Use linear systems to find if a Vector u is in the span of a set of vectors. For example:
Determine if (10, 5, 7) is in span {(3, -1, -3), (-5, -5, 1), (6, 8, 12)}.
Look for a non trivial solution to : x1(3, -1, -3) + x2(-5, -5, 1) + x3(6, 8, 12) = (10, 5, 7) equiv 3x1 - 5x2 + 6x3 = 10 -x1 -5x2 + 8x3 = 5 -3x1 + x2 + 12x3 = 7
Gaussian elimination: Swap the first row and the second row?
x1 + 5x2 -8x3 = -5 3x1 - 5x2 + 6x3 = 10 -3x1 + x2 + 12x3 = 7
R1 = R2 - 3R1 R3 = R3 + 3R1 R2 = R2/(-5)
x1 + 5x2 -8x3 = -5 -20x2 + 30x3 = 25 16x2 - 12x3 = -8 4X2 - 6X3 = -5 16x2 -12x3 = -8 12x3 = 12
R3 = R3 - 4R2
x1 + 5x2 - 8x3 = -5 4x2 -6x3 = -5 12x3 = 12
Back subsitution 4x2 -6 (1) = -5 4x2 = 1 = 1/4
x1 + (5/4) -8 = -5 ⇒ x1 = 7/4
Augmented matrix
Is a simplified way to write a system of linear equations. In our previous example,
3x1 - 5x2 + 6x3 = 10 -x1 -5x2 + 8x3 = 5 -3x1 + x2 + 12x3 = 7
x1 x2 x3 = 3 -5 6 | 10 -1 -5 8 | 5 -3 1 12 | 7
swap R1 and -R2
x1 x2 x3 = 1 5 8 | - 5 3 -5 6 | 10 -3 1 12 | 7
R3 = R3 +R2
x1 x2 x3 = 1 5 8 | - 5 3 -5 6 | 10 0 -4 18 | 17
R2 = R2 -3R1 x1 x2 x3 = 1 5 - 8 | - 5 0 -20 30 | 25 0 -4 18 | 17
R2 = R2 / -5 x1 x2 x3 = 1 5 -8 | - 5 0 4 -6 | -5 0 -4 18 | 17
R3 = R3 + R2 x1 x2 x3 = 1 5 -8 | - 5 0 4 -6 | -5 0 0 12 | 12
Row Echelon form
The form of the augmented matrix of a linear system at which back substitution ca be applied readily. It is characterized by the following features:
- All zero rows are at the bottom
- The first row is the longest non=zero row,
- each subsequent row contains at least 1 less non-zero entry and 1 more zero entry,
- the all-zero entries form a downward echelon structure.
Reduced Row echelon form
RREF is the reduced row echelon form. When you have all the numbers in the table, divide the leftmost number by itself to get 1. Every number above a pivot can also be turned into a zero by… ?
Divide second row by 4 Divide third row by 12
1 5 -8 | - 5 0 1 -3/2 | -5/4 0 0 1 | 1
First row = R1 - 5R2 + 8R3 Second row = R2 + R2 +3/2R3
1 0 0 | 7/4 0 1 0 | 1/4 0 0 1 | 1
21
Do the vectors (3, -1, -3), (-5, -5, 1), (6, 8, 12) span R3? What about (-2, 1, -1), (+8, -4, 5), (18, -9, 11), (-26, 13, -16)
22
x1 = (2, 1, 1, 1) x2 = (1, 1, 1, 1) x3 = (1, 1, 2, 1)
and
y1 = x1 + x2 + x3 y2 = x1 + 2x2 + x3 y3 = x1 + 4x2 + x3