Page 372 - 35Linear Algebra
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Any value of µ will give a solution of the system, and any system can be written
in this form for some value of µ. Since there are multiple solutions, we can
also express them as a set:
x 1 2 −3
x 2 = 1 + µ 0 µ ∈ R .
x 3 0 1
Worked Examples of Gaussian Elimination
Let us consider that we are given two systems of equations that give rise to
the following two (augmented) matrices:
2 5 2 0 2 5 2 9
1 1 1 0 1 0 5 10
1 4 1 0 1 0 3 6
and we want to find the solution to those systems. We will do so by doing
Gaussian elimination.
For the first matrix we have
2 5 2 0 2 1 1 1 0 1
∼
1 1 1 0 1 R 1 ↔R 2 2 5 2 0 2
1 4 1 0 1 1 4 1 0 1
1 1 1 0 1
R 2 −2R 1 ;R 3 −R 1
∼ 0 3 0 0 0
0 3 0 0 0
1 1 1 0 1
1
3 R 2
∼ 0 1 0 0 0
0 3 0 0 0
1 0 1 0 1
R 1 −R 2 ;R 3 −3R 2 0
∼ 1 0 0 0
0 0 0 0 0
1. We begin by interchanging the first two rows in order to get a 1 in the
upper-left hand corner and avoiding dealing with fractions.
2. Next we subtract row 1 from row 3 and twice from row 2 to get zeros in the
left-most column.
3. Then we scale row 2 to have a 1 in the eventual pivot.
4. Finally we subtract row 2 from row 1 and three times from row 2 to get it
into Reduced Row Echelon Form.
Therefore we can write x = 1 − λ, y = 0, z = λ and w = µ, or in vector form
x 1 −1 0
0
y
0
0
= + λ + µ .
z 0 1 0
w 0 0 1
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