TMA4110
Matematikk 3
Autumn 2013
Norwegian University of Science
and Technology
Institutt for matematiske fag
Solutions to exercise set 5
Exercises from the textbook, p. 32-34
11 As we know, b is a linear combination of a1 , a2 and a3 if and only if the linear system
with the augmented matrix


1 0 5
2
[a1 a2 a3 b] = −2 1 −6 −1
0 2 8
6
has a solution. The reduced echelon form

1 0
0 1
0 0
of this matrix is

5 2
4 3
0 0
The matrix is consistent, hence b is a linear combination of a1 , a2 and a3 .
15 We consider the matrix

1 −5 3
[a1 a2 b] =  3 −8 −5
−1 2
h

and reduce it to an echelon form








1 0 −7
1 −5
3
1 −5
3
1 −5 3
 3 −8 −5 → 0 7
−14  → 0 1
−2  → 0 1 −2 
0 0 h−3
−1 2
h
0 −3 h + 3
0 −3 h + 3
The matrix is consistent if and only if h − 3 = 0, hence b is a linear combination of
a1 and a2 if and only if h = 3.
17 Some of the vectors in Span(v1 , v2 ) are 0 · v1 + 0 · v2 = 0,
 
 
 
3
−4
−1
1 · v1 + 0 · v2 = 1 , 0 · v1 + 1 · v2 =  0  , 1 · v1 + 1 · v2 =  1  ,
2
1
3




7
0
1 · v1 + (−1) · v2 =  1  , 1/3 · v1 + 1/4 · v2 =  1/3  .
−1
11/12
September 18, 2013
Page 1 of 4
Solutions to exercise set 5
Exercises from the textbook, p. 40-42
3 By the definition,
 
   

1
2
4
1 2 −3 1 −2 = (−2) −3 + 3 1 =  9 
3
1
6
16
1 6

Using the row-vector rule, we get

  


1 · (−2) + 2 · 3
4
1 2 −3 1 −2 = (−3) · (−2) + 1 · 3 =  9 
3
1 · (−2) + 6 · 3
16
1 6
13 u is in the plane spanned by the columns of A if and only if the system Ax = u is
consistent. We write down the augmented matrix and reduce it to echelon form




3 −5 0
1 1 4
−2 6 4 → 0 8 12
1
1 4
0 0 0
We see that the system is consistent, hence u is in the plane spanned by columns of
A.
21 We consider the matrix [v1 v2 v3 ]. This matrix has three columns and four rows. It
can not hvae a pivot position in every row, since this position should be in distinct
columns and the matrix has more rows than columns. Thus v1 v2 v3 do not span
R4 .
29 We start with a matrix in the echelon form that has pivot position in each column
and perform row operations to get a row
form. For example,



1 0 0
1 0
0 1 0 →  2 1
0 0 1
−1 0
equivalent matrix that is not in echelon



0
2 0 −1
0 →  3 1 −1
1
−1 0 1
30 We start with a 3 × 3 matrix in the echelon form that has a row of zeros and perform
row operations to get
example,

1
0
0
a row equivalent matrix that is not in echelon form. For





0 0
1 0 0
−1 −1 −2
1 2 →  2 1 2 →  2
1
2
0 0
−3 0 0
1
2
4
Exercises from the textbook, p. 47-48
11 We consider the coefficient matrix and find its reduced echelon form

 

1 −4 −2 0 3 −5
1 −4 0 0 0 5
0 0


1 0 0 −1

 ∼ 0 0 1 −0 0 −1
0 0
0 0 1 −4 0 0 0 0 1 −4
0 0
0 0 0 0
0 0 0 0 0 0
September 18, 2013
Page 2 of 4
Solutions to exercise set 5
The pivot elements are in the 1st, 3rrd and 5th columns, thus variables x2 , x4 , x6 are
free. The other variables can be expressed as x1 = 4x2 − 5x6 , x3 = x6 and x5 = 4x6 .
The solutions are parametrized by

 
 
 
  
4
0
−5
x1
4x2 − 5x6

1
0
0
x2  
x
2

 
 
 
  

 
 
 
x3  
x6
 = x2 0 + x4 0 + x6  1  .
 =





0
x4  
x4

0
1
 
  
0
0
4
x5   4x6 
0
0
1
x6
x6


1 −1 0
33 Take for example A = 0 1 −1 .
1 1 −2
40 Consider the augmented matrix of the system Ax = b. After reducing it to an
echelon form we get a row [0 0 0 0 c]. It means that A has at most two pivot
positions, then for any vector y the reduced matrix of the system Ax = y has either
no solutions or infinitely many (there is at least one free variable).
Exercises from the textbook, p. 54-55
3
a) The exchange table is
Fuels and Power
0.1
0.8
0.1
Manufacturing
0.1
0.1
0.8
Services
0.2
0.4
0.4
Purchased by:
Fuels and Power
Manufacturing
Services
b) Let pF P , pM and pS be the total annual outputs of these three sectors. We get
the following system of equations
pF P = 0.1pF P + 0.1pM + 0.2pS
pM = 0.8pF P + 0.1pM + 0.4pS
pS = 0.1pF P + 0.8pM + 0.4pS
The augmented matrix of the system is


0.9 −0.1 −0.2 0
−0.8 0.9 −0.4 0
−0.1 −0.8 0.6 0
c) We use for example rref command
reduced echelon form. The solution

1 0
0 1
0 0
in MATLAB to reduced the matrix to the
(up to a rounding error) is

−0.3014 0
−0.7123 0
0
0
The ratios of the outputs are pF P = 0.3014pS and pM = 0.7123pS , pS is a free
variable. If pS = 100 then PF P = 30.14 and pM = 71.23.
September 18, 2013
Page 3 of 4
Solutions to exercise set 5
9 We consider the equation x1 B2 S3 + x2 H2 O = x3 H3 BO3 + x4 H2 S and balance boron,
sulfur, hydrogen, and oxygen. We get four equations:
2x1
= x3
3x1
= x4
2x2 = 3x3 + 2x4
x2
= 3x3
It is a homogeneous system of liner equations

2 0 −1
3 0 0

0 2 −3
0 1 −3
The reduced echelon form is

1
0

0
0
with the coefficient matrix

0
−1

−2
0

0 −1/3
0 −2 

1 −2/3
0
0
0
1
0
0
The general solution is x1 = x4 /3, x2 = 2x4 and x3 = 2/3x4 , where x4 is free. We
take x4 = 3 and get a balaced equation B2 S3 + 6H2 O = 2H3 BO3 + 3H2 S.
12 We write an equation for each intersection and get the following system
x1 + x4 = x2
x2 = x3 + 100
x3 + 80 = x4
The augmented matrix of this system

1 −1
0 1
0 0
is

0
1
0
−1 0 100 
1 −1 −80
The reduced echelon form is


1 0 0 0
20
0 1 0 −1 20 
0 0 1 −1 −80
Then x4 is free, x1 = 20, x2 = x4 + 20, x3 = x4 − 80 and since all variables are
non-negative we get x4 ≥ 80.
September 18, 2013
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