MATH136 Lecture Notes - Lecture 11: Augmented Matrix, Coefficient Matrix, Free Variables And Bound Variables
14 May 2018
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Wednesday, May 24
−
Lecture 11 : Homogeneous systems. (Refers to 2.2)
Concepts:
1. homogeneous systems
2. trivial solution
11.1 Definition − Homogeneous system. A system of the form
a11x1 + a12 x2 + ... + a1nxn = 0
a21x1 + a22 x2 + ... + a2n xn = 0
:
:
am1x1 + am2 x2 + ... + amn xn = 0
is called a homogeneous system. It can be represented more succinctly as [A | 0 ]
where A is it coefficient matrix. Given a system of linear equations
a11x1 + a12 x2 + ... + a1nxn = b1
a21x1 + a22 x2 + ... + a2n xn = b2
:
:
am1x1 + am2 x2 + ... + amn xn = bm
represented as [A | b] we will say that the system [A | 0 ] is its “associated
homogeneous system”.
11.1.1 Note − A homogeneous system always has at least one solution: the trivial
solution x1 = x2 = ... = xn = 0 and so is always consistent.
11.1.2 Claim : A homogeneous system with fewer equations than unknowns must
have infinitely many (non-trivial) solutions.”
Proof of this claim: Suppose a system has m equations and n unknowns where m < n.
Suppose we have transformed the associated augmented matrix [Am × n | 0] into a
RREF matrix [ARREF | 0].
- Each the n columns of the matrix ARREF points to precisely one of the n variables.
- It is possible that ARREF has a row of zeros, but each row which contains non-zero
entries must have a leading-1. So
# basic variables = # leading-1’s ≤ m < # columns = n = # variables
- So “# basic variables < # variables” implies there are free unknowns.
The existence of at least one free variable guarantees that there are infinitely many
solutions, as claimed.
Question: If we have a non-homogeneous system where the number of rows m
is less than the number of variables n are we guaranteed to have infinitely many
solutions (just like homogeneous systems)? The answer is “No”. This is because
non-homogeneous systems may be inconsistent. This can never happen for
homogeneous systems.
11.2 Examples.
We verify that for the system
3x1 + 5x2 − 4x3 = 0
−3x1 − 2x2 + 4x3 = 0
6x1 + x2 − 8x3 = 0
Row reducing to reduced row echelon form we get x1 and x2 as basic variables and x3
as a free variable and obtain the solution set and
x = t ( 4/3, 0, 1) .
Note: It is important to be able to express the complete set of solutions to a system in
the form of a vector equation.
11.2.1 Question: Are non-homogeneous systems with more equations than unknowns
necessarily inconsistent?
Answer: No; we may still end up with an RREF with rows of zeroes at the bottom
which produces free unknowns.
11.2.2 Question: Does a homogeneous system with the same number of equations as
unknowns always have only the trivial solutions? Answer: No. The RREF of the
matrix may have rows of zeroes at the bottom.
11.3 Exercise question − For which values of a, b, c will the following system of
equations be consistent? For those values, give the general solution and state whether or
not it will be unique.
x + z = a
2x + 6y + 4z = b
3y + z = c
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Document Summary
Wednesday, may 24 lecture 11 : homogeneous systems. (refers to 2. 2) It can be represented more succinctly as [a | 0 ] where a is it coefficient matrix. 11. 1. 1 note a homogeneous system always has at least one solution: the trivial solution x1 = x2 = = xn = 0 and so is always consistent. 11. 1. 2 claim : a homogeneous system with fewer equations than unknowns must have infinitely many (non-trivial) solutions. Proof of this claim: suppose a system has m equations and n unknowns where m < n. Suppose we have transformed the associated augmented matrix [am n | 0] into a. Each the n columns of the matrix arref points to precisely one of the n variables. It is possible that arref has a row of zeros, but each row which contains non-zero entries must have a leading-1. # basic variables = # leading-1"s m < # columns = n = # variables.
