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THE ECHELON FORM [1.2]
The standard echelon form A rectangular matrix is in echelon form (or row echelon form) if it has the following three properties:
- All nonzero rows are above any rows of all zeros.
- Each leading entry of a row is in a column to the right of the leading entry of the row above it.
- All entries in a column below a leading entry are zeros.
ä Each is a nonzero (lead- ing) entry. ä A ∗ can be a non-zero or a zero entry.
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Terminology: Pivots, and pivot columns
PIVOTS
PIVOT COLUMNS
ä Important in capturing the span of the columns of A (called the range of A - to be covered in detail later)
The reduced row echelon form
If a matrix in echelon form satisfies the following additional conditions, then it is in reduced echelon form (or reduced row echelon form): Matlab: rref
- The leading entry in each nonzero row is 1.
- Each leading 1 is the only nonzero entry in its column.
ä Any nonzero matrix may be row reduced (i.e., trans- formed by elementary row operations) into more than one matrix in echelon form, using different sequences of row operations.
ä However, the reduced echelon form one obtains from a matrix is unique:
Each matrix is row equivalent to one and only one reduced echelon matrix.
ä Remember that the permissible row operations are:
- Interchange; 2) addition; 3) scaling.
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Pivot position
ä A pivot position in a matrix A is a location in A that corresponds to a leading 1 in the reduced echelon form of A. A pivot column is a column of A that contains a pivot position.
The pivot columns are 1, 3, 7, and 10
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Pivot position : Example
Example: Row reduce the matrix A below to echelon form, and locate the pivot columns of A.
Solution: The top of the leftmost nonzero column is the first pivot position. A nonzero entry, or pivot, must be placed in this position.
ä Interchange rows 1 and 4 (note: in reality we should interchange rows 1 and 3. Why?)
↓ Pivot
↑ Pivot Column ä Create zeros below the pivot, 1, by adding multiples of the first row to the rows below −→ Next matrix:
Next pivot column
Step 3 Write solutions: solutions depend on parameters which are the free variables.
ä Express basic variables in terms of the free variables
ä For any values given to the free variables you will get a solution
ä For example for the above picture: x 10 = b 4 ; x 7 = b 3 − scalar.x 8 − scalar.x 9 etc..
- Find general solution when augmented matrix is: 1 2 0 0 1 0 0 − 2 1 2 − 1 − 2 2 2 3 − 3 1 2 2 4 − 3 − 4 − 4 − 6 0 0 1 2 − 3 0 − 5 − 3 1 2 0 0 2 0 − 1 1
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ä Get the reduced echelon form [use matlab!]
1 2 0 0 0 0 1 − 5 0 0 1 2 0 0 − 8 6 0 0 0 0 1 0 − 1 3 0 0 0 0 0 1 − 2 1 0 0 0 0 0 0 0 0
ä Basic variables: 1, 3, 5, 6 ä Free variables: 2, 4, 7 ä Right-hand side : 8
(1): x 1 +2x 2 +x 7 = − 5 (2): x 3 +2x 4 − 8 x 7 = 6 (3): x 5 − x 7 = 3 (4): x 6 − 2 x 7 = 1 (5): 0 = 0 (vacuous)
x 1 = − 5 − 2 x 2 − x 7 x 3 = 6 − 2 x 4 + 8x 7 x 5 = 3 + x 7 x 6 = 1 + 2x 7
Note: It is also possible to use the standard (non-reduced) row-echelon form - Requires back substitution. Result is the same.
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- Below is the standard echelon form for the previous example. Find all solutions.
- Find all solutions for which x 4 and x 7 are zero.
- Among these find all solutions for which x 1 is zero.
- We seek 5 numbers x 1 , · · · , x 5 such that their sum is 50, the sum of 3 of them (e.g. the odd-labeled ones) is 25, and the difference between the other 2 is 5. Write the equations to be satisfied and find the general solution.
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