Download Algorithm for Solving Linear Equations: §1.2, §1.3 Professor Karen

Algorithm for Solving Linear Equations: §1.2, §1.3
Professor Karen Smith
Winter 2017
(c)2015 UM Math Dept
licensed under a Creative Commons
By-NC-SA 4.0 International License.
Vocabulary: Augmented matrix, row reduced echelon form, pivot (or leading 1), free variables,
dependent variables, Gauss-Jordan elimination, matrix multiplication, matrix addition.
A. When propane gas burns, the resulting chemical reaction transforms propane (C3 H8 ) and oxygen
(O2 ) to form carbon dioxide (CO2 ) and water (H2 O). In the reaction, the numbers of each atom
must remain constant, which allows us to find the numbers of each of the molecules in the reaction
(that is, we have (x1 )C3 H8 + (x2 )O2 = (x3 )CO2 + (x4 )H2 O). This leads to the system of equations
3x1
= x3
8x1
=
2x4
2x2 = 2x3 + x4 .
1. Explain the physical meaning of these equations. What does each variable xi represent?
What does each equation represent?
2. Write your system of linear equations in the form Ax = b where A, x and b are matrices.
Find an augmented matrix representing this system of linear equations.
3. Row-reduce your augmented matrix to reduced row-echelon form (rref). Identify the pivots,
the free variables, and the dependent variable. Solve the system for the xj . Write your answer
 
a
b

in the form {
 c  t | t ∈ N}.
d
4. What is the physical interpretation of your answer? Does it make physical sense that there
is more than one solution?
Solution note: We have
3x1
8x1
− x3
=0
− 2x4 = 0
2x2 − 2x3 − x4 = 0.
The augmented matrix is thus

3 0 −1 0
A = 8 0 0 −2
0 2 −2 −1
We reduce as follows.



0
3 0 −1 0
1
8 0 0 −2
0 → 8
0 2 −2 −1
0
0

1
→ 0
0

1

→ 0
0

0
0 .
0


0
1 0 −1/3 0
0 → 0 0 8/3 −2
0
0 2 −2 −1


0
1 0 0 −1/4
−1/3
0
1
−3/4
0 → 0 0 1 −3/4
−2
−1
0
0 2 0 −5/2



1 0 0 −1/4
0 −1/4
0
0
1 −3/4
0 → 0 1 0 −5/4
0
0 −5/4
0
0 0 1 −3/4
0
0 −1/3 0
0
0
−2
2 −2 −1
0
0
2
0
0
1

0
0
0

0
0
0
We have one free variable, namely x4 , and solving for the dependent variables in
terms of the free variable, we have x3 = 3/4x4 , x2 = 5/4x4 and x1 = 1/4x4 . Since
the xi represent numbers of molecules,
  they
 can not be fractional quantities, so we
x2
1
x2  5
  
scale by 4. The answer is that 
x3  = 3 t where t can be any whole number.
x4
4
B. Consider a matrix A, which we transform to the matrix


0 a 0 0 b
B = c 0 d 0 e 
0 0 0 1 f
by a series of row operations. Assume B = rref(A).
1. What can we say about the constants a–f ?
Solution note: c = 0; a = 1; d = 1; the remaining constants are arbitrary.
2. We define (definition 1.3.2) the rank of a matrix A as the number of leading 1s in rref(A).
What is the rank of the matrix A?
Solution note: rank = 3.
3. If A | ~0 is the augmented matrix for a linear system, is the system consistent? If it is, how
many solutions are there? How many degrees of freedom?
Solution note: Yes, infinitely many. Two degrees of freedom.
4. Now, if we want to go further, suppose that A | ~v (where ~v 6= ~0) is the augmented matrix
for a linear system, is the system consistent? If it is, how many solutions are there?
Solution note: Yes, infinitely many.
5. Continue to suppose that A | ~v (where ~v 6= ~0) is the augmented matrix for a linear system.
How can you change the last row of B to allow the possibility that there could be no solutions
to the system? Can you change the last row of B to ensure a unique solution?
Solution note: Change the 1 in the third row/fourth column to a zero, and assume
also that f = 0. Then for certain choices of ~v (whose last entry after row reducing is
non-zero), the system is inconsistent. You can never get a unique solution because x
can be any real number.
C. Let S be a system of 3 linear equations in 5 variables v, w, x, y, z.
1. What do you expect for the shape of the solution set of S? What do you expect probabilistically?
2. Suppose the augmented matrix of S has row

0 1 0
0 0 1
0 0 0
reduced echelon form

2 0 3
2 0 4 .
0 1 5
3. Identify the pivots, the free variables, and the dependent variables. What is the rank of S?
4. Write the solutions to S in the form {~v1 s + ~v2 t + ~v0 | s, t ∈ R} for some ~v1 , ~v2 , ~v0 ∈ R5 .
Solution note: We expect two free variables, so we expect a “plane” in R5 . The pivots
are the three leading ones in columns 2, 3, 5. Thus the free variables are x1 and x4 .
The dependent variables are x2 , x3 and x5 . The equivalent system we get from the
rref is
x2
+2x4
= 3
x3 +2x4
= 4 ,
x5 = 5
so moving the free variables to the right side of the equations we have
x2 = −2x4 +3
x3 = −2x4 +4 ,
x5 =
5


x1
−2x4 + 3


 where x1 and x4 can be any real numbers. In the desired
−2x
+
4
So the answer is 
4


 x4

5
 
 
 
0
1
0
3
0
−2
 
 
 

 
 
format, we see ~v0 = 
4 , ~v1 = 0 and ~v2 = −2 .
0
0
1
5
0
0
D. DeMorgan’s First Law states if A and B are sets, then (A ∪ B)C = AC ∩ B C .
1. Draw a Venn diagram to understand DeMorgan’s Law and why it should be true.
2. Proof Technique: To show two sets X and Y are equal, show separately that X ⊂ Y and
that Y ⊂ X.
Scaffold a proof of Demorgan’s Law. That is, write an outline of a possible proof, which starts
with a phrase like ”Given..., we need to show...” and continues ”First we show...”. ”Next we
show...” and finally should conclude with a phrase like ”We conclude that ...”.
3. Proof Techniques:
(a) A standard way to start a proof that X ⊂ Y is to say “Let x be an element in X” and
then show somehow that x is also in Y .
(b) A standard way to show an element x is in Z ∩ W is to show that x ∈ Z and that x ∈ W
separately.
Use these techniques to complete a proof of Demorgan’s Laws.
Solution note: Proof: First we show that (A ∪ B)C ⊂ AC ∩ B C . Take x ∈ (A ∪ B)C .
This means that x ∈
/ A ∪ B. This means that x ∈
/ A and also that x ∈
/ B. That is,
C
C
C
C
x ∈ A and x ∈ B . We conclude that x ∈ A ∩ B , as needed.
Next we show AC ∩ B C ⊂ (A ∪ B)C . Take x ∈ AC ∩ B C . This means that x ∈ AC
and also that x ∈ AC . So x ∈
/ A and x ∈
/ B. That is, x ∈ (A ∪ B)C . QED
Bonus: Is it also true that (A ∩ B)C = AC ∪ B C ? Prove or give a counterexample.