Gaussian elimination is a powerful and widely-used method for solving systems of linear equations. It is used across many fields, from engineering to economics, for finding the values of variables that satisfy a system of equations. Whether you’re working with data science models, electrical circuits, or optimization problems, understanding this technique is crucial.

Row Echelon Form

For each row in a matrix, if the row does not consist of only zeros, then the leftmost nonzero entry is called the pivot of that row.

If two leading entries are in the same column, then a row replacement operation can be used to make one of those entries zero.

Then by using the row swapping operation, one can always order the rows so that for every non-zero row, the leading entry is to the right of the leading entry of the row above.

The following matrix is in echelon form

(112305670010110000)\begin{pmatrix} 1 & 1 & 2 & 3 \\ 0 & 5 & 6 & 7 \\ 0 & 0 & 10 & 11 \\ 0 & 0 & 0 & 0 \end{pmatrix}

Reduced Row Echelon Form (RREF)

A matrix is said to be in reduced row echelon form if

  1. Leading entries are 1’s
  2. Leading 1’s are to the right of those above
  3. Rows with all zeroes are at the bottom
  4. Each leading 1’s must be the only non-zero entry in its column

(100122501657500111100000)\begin{pmatrix} 1 & 0 & 0 & -\frac{12}{25} \\ 0 & 1 & \frac{6}{5} & \frac{7}{5} \\ 0 & 0 & 1 & \frac{11}{10} \\ 0 & 0 & 0 & 0 \end{pmatrix}

Row Echelon Form Example

Suppose the goal is to find and describe the set of solutions to the following system of linear equations

2x+yz=8(L1)3xy+2z=11(L2)2x+y+2z=3(L3)\begin{aligned} 2x + y – z &= 8 \quad & (L_1) \\ -3x – y + 2z &= -11 \quad & (L_2) \\ -2x + y + 2z &= -3 \quad & (L_3) \end{aligned}

Then it’s associated augmented matrix is

[2118312112123]\left[ \begin{array}{ccc|c} 2 & 1 & -1 & 8 \\ -3 & -1 & 2 & -11 \\ -2 & 1 & 2 & -3 \\ \end{array} \right]

Applying R2+32R1R2R_2 + \frac{3}{2} R_1 \rightarrow R_2 and R3+R1R3R_3 + R_1 \rightarrow R_3

[21180121210215]\left[ \begin{array}{ccc|c} 2 & 1 & -1 & 8 \\ 0 & \frac{1}{2} & \frac{1}{2} & 1 \\ 0 & 2 & 1 & 5 \\ \end{array} \right]

Applying R34R2R3R_3 – 4 R_2 \rightarrow R_3

[21180121210011]\left[ \begin{array}{ccc|c} 2 & 1 & -1 & 8 \\ 0 & \frac{1}{2} & \frac{1}{2} & 1 \\ 0 & 0 & -1 & 1 \\ \end{array} \right]

Row Echelon Form Example

We will continue from the previous example

Applying 12R1R1\frac{1}{2}R_1 \rightarrow R_1

[1121240121210011]\left[ \begin{array}{ccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 4 \\ 0 & \frac{1}{2} & \frac{1}{2} & 1 \\ 0 & 0 & -1 & 1 \\ \end{array} \right]

Applying 2R2R22R_2 \rightarrow R_2 and R112R2R1R_1 – \frac{1}{2} R_2 \rightarrow R_1

[101301120011]\left[ \begin{array}{ccc|c} 1 & 0 & -1 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & -1 & 1 \\ \end{array} \right]

Applying R2R3R2 R_2 – R_3 \rightarrow R_2 and R1+R3R1R_1 + R_3 \rightarrow R_1

[100201030011]\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -1 \\ \end{array} \right]

Therefore the solution is x=2,y=3,z=1x = 2, y = 3, z = -1

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