Questions about the assignment may be sent to the TA, Lauren D’Elia at rldelia@math.ncsu.edu

There are several different ways to solve these problems; I have shown one approach for each.
Each problem is separately worth 25 points; everything else is broken down below.
1. Find all solutions to the given linear system.

a.  x - 4y =  5
-4x + 6y  =  8
Copying down the problem correctly is worth 2 points.
A = [ 1 -4 | 5] à [1 -4 | 5] à [1 -4 | 5 ] à
[-4 6 | 8] [0 -10 | 28] [0 1 |-14/5]
4*r1+R2 -1/10*R2 4*r2+R1
[1 0 | -6.2]
[0 1 | -2.8]
Working out this portion correctly is worth 5 points.
Thus, this system is consistent and there is only one solution.
x = -6.2; y = -2.8
Recognizing that this is consistent or that there is only one solution is worth 1 point where the actual values for x and y are worth 2 points each.

  b.  3x + 2y -  z = 4
x - 2y + 2z = 1
11x + 2y + z  = 14
Copying down the problem correctly is worth 3 points.

B = [ 3 2 -1 | 4] à [1 -2 2 | 1] à [1 -2 2 | 1]à
[1 -2 2 | 1] [3 2 -1 | 4] [0 8 -7 |1]
[11 2 1 | 14] [11 2 1 | 14] [0 24 -21|3]
R1<--> R2 -3r1+R2 -3*r2+R3
-11r1 + R3
[1 -2 2 | 1]
[0 8 -7 | 1]
[0 0 0 | 0]

Working out this portion correctly is worth 5 points.
In this case, if z = t (ie, z is a free variable), y = (1+7t)/8, and
x=1-2t+(1+7t)/4
S = {(1-2t+(1+7t)/4, (1+7t)/8,t)| where t is any real number}
Recognizing that this system has infinite solutions is worth 1 point. Having solutions for x, y, and z in terms of only one variable is worth 4 points.

2. Find all the values of a for which the resulting linear system has (a) no solutions, (b) a unique solution, and (c) infinitely many solutions.

x + y + 3z = 2
x + 2 y + 4z = 3
x + 3y + (a)z = (b)

Copying down the problem correctly is worth 2 points.
C = [1 1 3 | 2] [1 1 3 | 2] [1 0 2 | 1]
[1 2 4 | 3] à [0 1 1 | 1] à [0 1 1 |1 ]
[1 3 a | b] [0 2 a-3 | b-2] [0 0 a-5 | b-4]
-r1 + R2 -2*r2+R3
-r1 + R3
Working out this portion correctly is worth 10 points.
i) For a system to have no solutions means the system is inconsistent. A system is inconsistent if there is a pivot element in the last (solution) column. In order for the system to have any type of solution, the a-5 = b-4 must = 0. Therefore, we find that no solutions will exist for any real numbers when a = 5 and b DOES NOT equal 4.

5 points

ii) In this case, only infinitely many solutions exist. However, depending on defense, some answers may differ here.
3 points

iii) For the system to have infinitely many solutions means that a = 5 and b = 4.

5 points

3. Write the augmented matrix for the following linear system and write it in row echelon form.

  2 x + 3y + z = 1
x + y + z = 3
3x + 4 y + 2 z = 4

Copying down the problem correctly is worth 2 points.

D = [2 3 1 | 1] à [1 1 1 | 3] à [1 1 1 | 3]
[1 1 1 | 3] [0 1 -1 | -5] [0 1 -1 |1]
[3 4 2 | 4] [0 1 -1 | -5] [0 0 0 | 0]
R1<--> R2 R2 - R3
-2*r1 + R2
-3*r1 + R3
Working out this portion correctly is worth 15 points.

Is the system consistent or inconsistent?
This system is consistent because 0 = 0 and there are infinitely many solutions.
Consistency is worth 8 points.

4. Let the matrix A be an augmented matrix for a system of linear equations.

A =	[	m	n	p	0	|	0	]
	[	0	m	n	p	|	0	]
	[	0	0	m	n	|	p	]
	[	0	0	0	m	|	n	]

Find conditions on m, n and p such that

a) the matrix is in reduced row echelon form.

To be in reduced row echelon form, the pivot points must be equal to 1 while everything above and below a pivot point must be 0. Thus, our answers include: (m = 1 and n = 0 and p = 0) OR (m = 0 and n = 1, and p = 0) OR (m = 0, n = 0, p = 1) OR (m = 0 and n = 0 and p = 0).

10 points
b) the matrix is in reduced row echelon form and the system is consistent.

For the matrix to be consistent, there can be no pivot element or 1 in the answer column.  So we start with the possibilities from part 4a, and we see that n cannot be equal to 1 since then there would be a pivot element in the last (answer) column and p cannot be equal to 1 for the same reason. So the only possible answers are 
(m=1 and n=0 and p=0) OR (m=0 and n=0 and p=0).

15 points