math111_logo Existence


Exercise For what choices of the parameters are the following systems consistent?

1)

x + 2y = a
3x + 4y = b

Answer The augmented matrix is

[ 1 2 a ]
3 4 b

By -3[row 1] + [row 2], we get

[ 1 2 a ]
0 -2 b - 3a

For any a and b, this is a row echelon form with no rows of the form [0 0 ≠0]. Therefore the system is consistent for all choice of a and b.

2)

x + y = a
2x + 2y = b
3x + 3y = c
4x + 4y = d

Answer The augmented matrix is

[ 1 1 a ]
2 2 b
3 3 c
4 4 d

By row operations, we get

[ 1 1 a ]
0 0 b - 2a
0 0 c - 3a
0 0 d - 4a

The condition for no rows of the form [0 0 0] is b = 2a, c = 3a, d = 4a. This is exactly the condition for the system to be consistent.

3)

x1 + x2 + 2x3 + x4 = 1
x1   + 2x3   = 0
2x1 + 2x2 + 3x3   = h
  x2 + x3 + 3x4 = h

Answer The augmented matrix is

[ 1 1 2 1 1 ]
1 0 2 0 0
2 2 3 0 h
0 1 1 3 h

By -[row 1] + [row 2] and 2-[row 1] + 3[row 2], we get

[ 1 1 2 1 1 ]
0 -1 0 -1 -1
0 0 -1 -2 h - 2
0 1 1 3 h

By [row 2] + [row 4], we get

[ 1 1 2 1 1 ]
0 -1 0 0 -1
0 0 -1 -2 h - 2
0 0 1 2 h - 1

By [row 3] + [row 4], we get

[ 1 1 2 1 1 ]
0 -1 0 0 -1
0 0 -1 -2 h - 2
0 0 0 0 2h - 3

The condition for the system to be consistent is 2h - 3 = 0. In other words, h = 3/2.

4)

    u + 2v = 1
- 2x + 2y - u + 4v = h
x - y + u - v = 1

Answer The augmented matrix is

[ 0 0 1 2 1 ]
-2 2 -1 4 h
1 -1 1 -1 1

By row operations, we get

[ 1 -1 0 -3 0 ]
0 0 1 2 1
0 0 0 0 h + 1

The condition for the system to be consistent is h = -1.

5)

x1 + x2 - x3 = -2
x1   - ax3 = -1
x1 + ax2   = -1

Answer The augmented matrix is

[ 1 1 -1 -2 ]
1 0 -a -1
1 a 0 -1

By -[row 1] + [row 2] and -[row 1] + [row 3], we get

[ 1 1 -1 -2 ]
0 -1 1 - a 1
0 a - 1 1 1

Then by (a+1)[row 2] + [row 3], we have

[ 1 1 -1 -2 ]
0 -1 1 - a 1
0 0 2a - a2 a

If 2a - a2 ≠ 0, i.e., a ≠ 0 or 2, then the system is clearly consistent.

If a = 0, then the last row consists of zeros. Thus the system is also consistent.

If a = 2, then the last row is [0 0 0 2]. Thus the system is inconsistent.

6)

x1 + x2 + x3 + x4 = 1
x1     + ax4 = 1
  x2 + ax3   = 1

Answer The augmented matrix is

[ 1 1 1 1 1 ]
1 0 0 a 1
0 1 a 0 1

By row operations, we get

[ 1 1 1 1 1 ]
0 1 1 1 - a 0
0 0 a - 1 a - 1 1

The condition for the system to be consistent is a ≠ 1.

7)

  v + w = a
u   + ax = 1
u + v   = 0
  av - w = 1

Answer The augmented matrix is

[ 0 1 1 a ]
1 0 a 1
1 1 0 0
0 a -1 1

By row operations, we get

[ 1 0 a 1 ]
0 1 1 a
0 0 - a - 1 - a - 1
0 0 0 - a2 + a + 2

The condition for the system to be consistent is -a2 + a + 2 = 0. In other words, the system has solutions only for a = -1 or 2.