Download methods of solving a linear system – echelon form

1
● METHODS OF SOLVING A LINEAR SYSTEM – ECHELON FORM
𝑥 + 3𝑦 − 2𝑧 = 5
3𝑥 + 5𝑦 + 6𝑧 = 7
2𝑥 + 4𝑦 + 3𝑧 = 8
Consider the following linear system :
There are several algorithms for solving a system of linear equations.
1. Elimination of variables – no matrices
Substitution: The simplest method for solving a system of linear equations is to repeatedly eliminate variables.
2. Row reduction (Gaussian elimination) – Augmented matrix
In row reduction, the linear system is represented as an augmented matrix:
1
[3
2
3
5
4
−2 5
6 | 7]
3 8
This matrix is then modified using elementary row operations until it reaches reduced echelon form.
There are three types of elementary row operations:
Type 1: Interchange equations → Swap the positions of two rows.
Type 2: Multiply equation → Multiply a row by a nonzero scalar.
Type 3: Any of these equations can be replaced by a multiple of itself ± a multiple of another equation
without changing the outcome → Add to one row a scalar multiple of another.
The augmented matrix produced always represents a linear system that is equivalent to the original.
The following computation shows Gauss-Jordan elimination applied to the matrix above:
Combine two at the time to eliminate one variable (for example 𝑥 from two equations) and then combine these
two to eliminate the second variable from one of them (for example 𝑦)
𝑥 + 3𝑦 − 2𝑧 = 5
3𝑥 + 5𝑦 + 6𝑧 = 7
2𝑥 + 4𝑦 + 3𝑧 = 8
1
[3
2
[𝑅2 − 3𝑅1 ] ~
3 −2 5
5 6 | 7]
4 3 8
[𝑅2 − 2𝑅3 ] ~
~
𝑥 + 3𝑦 − 2𝑧 = 5
−2𝑧 = −4
−2𝑦 + 7𝑧 = −2
1
[0
0
3
0
−2
𝑥 + 3𝑦 − 2𝑧 = 5
−4𝑦 + 12𝑧 = −8
2𝑥 + 4𝑦 + 3𝑧 = 8
1
[0
2
3
−4
4
[𝑅3 − 2𝑅1 ]
−2 5
12 | −8]
3 8
𝑅2 = 𝑅3
𝑅3 = −𝑅2
−2 5
−2| −4]
7 −2
~
~
~
~
𝑥 + 3𝑦 − 2𝑧 = 5
−4𝑦 + 12𝑧 = −8
−2𝑦 + 7𝑧 = −2
1
[0
0
3
−4
−2
−2 5
12 | −8]
7 −2
𝑥 + 3𝑦 − 2𝑧 = 5
−2𝑦 + 7𝑧 = −2
2𝑧 = 4
1
[0
0
3
−2
0
−2 5
7 | −2]
2 4
The last matrix is in reduced row echelon form – augmented matrix with the zeroes in the bottom left corner.
It represents the solution of the system of three linear equations.
𝒛=𝟐
𝐲=𝟖
𝒙 = −𝟏𝟓
(−2y + 2 ∙ 7 = −2 )
(𝑥 + 3 ∙ 8 − 2 ∙ 2 = 5)
2
● THE MEANING OF THE SOLUTIONS
THE A LINEAR SYSTEM MAY BEHAVE IN ANY OF THREE POSSIBLE WAYS
1. The system has a single unique solution.
2. The system has no solution.
3. The system has infinitely many solutions.
Geometric interpretation
2–D
∎ For a system involving two variables (x and y), each linear equation determines a line on the xy-plane. Because a
solution to a linear system must satisfy all of the equations, the solution set is the intersection of these lines, and is
hence either
unique solution: a single point
no solution: the empty set
infinitely many solutions: a line
parallel lines
The slopes are equal – no common points
coincident lines
The slopes are equal – one common point
⇒ all points are common
3–D
∎ For three variables, each linear equation determines a plane in 3-D space, and the solution set is the intersection
of these planes. Thus the solution set may be a plane, a line, a single point, or the empty set.
∎ For n variables, each linear equation determines a hyperplane in n-D space. The solution set is the intersection
of these hyperplanes, which may be a flat of any dimension.
A 𝒏 × 𝒏 homogeneous system of linear equations has a unique solution (the trivial solution) if and only if its
determinant is non-zero. If this determinant is zero, then the system has an infinite number of solutions.
3
● SOLUTIONS OF A LINEAR SYSTEM USING AUGMENTED MATRIX IN ECHELON FORM
using row reduction (Gaussian elimination) for augmented matrix to get echelon form
System
a11x1 + a12x2 + a13x3 = b1
a 21x1 + a 22x2 + a23x3 = b2
a 31x1 + a32x2 + a33x3 = b3
Augmented form of the matrix of coefficients of the system
𝑎11
𝑎
[ 21
𝑎31
𝑎12
𝑎22
𝑎32
𝑎13 𝑏1
𝑎23 | 𝑏2 ]
𝑎33 𝑏3
Echelon form
𝑎
[0
0
𝑏 𝑐 𝑑
𝑘
→ 𝑦&𝑥
𝑒 𝑓 | 𝑔] → 𝑧 =
ℎ
0 ℎ 𝑘
𝒖𝒏𝒊𝒒𝒖𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏:
𝒏𝒐 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏:
𝑘
ℎ ≠0 → 𝑧 =ℎ → 𝑦&𝑥
(𝑘 𝑚𝑎𝑦 𝑜𝑟 𝑚𝑎𝑦 𝑛𝑜𝑡 𝑏𝑒 0)
ℎ = 0 𝑎𝑛𝑑 𝑘 ≠ 0 → 0 ∙ 𝑧 = 𝑘 ≠ 0 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑎𝑏𝑠𝑢𝑟𝑑 → 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑛𝑜 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑠𝑦𝑠𝑡𝑒𝑚 𝑖𝑠 𝒊𝒏𝒄𝒐𝒏𝒔𝒊𝒔𝒕𝒆𝒏𝒕
𝒊𝒏𝒇𝒊𝒏𝒊𝒕𝒆𝒍𝒚 𝒎𝒂𝒏𝒚 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏𝒔: ℎ = 0 𝑎𝑛𝑑 𝑘 = 0 → 𝑧 𝑐𝑎𝑛 𝑏𝑒 𝑎𝑛𝑦 𝑛𝑢𝑚𝑏𝑒𝑟, 𝑠𝑜 𝑤𝑒 𝑤𝑟𝑖𝑡𝑒: 𝑧 = 𝑡 𝑤ℎ𝑒𝑟𝑒 𝑡 ∈ 𝑅
𝑥 = 𝑥(𝑡)
𝑦 = 𝑦(𝑡)
𝑧=𝑡
Parametric representation of infinitely many solutions is not unique. We expressed the variable
which was not free (z) in terms of the parameter. We eliminated from row 3 variables x and y. It
does not have to be that way. We could eliminate z and y to get x, and then express y and z.
One can solve for any of the variables. Of course, the solution set will look different. However, it
will still represent the same solutions.
𝑢𝑛𝑖𝑞𝑢𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛:
1 1
[0 2
0 0
𝑛𝑜 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛:
1
[0
0
𝑖𝑛𝑓𝑖𝑛𝑒𝑡𝑒𝑙𝑦 𝑚𝑎𝑛𝑦 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠:
1 1
[0 2
0 0
1 1
2| 2] → 𝑧 = 2
3 6
1 1 1
2 2| 2] → 𝑧 ∙ 0 = 3
0 0 3
𝑦 = −1 𝑥 = 0 𝑃(0, −1,2)
𝑎𝑏𝑠𝑢𝑟𝑑
1 1
𝑧∙0=0 ⇒ 𝑧 =𝑡𝜖𝑅
2| 2] → ⏟
𝑡𝑟𝑢𝑒
𝑓𝑜𝑟 𝑎𝑛𝑦 𝑧
0 0
𝑦 =1−𝑡
𝑥=0
4
● INTERSECTION OF TWO or MORE PLANES