### 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.