the linear system $A {\mathbf x}= {\mathbf b}$ is consistent. The answer to this question is in the affirmative. To proceed further, we need a few definitions and remarks.
Recall that the row reduced echelon form of a matrix is unique and therefore, the number of non-zero rows is a unique number. Also, note that the number of non-zero rows in either the row reduced form or the row reduced echelon form of a matrix are same.

DEFINITION 2.4.2 (Row rank of a Matrix) The number of non-zero rows in the row reduced form of a matrix is called the row-rank of the matrix.

By the very definition, it is clear that row-equivalent matrices have the same row-rank. For a matrix

we write ` ${\mbox{row-rank }}(A)$ ' to denote the row-rank of

EXAMPLE 2.4.3

Determine the row-rank of $A = \begin{bmatrix}1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 2 \end{bmatrix}.$
Solution: To determine the row-rank of we proceed as follows.
1. $\begin{bmatrix}1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 2 \end{bmatrix} \overrightarr... ..._{31}(-1)} \begin{bmatrix}1 & 2 & 1 \\ 0 & -1 & -1 \\ 0 & -1 & 1 \end{bmatrix}.$
2. $\begin{bmatrix}1 & 2 & 1 \\ 0 & -1 & -1 \\ 0 & -1 & 1 \end{bmatrix} \overright... ..., R_{32}(1) } \begin{bmatrix}1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{bmatrix}.$
3. $\begin{bmatrix}1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{bmatrix}\overrightarro... ...R_{12}(-2) } \begin{bmatrix}1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}.$
4. $\begin{bmatrix}1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}\overrightarr... ...1), R_{13}(1)}\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
The last matrix in Step 1d is the row reduced form of which has non-zero rows. Thus, ${\mbox{row-rank}}(A)~=~3.$ This result can also be easily deduced from the last matrix in Step 1b.
Determine the row-rank of $A = \begin{bmatrix}1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 0 \end{bmatrix}.$
Solution: Here we have
1. $\begin{bmatrix}1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 0 \end{bmatrix} \overrightarr... ...31}(-1) } \begin{bmatrix}1 & 2 & 1 \\ 0 & -1 & -1 \\ 0 & -1 & -1 \end{bmatrix}.$
2. $\begin{bmatrix}1 & 2 & 1 \\ 0 & -1 & -1 \\ 0 & -1 & -1 \end{bmatrix} \overrigh... ..., R_{32}(1) } \begin{bmatrix}1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}.$
From the last matrix in Step 2b, we deduce ${\mbox{row-rank}}(A)=2.$

Remark 2.4.4 Let $A {\mathbf x}= {\mathbf b}$ be a linear system with equations and unknowns. Then the row-reduced echelon form of agrees with the first columns of $[A \; \; {\mathbf b}],$ and hence

$\displaystyle {\mbox{row-rank}} (A) \leq {\mbox{row-rank}} ([A \; \; {\mathbf b}]).$

The reader is advised to supply a proof.

Remark 2.4.5 Consider a matrix After application of a finite number of elementary column operations (see Definition 2.3.16) to the matrix we can have a matrix, say which has the following properties:

The first nonzero entry in each column is
A column containing only 0 's comes after all columns with at least one non-zero entry.
The first non-zero entry (the leading term) in each non-zero column moves down in successive columns.