My Note

Echelon Form

(1) All nonzero rows are above any rows of all zeros.

(2) Each leading entry of a row is in a column to the right of the leading entry of the row above it.

eg:

\(\begin{bmatrix} 1&3&-1&0\\ 0&0&2&4\\ 0&0&0&3\\ 0&4&0&0 \end{bmatrix}\)

(3) All the number under the leading entries are zeros.

(4) All the leading entries are ones.

(5) Leading ones are the only nonzeros in its columns.

All,

a reduced echelon form!

\(\begin{bmatrix} (1)&0&-1&0&0&0\\ 0&0&2&4&(1)&0\\ 0&(1)&3&3&0&0\\ 0&0&0&0&0&(1) \end{bmatrix}\)

and (1)s are the pivots! (only interchange and replacement)

\(\begin{bmatrix} 0&0&\ldots&0\\ 0&0&\ldots&0\\ \vdots&\vdots&\ddots&0\\ 0&0&0&0 \end{bmatrix}\)

eg:

\(\begin{bmatrix} 0 & -3 & -6 & 4 & 9 \\ -1 & -2 & -1 & 3 & 1 \\ -2 & -3 & 0 & 3 & -1 \\ 1 & 4 & 5 & -9 & -7 \end{bmatrix} \;\xrightarrow{R_1 \leftrightarrow R_4}\; \xrightarrow{R_1+R_2,\; 2R_1+R_3}\; \xrightarrow{ -\frac{5}{2} R_2 + R_3,\; \frac{3}{2} R_2 + R_4 }\; \xrightarrow{R_3 \leftrightarrow R_4} \begin{bmatrix} 1 & 4 & 5 & -9 & -7 \\ 0 & 2 & 4 & -6 & -6 \\ 0 & 0 & 0 & -5 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\)

All in all,

• The algorithm that follows consists of five steps.
  

• The first four steps produces a matrix in echelon form.
  
  • The combination of steps 1–4 is called the forward phase.
    
• The fifth step produces a matrix in reduced echelon form.
  
  • Step 5 is called the backward phase.
    
  • The algorithm is pretty much the same as in Lecture.
    
    

Solution of Linear System:

\(\begin{bmatrix} 1 & 6 & 0 & 3 & 0 & 0 \\ 0 & 0 & 1 & -4 & 0 & 5 \\ 0 & 0 & 0 & 0 & 1 & 7 \end{bmatrix} \rightarrow \begin{cases} x_1 + 6x_2 + 3x_4 = 0 \\ x_3 - 4x_4 = 5 \\ x_5 = 7 \end{cases}\)

Therefore,

\(\begin{cases} x_1 = -6x_2 - 3x_4 \\ x_2\;is\;free \\ x_3 = 5 + 4x_4 \\ x_4\;is\;free \\ x_5 = 7 \end{cases}\)

A linear system is consistent if and only if the rightmost column of the augmented matrix is not a pivot column – that is, if and only if an echelon form of the augmented matrix has no row of the form:

\(\begin{bmatrix} \ldots&\ldots&b \end{bmatrix}\)

and b is nonzero.