Download Solutions for Assignment 2

Math 266 - Assignment 2 - Winter 2013.
1. Determine if each of the following matrices are in reduced row-echelon form,
row-echelon form, or neither. In each case, find the rank of the matrix.
1 0 0 −1
(i)
0 2 1 0
1 −2 3 −1
(ii)
0 0 0 0
0 0 1
(iii)
0 0 1


1 6 8 −1
(iv)  0 0 1 0 
0 0 0 1
Solution to (i) The matrix is neither REF nor RREF. We can reduce it to an
RREF matrix in the following manner:
1 0 0 −1
1 0 0 −1
1/2(R2) 7→ (R2)
,
0 2 1 0 −−−−−−−−−−−−→ 0 1 1/2 0
therefore the rank is 2.
Solution to (ii) The matrix is an RREF matrix, and the rank is 1.
Solution to (iii) The matrix is neither REF nor RREF. We can reduce it to
an RREF matrix in the following manner:
0 0 1
0 0 1
−(R1) + (R2) 7→ (R2)
,
0 0 1 −−−−−−−−−−−−−−−−→ 0 0 0
therefore the rank is 1.
Solution to (iv) The matrix a REF matrix. We can reduce it to
matrix in the following manner:





1 6 8 −1
1 6 0 −1
1
 0 0 1 0  −8(R2) + (R1) 7→ (R1)  0 0 1 0  (R3) + (R1) 7→ (R1)  0
−−−−−−−−−−−−−−−−−→
−
−−−−−−−−−−−−−−
→
0 0 0 1
0 0 0 1
0
therefore the rank is 3.
an RREF

6 0 0
0 1 0 ,
0 0 1
2. Find all the solutions to each of the following systems of linear equations by
applying the Gaussian algorithm to the corresponding augmented matrix. At
each step state which operations you are performing.

3x1 + 8x2 − 3x3 − 14x4 = 1



2x1 + 3x2 − x3 − 2x4 = 2
(i)
x1 − 2x2 + x3 + 10x4 = 3



x1 + 5x2 − 2x3 − 12x4 = −1

3x1 + 8x2 − 3x3 − 14x4 = 1



2x1 + 3x2 − x3 − 2x4 = −6
(ii)
x1 − 2x2 + x3 + 10x4 = 0



x1 + 5x2 − 2x3 − 12x4 = 8


 x1 − x2 + x3 − 2x4 = 1

−x1 + x2 + x3 + x4 = 0
(iii)
−x1 + 2x2 + 3x3 − x4 = 0



x1 − x2 + 2x3 + x4 = 0


 x1 − x2 + x3 − 2x4 = 0

−x1 + x2 + x3 + x4 = 1
(iv)
−x1 + 2x2 + 3x3 − x4 = 0



x1 − x2 + 2x3 + x4 = 0
Solution to (i) The augmented matrix of this system is


3 8 −3 −14 1
 2 3 −1 −2 2 
.

 1 −2 1
10
3 
1 5 −2 −12 −1
We now apply the Gaussian algorithm to this matrix in order to obtain a RREF
2
matrix.

3 8 −3 −14 1
 2 3 −1 −2 2

 1 −2 1
10
3
1 5 −2 −12 −1





1 8/3 −1 −14/3 1/3
 2 3 −1
2 
−2


 1 −2 1
10
3 
1 5 −2 −12 −1


1 8/3 −1 −14/3 1/3
 0 −7/3 1
22/3 4/3 


 1 −2
1
10
3 
1
5
−2 −12 −1


1 8/3 −1 −14/3 1/3
 0 −7/3 1
22/3 4/3 


 0 −14/3 2
44/3 8/3 
1
5
−2 −12 −1


1 8/3 −1 −14/3 1/3
 0 −7/3 1
22/3
4/3 


 0 −14/3 2
44/3
8/3 
0 7/3 −1 −22/3 −4/3

1 8/3
−1 −14/3 1/3
 0
1
−3/7 −22/7 −4/7

 0 −14/3
8/3
2
44/3
0 7/3
−1 −22/3 −4/3


1 8/3 −1 −14/3 1/3
 0 1 −3/7 −22/7 −4/7 


 0 0
0
0
0 
0 7/3 −1 −22/3 −4/3


1 8/3 −1 −14/3 1/3
 0 1 −3/7 −22/7 −4/7 


 0 0
0
0
0 
0 0
0
0
0


1 0 1/7
26/7 13/7
 0 1 −3/7 −22/7 −4/7 

.
 0 0
0
0
0 
0 0
0
0
0

1/3(R1) 7→ (R1)
−
−−−−−−−−−−−
→
−2(R1) + (R2) 7→ (R2)
−−−−−−−−−−−−−−−−−→
−(R1) + (R3) 7→ (R3)
−−−−−−−−−−−−−−−−→
−(R1) + (R4) 7→ (R4)
−−−−−−−−−−−−−−−−→
−3/7(R2) 7→ (R2)
−−−−−−−−−−−−−→
14/3(R2) + (R3) 7→ (R3)
−−−−−−−−−−−−−−−−−−→
−7/3(R2) + (R4) 7→ (R4)
−
−−−−−−−−−−−−−−−−−−
→
−8/3(R2) + (R1) 7→ (R1)
−
−−−−−−−−−−−−−−−−−−
→
3




The equation corresponding to the above augmented matrix is

x1 + 17 x3 + 26
x = 13

7 4
7


3
22
x2 − 7 x3 − 7 x4 = − 47
.
0=0



0=0
Let s and t be parameters. The solution to the above system an be written in
the parametric form:

s + 13
x1 = − 71 t − 26

7
7


3
22
x2 = 7 t + 7 s − 47
.
x3 = t



x4 = s
Solution to (ii) The augmented matrix of this system is


3 8 −3 −14 1
 2 3 −1 −2 −6 

.
 1 −2 1
10
0 
1 5 −2 −12 8
We now apply the Gaussian algorithm to this matrix in order to obtain a RREF
4
matrix.


3 8 −3 −14 1
 2 3 −1 −2 −6 


 1 −2 1
10
0 
1 5 −2 −12 8

1/3(R1) 7→ (R1)
−
−−−−−−−−−−−
→
−2(R1) + (R2) 7→ (R2)
−−−−−−−−−−−−−−−−−→
−(R1) + (R3) 7→ (R3)
−−−−−−−−−−−−−−−−→
−(R1) + (R4) 7→ (R4)
−−−−−−−−−−−−−−−−→
−3/7(R2) 7→ (R2)
−−−−−−−−−−−−−→
14/3(R2) + (R3) 7→ (R3)
−−−−−−−−−−−−−−−−−−→
−7/3(R2) + (R4) 7→ (R4)
−
−−−−−−−−−−−−−−−−−−
→
1/13(R3) 7→ (R3)
−−−−−−−−−−−−→
−(R3) + (R4) 7→ (R4)
−−−−−−−−−−−−−−−−→
5

1 8/3 −1 −14/3 1/3
 2 3 −1
−6 
−2


 1 −2 1
10
0 
8
1 5 −2 −12


1 8/3 −1 −14/3 1/3
 0 −7/3 1
22/3 −20/3 


 1 −2

0
1
10
1
5
−2 −12
8


1 8/3 −1 −14/3 1/3
 0 −7/3 1
22/3 −20/3 


 0 −14/3 2
44/3 −1/3 
1
5
−2 −12
8


1 8/3 −1 −14/3 1/3
 0 −7/3 1
22/3 −20/3 


 0 −14/3 2
44/3 −1/3 
0 7/3 −1 −22/3 23/3


1 8/3
−1 −14/3 1/3
 0
1
−3/7 −22/7 20/7 


 0 −14/3
2
44/3 −1/3 
0 7/3
−1 −22/3 23/3


1 8/3 −1 −14/3 1/3
 0 1 −3/7 −22/7 20/7 


 0 0
13 
0
0
0 7/3 −1 −22/3 23/3


1 8/3 −1 −14/3 1/3
 0 1 −3/7 −22/7 20/7 


 0 0
0
0
13 
0 0
0
0
1


1 8/3 −1 −14/3 1/3
 0 1 −3/7 −22/7 20/7 


 0 0
0
0
1 
0 0
0
0
1


1 8/3 −1 −14/3 1/3
 0 1 −3/7 −22/7 20/7 


 0 0
0
0
1 
0 0
0
0
0

0 1/7
26/7 −51/7
1 −3/7 −22/7 20/7 
.

0
0
0
1
0
0
0
0

−8/3(R2) + (R1) 7→ (R1)
−
−−−−−−−−−−−−−−−−−−
→
1
 0

 0
0
The equation corresponding to the above augmented matrix is

x1 + 71 x3 + 26
x = −51

7 4
7


3
22
x2 − 7 x3 − 7 x4 = − 20
7 .
0
=
1



0=0
Thus, the system has no solution, since the augmented matrix has an inconsistent row.
Solution for (iii): We proceed
mented matrix is

1
 −1

 −1
1
as shown in the above two cases. The aug
−1 1 −2 1
1 1 1 0 
,
2 3 −1 0 
−1 2 1 0
and applying the Gaussian algorithm

1 0 0
 0 1 0

 0 0 1
0 0 0
will reduce this to the matrix

0 −11/7
0 −10/7 
.
0 2/7 
1 −3/7
(This reduction is omitted from the solution set. It however, needs to be shown
in your work.) Thus the solution to the above system is

x1 = − 11

7


x2 = − 10
7 .
2
x
=

3
7


x4 = − 37
Solution for (iv): We proceed
mented matrix is

1
 −1

 −1
1
as shown in the above two cases. The aug
−1 1 −2 0
1 1 1 1 
,
2 3 −1 0 
−1 2 1 0
6
and applying the Gaussian algorithm

1 0 0
 0 1 0

 0 0 1
0 0 0
will reduce this to the matrix

0 −20/7
0 −15/7 
.
0 3/7 
1 −1/7
Thus the solution to the above system is

x1 = − 20

7


x2 = − 15
7 .
3
x
=

3
7


x4 = − 17
3. In the following system, find conditions (if possible) on a, b, and c such that
the system has no solution, one solution, or infinitely many solutions.

y + 2z = a

x + 2y + 3z = b

2x + 3y + 4z = c
Solution: The augmented matrix for this equation is


0 1 2 a
 1 2 3 b ,
2 3 4 c
and applying

0
 1
2
the Gaussian algorithm will reduce this to the matrix



1 2 a
1 2 3 b
 0 1 2 a 
2 3 b 
(R1) ↔ (R2)
−
−−−−−−−−
→
3 4 c
2 3 4 c


1 2
3
b
 0 1
2
a 
−2(R1) + (R3) ↔ (R3)
−−−−−−−−−−−−−−−−−→
0 −1 −2 c − 2b


1 2 3
b
 0 1 2
a 
−(R3) ↔ (R3)
−−−−−−−−−−→
0 1 2 2b − c


1 2 3
b
 0 1 2

a
−(R2) + (R3) ↔ (R3)
−−−−−−−−−−−−−−−−→
0 0 0 2b − c − a


b − 2a
1 0 −1
.
 0 1 2
a
−2(R1) + (R2) ↔ (R1)
−−−−−−−−−−−−−−−−−→
0 0 0 2b − c − a
7
Therefore we have the following cases:
1. if 2b−c−a 6= 0 then the RREF of the augmented matrix has an inconsistant
row, therefore, the system has no solution.
2. If 2b − c − a = 0 then {(b − 2a + s, a − 2s, s) : s ∈ R} is the solution set
for the system. So the system has infinitely many solutions in this case.
Note that there is no case in which the system has exactly one solution.
4. Write down all the 3 × 3 matrices in the reduced row-echelon form.
Solution: There is only one RREF 3 × 3 matrix of rank 3, i.e.


1 0 0
 0 1 0 .
0 0 1
The 3 × 3 RREF matrices of rank 2 are:

 
 

1 0 ∗
1 ∗ 0
0 1 0
 0 1 ∗ ,  0 0 1 ,  0 0 1 .
0 0 0
0 0 0
0 0 0
The 3 × 3 RREF matrices of rank 1 are:
 


 
0 1 ∗
0 0 1
1 ∗ ∗
 0 0 0 ,  0 0 0 ,  0 0 0 .
0 0 0
0 0 0
0 0 0
The 3 × 3 RREF matrices of rank zero are:


0 0 0
 0 0 0 .
0 0 0
5. Consider a system of linear equations with augmented matrix A and coefficient
matrix C. In each case either prove the statement or give an example showing
that it is false.
(i) If there is more than one solution, A has a row of zeros.
(ii) If A has a row of zeros, there is more than one solution.
(iii) If there is no solution, the row-echelon form of C has a row of zeros.
8
(iv) If the row-echelon form of C has a row of zeros there is no solution.
x + 2y + z = 4
Solution of (i): False. Example: The system
has infinitely
x + 3y + 3z = 1
many solutions, but the augmented matrix does not have a row of zeros.

 x + 2y = 4
x + 3y = 1 has a unique
Solution of (ii): False. Example: The system

0x + 0y = 0
solution, but the augmented matrix has a row of zeros.
Solution of (iii): True. To have no solution, the row-echelon form of the
augmented matrix should have an inconsistant row. Therefore the row-echelon
form of the coefficient matrix should have a row of zeros.

 x + 2y + 3z = 4
x + 3y − z = 1 has inSolution of (iv): False. Example: The system

2x + 6y − 2z = 2
finitely many solutions, but the row-echelon form of the coefficient matrix has
a row of zeros.
9