PROBLEM SET
Assignment #1
Math 2360, Fall 2005
Sept. 13, 2005
• Write all of your answers on separate sheets of paper.
You can keep the question sheet.
• You must show enough work to justify your answers.
Unless otherwise instructed,
give exact answers, not
√
approximations (e.g., 2, not 1.414).
• This problem set has 3 problems.
• Due Wednesday, Sept. 21
Good luck!
Problem 1. In each part you are given the augmented matrix of a system
of linear equations, with the coefficient matrix in reduced row
echelon form. Determine if the system is consistent and, if it is consistent, find
all solutions.
A.








B.








C.
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
2
−3
5
7
0
1

1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
1
−2
4
3
0
0


1
 0

 0
0
D.

1
 0

 0
0
E.






−2 0 0
0 1 0
0 0 1
0 0 0















2
3 

5 
0

−2 3 0 0 −1
1
2 
0 0 1 0
2

0 0 0 1
3 −5 
0
0 0 0 0
0
1
0
0
0
0
0
1
0
0
0
−2 0 −3
1 0
2
0 1 −1
0 0
0
0 0
0
0
0
0
0
0






Problem 2. In each part you are given the augmented matrix of a system of
linear equations, with the coefficient matrix in row echelon form
(but not reduced row echelon form). Determine if the system is consistent and,
if it is consistent, find all solutions using back substitution.
1
A.


−1 −5
2
2 
1 −1

1
 0
0

−1 2 −2
5 
1 3
0 0
3


−1 2 0
1 3 2 
0 0 0
1 2
 0 1
0 0
B.
C.
1
 0
0
D.

1
 0

 0
0

−1 2 3 −2
3
0 1 2 −1 −2 

1 
0 0 0
1
0 0 0
0
0
Problem 3. In each part, you are given a system of linear equations. Find the
augmented matrix of the system and perform row operations to
get the coefficient matrix in RREF. Then determine if the system is consistent
and, if it is, find all solutions. You can use a calculator to do the row operations,
but show each operation you are performing and the resulting matrix.
A.
x1 + 2x2 − 2x3 + x4 = 8
x1 + 3x2 − x3 + 2x4 = 14
2x1 + 4x2 − 3x3 + 4x4 = 16
x1 + 3x2 − x3 + 3x4 = 13
B.
x1 + x2 − 3x3 + 5x5 = 4
x2 − x3 − 2x4 + 4x5 = 5
x1 − 2x2 + x4 − 2x5 = −1
x1 + 3x2 − 5x3 − 6x4 + 15x5 = 18
x1 + 4x2 − 6x3 − x4 + 12x5 = 9
2
C.
x1 + 2x2 − 4x3 + 4x4 = 11
−x1 − x2 + 3x3 − 4x4 = −12
x1 + 2x2 − 4x3 + 3x4 = 9
x1 + x2 − 3x3 + 2x4 = 8
D.
x1 + 2x2 − 4x3 + 4x4 = 11
−x1 − x2 + 3x3 − 4x4 = −10
x1 + 2x2 − 4x3 + 3x4 = 8
x1 + x2 − 3x3 + 2x4 = 5
Hint: The last two can be done in one computation.
3