linear-guest

66 Systems of Linear Equations
Example 32 (Non-pivot variables determine the gemometry of the solution set)
⎛ ⎞
⎛
⎞ x
1 0 1 −1 1
⎛ ⎞ ⎧
⎝0 1 −1 1⎠
⎜x 2
1 ⎨ 1x 1 + 0x 2 + 1x 3 − 1x 4 = 1
⎟
⎝x 0 0 0 0 3
⎠ = ⎝1⎠ ⇔ 0x 1 + 1x 2 − 1x 3 + 1x 4 = 1
⎩
0 0x
x 1 + 0x 2 + 0x 3 + 0x 4 = 0
4
Following the standard approach, express the pivot variables in terms of the non-pivot
variables and add “empty equations”. Here x 3 and x 4 are non-pivot variables.
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
x 1 = 1 − x 3 + x 4
⎫⎪
x 1 1 −1 1
⎬
x 2 = 1 + x 3 − x 4
⇔ ⎜x 2
⎟
x 3 = x 3
⎝
⎪ x ⎭ 3
⎠ = ⎜1
⎟
⎝0⎠ + x ⎜ 1
⎟
3 ⎝ 1⎠ + x ⎜−1
⎟
4 ⎝ 0⎠
x 4 = x 4 x 4 0 0 1
The preferred way to write a solution set S is with set notation;
⎧⎛
⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎫
x 1 1 −1 1
⎪⎨
S = ⎜x 2
⎟
⎝x ⎪⎩ 3
⎠ = ⎜1
⎟
⎝0⎠ + µ ⎜ 1
⎟
1 ⎝ 1⎠ + µ ⎜−1
⎪⎬
⎟
2 ⎝ 0⎠ : µ 1, µ 2 ∈ R .
⎪⎭
x 4 0 0 1
Notice that the first two components of the second two terms come from the non-pivot
columns. Another way to write the solution set is
where
S = { x P + µ 1 x H 1 + µ 2 x H 2 : µ 1 , µ 2 ∈ R } ,
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1
−1
1
x P = ⎜1
⎟
⎝0⎠ , xH 1 = ⎜ 1
⎟
⎝ 1⎠ , xH 2 = ⎜−1
⎟
⎝ 0⎠ .
0
0
1
Here x P is a particular solution while x H 1 and xH 2
The solution set forms a plane.
2.5.3 Solutions and Linearity
are called homogeneous solutions.
Motivated by example 32, we say that the matrix equation Mx = v has
solution set {x P + µ 1 x H 1 + µ 2 x H 2 | µ 1 , µ 2 ∈ R}. Recall that matrices are linear
operators. Thus
M(x P + µ 1 x H 1 + µ 2 x H 2 ) = Mx P + µ 1 Mx H 1 + µ 2 Mx H 2 = v ,
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