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4.7 Echelon and Reduced Echelon Forms of Matrices

Echelon form, reduced echelon form, finding matrix inverse using row operations, and matrix rank.

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4.7 Echelon and Reduced Echelon Forms of Matrices

In any non-zero row of a matrix, the first non-zero entry is called the leading entry of that row.

Echelon Form of a Matrix
An m×nm \times n matrix AA is called in echelon form if:

(i) The number of zeros before the leading entry is greater than the number zeros in the preceding row.
(ii) Every non-zero row in AA precedes every zero row (if any).
(iii) The first non-zero entry (or leading entry) in each row is 1.

The matrices

[01−2400120000]and[12−3400120001]\begin{bmatrix} 0 & 1 & -2 & 4 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{bmatrix} \quad \text{and} \quad \begin{bmatrix} 1 & 2 & -3 & 4 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \end{bmatrix}

are in echelon form.

Reduced Echelon Form of a Matrix: An m×nm \times n matrix AA is said to be in reduced (row) echelon form if the first non-zero entry (or leading entry) in RiR_i lies in CjC_j, then all other entries of CjC_j are zero.

The matrices

[010400120000]and[120000100001]\begin{bmatrix} 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{bmatrix} \quad \text{and} \quad \begin{bmatrix} 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}

are in (row) reduced echelon form.

Example 8

Reduce

[23−191−12−33132]\begin{bmatrix} 2 & 3 & -1 & 9 \\ 1 & -1 & 2 & -3 \\ 3 & 1 & 3 & 2 \end{bmatrix}

to (row) echelon and reduced (row) echelon form.

Solution

[23−191−12−33132]≅[1−12−323−193132]By R1↔R2\begin{bmatrix} 2 & 3 & -1 & 9 \\ 1 & -1 & 2 & -3 \\ 3 & 1 & 3 & 2 \end{bmatrix} \cong \begin{bmatrix} 1 & -1 & 2 & -3 \\ 2 & 3 & -1 & 9 \\ 3 & 1 & 3 & 2 \end{bmatrix} \quad \text{By } R_1 \leftrightarrow R_2 ≅[1−12−305−51504−311]By R2+(−2)R1→R2′ and R3+(−3)R1→R3′\cong \begin{bmatrix} 1 & -1 & 2 & -3 \\ 0 & 5 & -5 & 15 \\ 0 & 4 & -3 & 11 \end{bmatrix} \quad \text{By } R_2 + (-2)R_1 \rightarrow R'_2 \text{ and } R_3 + (-3)R_1 \rightarrow R'_3 ≅[1−12−301−1304−311]By 15R2→R2′\cong \begin{bmatrix} 1 & -1 & 2 & -3 \\ 0 & 1 & -1 & 3 \\ 0 & 4 & -3 & 11 \end{bmatrix} \quad \text{By } \frac{1}{5}R_2 \rightarrow R'_2 ≅[1−12−301−13001−1]By R3+(−4)R2→R3′\cong \begin{bmatrix} 1 & -1 & 2 & -3 \\ 0 & 1 & -1 & 3 \\ 0 & 0 & 1 & -1 \end{bmatrix} \quad \text{By } R_3 + (-4)R_2 \rightarrow R'_3 ≅[10010102001−1]By R1+(1)R3→R1′ and R2+1R3→R2′\cong \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -1 \end{bmatrix} \quad \text{By } R_1 + (1)R_3 \rightarrow R'_1 \text{ and } R_2 + 1R_3 \rightarrow R'_2

Thus

[1−12−301−13001−1]and[10010102001−1]\begin{bmatrix} 1 & -1 & 2 & -3 \\ 0 & 1 & -1 & 3 \\ 0 & 0 & 1 & -1 \end{bmatrix} \quad \text{and} \quad \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -1 \end{bmatrix}

are (row) echelon and reduced (row) echelon forms of the given matrix respectively.

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