TEXT 2.3
#4
We want to find scalars x and y such that:
0  3
 1
   
 
x  1 + y  1 =  2
 1   1
0
 
   
Expending, we obtain the system:
x
= 3
x + y= 2
y = –1
Since x = 3 and y = –1 implies x + y = 2, those values of x and y are clearly the solution.
So the solution is x = 3 , y = –1, and the linear combination is
0  3
 1
   
 
3  1 – 1  1 =  2 .
 1   1
0
 
   
#8
Applying Theorem 2.4, we check to see if [A│b] is consistent. Why?
Theorem 2.4 says [A│b] is consistent  b is linear combination of the columns of A.
That is exactly what is required for b to be in the span of the columns of A.
28 
1 0  1
 

3

 1 2 3 10 


 
Row reduce [A│b] =  4 5 6 11  0 1 2
29  to see that it is consistent .
 7 8 9 12 
3

 

0
0
0 0


What do we conclude? The vector b is in the span of the columns of A.
The solution tells us one possible linear combination is
 1
2 10
28   29    

4 
5  11 .
3   3    
7
8 12
 
# 10
 3
0  a 
We must show x   + y   =   can always be solved.
  2
 1  b 
 3 0 a
The augmented matrix is 
 , and row reduction produces:
  2 1 b
1
R
R2  2R1
3 1
 3 0 a


  2 1 b

 1 0 a / 3


b
 2 1

a / 3
1 0


0 1 (2a  3b) / 3
We see that x = a/3 and y = (2a+3b)/3, so for any choice of a and b we have
 a   3  2a  3b  0  a 
   +
   =   Check this!
 3    2  3   1  b 
# 16
We should describe the span of the given vectors (a) geometrically and (b) algebraically.
 1
  1
 
 
(a) Geometrically, we can see that the set of linear combination of  0 and  1 is just
  1
 0
 
 
 1
  1
 
 
the plane though the origin with  0 and  1 as direction vectors.
  1
 0
 
 
Q: Why do we get to ignore the third vector, v3 = [0, –1, 1] ?
A: Since v3 = –v1 + v2, it does not affect the span. Why not?
 1
  1
 
 
 0 + t  1 .
  1
 0
 
 
 1
  1
 x
 
 
That is just another way of saying that  y  is in the span of  0 and  1 .
 
  1
 0
 z 
 
 
Suppose we want to obtain the general equation of this plane.
One method is to use the system of equation arising from the vector equation:
s–t= x
1  1
x
 1  1 x




1
y
1 y   0
t= y   0




0 x  y  z
 1 0 z 
0
–s
= z
 x
(b) Algebraically, the vector equation of this plane is  y  = s
 
 z 
 1
 x
 


We know this system is consistent, since y is in the span of  0 and
 
  1
 z 
 
So, we must have x + y + z = 0, giving us the general equation we seek.
 1
  1
1
 
 

Note: Both  0 and  1 are orthogonal to 1 . Should they be?
  1
 0
1
 
 

  1
 
 1 .
 0
 
# 24
Since there is no obvious dependence relation here.
 1 0
2
 3
 
Find scalars c1, c2 and c3 such that c1  2  + c2 1  + c3   5 = 0 .
 
 
 
 2  0
1 
 2 
 
Form the linear system, its associated augmented matrix, and row reduce to solve:
2c1 + 3c2 + c3 = 0
1 0
2 3
 1 0  4 0




2c1 + c2 – c3 = 0  2 1  5 0  0 1
3 0
1 2
0 0
2 0
0 0


c1 + 2c2 + c3 = 0
Since c1 = 4c3 and c1 = –3 c3 is a solution, the vectors are linearly dependent.
2
 3  1 0
 
One dependence relationship is: 4  2  – 3 1  +   5 = 0 . Are there others?
 
 
 
1 
 2   2  0
# 34
We construct a matrix with these vectors as its rows and proceed to reduce it to echelon form.
Each time a row changes, we denote the new row by adding a prime symbol:
R1’=R3
R2’=R1
R3’=R2
R2’’=R2’–2R1’
R3’’=R3’–3R1’
R3’’’=R3’’–
4
3
R2’’
1  5 2 
 v1   2 2 1 
1  5 1 






0 12  3
A = v2 = 3 1 2
2 2 1


  



0 16  4
v 3  1  5 2
 3 1 2
1  5 2 
0 12  3


0 
0 0
We can stop. Why? We have created a zero row. What does that tell us?
Since the rank of a matrix is the number of nonzero rows in its row echelon form, rank (A) = 2.
What do we conclude? We conclude v1 , v2 and v3 are linearly depended. How?
Theorem 2.7 states that v1 , v2 …vm are linearly depended if and only if rank (A) < m.
So, since rank (A) =2< 3, Theorem 2.7 implies v1 , v2 and v3 are linearly depended.
4
Furthermore, from the row reduction above, we see that: 0 = R3’’’ = R3’’  R2’’.
3
Multiplying both sides by 3 implies: 0 = R3’’’ = 3R3’’ – 4 R2’’. Substituting, we have:
0 = 3 (R3’ – 3R1’) – 4 (R2’ – 2R1’) = 3(R2 – 3R3) – 4 (R1 – 2R3) = – 4 R1 + R2 – R3.
Multiplying both sides by –1 yields this dependence relation among the original vectors:




2
4 2 – 3
 
1 
 3  1   0 
1  +   5 = 0 .
     
 2   2  0
# 42
We will use theorem of Section 2.2 and 2.3 to support the conclusion below.
(a) Given A = [v1 , v2 …vn], we will show that rank (A) = n.
Theorem 2.6 Section 2.3 implies:
Vectors v1 , v2 …vn are linearly independent if and only if the solution of
[A│b] = [v1 , v2 …vn │0] is trivial solution.
Therefore, the number of free variables in the associated system is 0.
So, Theorem 2.2 Section 2.2 (The Rank Theorem) implies:
number of free variables = 0 = n – rank (A)  rank (A) = n, as claimed:
 v1 
v 
(b) Given A =  2  we will show that rank (A) = n.

 
v n 
Theorem 2.7 Section 2.3 implies:
Vectors v1 , v2 …vn are linearly independent if and only if
 v1 
v 
the rank of A =  2  is greater than or equal to n. That is, rank (A) ≥ n.

 
v n 
But A has n rows therefore we have n ≤ rank (A) ≤ n  rank (A) = n, as claimed.
TEXT 2.4
#2
Let x1, x2 and x3 be the number of bacteria of stains I, II and III, respectively.
x1 + 2x2
= 400
2 0
 1 2 0 400
1 0




2x1 + 2x2 + 3x3 = 500  2 1 3 500  0 1  1 0 .
 1 1 1 600
0 0
0 1



x1 + x2 + x3 = 600
This is an inconsistent system, so the bacteria cannot coexist.
#8
6CO2 + 6H2O  C6H12O6 + 6O2
# 16
(a) By applying the conservation of floe rule to each node we obtain the system of equations
f1 + f2 = 20
f1 + f3= 25 f2 + f4= 25
f3 + f4= 30
We form the augmented matrix and perform Gauss-Jordan elimination to get
1

1
0

0
1 0 0 20

0 1 0 25
1 0 1 25

0 1 1 30

1

0
0

0
0 0  1  5

1 0
1 25
0 1
1 30

0 0
0
0
Letting f4 = t, the possible flows are f1 = –5 + t, f2 = 25 – t, f3 = 30 – t, f4 = t.
(b) If f4 = t = 10, then average flows on other street will be f1 = 5, f2 = 15 and f3 = 20.
(c) Each flow must be nonnegative so the set of solutions gives the four constraints
t ≥ 5, t ≤ 25, t ≤ 30, t ≥ 0. To satisfy all four constraints t must satisfy 5 ≤ t ≤ 25.
Combining the constraints on t with the four solutions,
we see 0 ≤ f1 ≤ 20, 0 ≤ f2 ≤ 20, 5 ≤ f3 ≤ 25, 5 ≤ f4 ≤ 25.
(d) Reversing all directions would have no effect on the solution since this would be equivalent
to multiplying each row of the augmented matrix by –1. But this is an elementary row operation
so the new matrix would have the same reduced row echelon form, and the solution will be
same.