math111_logo Existence


Exercise Which of the following systems have solutions?

1)

2x1   + 2x3 + x4 = 2
x1 + x2 - x3   = -1
  x2 - 2x3 + x4 = -2

Answer The augmented matrix is

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

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

[ 1 1 -1 0 -1 ]
0 -2 4 1 3
0 1 -2 1 -2

Then by 2[row 3] + [row 2] and [row 2] ↔ [row 3], we get

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

Since there is no row of the form [0 0 0 0 ≠0], the system is consistent.

2)

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

Answer The augmented matrix

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

appeared in this exercise, in which we have computed its row echelon form.

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

Since [row 3] is of the form [0 0 0 ≠0], the system has no solution.

3&4)

x1 - x2 + x3   + 2x5 = 1
2x1 - 2x2   + 2x4 + 2x5 = 0
- x1 + x2 + 2x3 - 3x4 + x5 = 0
- 2x1 + 2x2 + x3 - 3x4 - x5 = 0
x1 - x2 + x3   + 2x5 = 3
2x1 - 2x2   + 2x4 + 2x5 = 2
- x1 + x2 + 2x3 - 3x4 + x5 = 3
- 2x1 + 2x2 + x3 - 3x4 - x5 = 0

Answer The two systems have the same coefficient matrices. We compute both systems at the same time. Specifically, for Ax = b and Ax = c, we form the double augmented matrix.

[A b c] = [ 1 -1 1 0 2 1 3 ]
2 -2 0 2 2 0 2
-1 1 2 -3 1 0 3
-2 2 1 -3 -1 0 0

Now we carry out row operations, keeping in mind that the last two columns are the right sides. By -2[row 1] + [row 2], [row 1] + [row 3], 2[row 1] + [row 4], and (1/2)[row 2], we get

[ 1 -1 1 0 2 1 3 ]
0 0 -1 1 -1 -1 -2
0 0 3 -3 3 1 6
0 0 3 -3 3 2 6

Then by 3[row 2] + [row 3] and 3[row 2] + [row 4], we get

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

Note that the coefficient matrix part has already become a row echelon form. If we delete the last column, we see that the augmented matrix [A b] has been simplified to

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

Because [row 3] is of the form [0 0 0 0 0 ≠0], the 3rd system has no solution. Similarly, the augmented matrix [A c] has been simplified to

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

From this we conclude that the 4th system has solutions.