math111_logo Row Operation


Remark Please do the problems by yourself and then compare with the given solution. Pay attention to the order the simplification is done: one column at a time, from left to right.

Please also note the three possible outcomes for the solutions: (1) no solution, (2) unique solution, (3) non-unique solution (with some arbitrary choice).

Exercise Solve systems of linear equations.

1)

x1 - x2 + 3x3 = 3
3x1 + x2 + x3 = 5
x1 + x2 - x3 = 1

Answer The augmented matrix is

[ 1 -1 3 3 ]
3 1 1 5
1 1 -1 1

We start the simplification with the [col 1]. By (-1)[row 3] + [row 1] and (-3)[row 3] + [row 2], [col 1] is as simple as we can get.

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

Then the exchange [row 1] ↔ [row 3] puts the most complicated row at the top.

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

Now we turn our attention to [col 2]. By (-1)[row 2] + [row 3], we create one more zero in [col 2].

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

Incidentally, [col 3] and [col 4] are also simplified. In fact, the shape is already the simplest we can get. Converted to a system of linear equations, we have

x1 + x2 - x3 = 1
  - 2x2 + 4x3 = 2
    0 = 0

Note that the last equation is trivial. The simplest nontrivial equation is the 2nd one, from which we get

x2 = -1 + 2x3, where x3 is arbitrary.

Substituting this into the 1st equation, we get

x1 = 1 - x2 + x3 = 3 - 1 + 2x3 - 3x3 = 2 - x3.

Thus the general solution is given by the formulae that express x1 and x2 in terms of x3, with x3 arbitrary.

2)

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

Answer The augmented matrix is

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

We apply (-2)[row 2] + [row 1] to get

[ 0 -5 -5 5 ]
1 1 2 -1

Then by [row 1] ↔ [row 2] and (-1/5)[row 2], we have

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

[row 2] corresponds to x2 + x3 = -1, which gives us x2 = -1 - x3, where x3 is arbitrary. [row 1] corresponds to x1 + x2 + 2x3 = -1. The general solution is

x1 = - 1 - x2 - 2x3 = - 1 - (-1 - x3) - 2x3 = - x3
x2 = -1 - x3
x3 is arbitrary

3)

x1 + 2x2 = 3
4x1 + 5x2 = 6
7x1 + 8x2 = 9

Answer The augmented matrix is

[ 1 2 3 ]
4 5 6
7 8 9

By (-4)[row 1] + [row 2] and (-7)[row 1] + [row 3], we get

[ 1 2 3 ]
0 -3 -6
0 -6 -12

Then we apply (-2)[row 2] + [row 3] and (-1/3)[row 2] to get

[ 1 2 3 ]
0 1 2
0 0 0

The last row corresponds to the trivial equality 0 = 0. The simplest nontrivial row is the second one, corresponding to the equation x2 = 2. Substituting to the equation x1 + 2x2 = 3, we have x1 = 3 - 4 = 1. Thus the system has a unique solution x1 = 1, x2 = 2.

4&5)

x1 +3 x2 + 2x3   + x5 = 0
- x1 - x2 - x3 + x4   = 1
  4x2 + 2x3 + 4x4 + 3x5 = 3
x1 + 3x2 + 2x3 - 2x4   = 0
x1 + 3x2 + 2x3   + x5 = 0
- x1 - x2 - x3 + x4   = 1
  4x2 + 2x3 + 4x4 + 3x5 = 2
x1 + 3x2 + 2x3 - 2x4   = 0

Answer The two systems have the same coefficient matrix A, with different right sides b and c. Therefore instead of doing row operations on the augmented matrices [A b] and [A c] separately, we may work on the combined augmented matrix.

[A b c] = [ 1 3 2 0 1 0 0 ]
-1 -1 -1 1 0 1 1
0 4 2 4 3 3 2
1 3 2 -2 0 0 0

By first applying [row 2] + [row 4] and then applying [row 1] + [row 2], we have

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

Next we apply (-2)[row 2] + [row 3] and (-1)[row 2] + [row 4] to get

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

Finally we apply [row 3] + [row 4] to get

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

If we take away the last column, then we see that the row operation changes the augmented matrix of the 4th system

x1 +3 x2 + 2x3   + x5 = 0
- x1 - x2 - x3 + x4   = 1
  4x2 + 2x3 + 4x4 + 3x5 = 3
x1 + 3x2 + 2x3 - 2x4   = 0

into

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

The last row corresponds to the equation 0 = 1. The contradiction shows that the fourth system has no solution.

Similarly, by taking away the 6th column, we see that the row operation changes the augmented matrix of the 5th system

x1 +3 x2 + 2x3   + x5 = 0
- x1 - x2 - x3 + x4   = 1
  4x2 + 2x3 + 4x4 + 3x5 = 2
x1 + 3x2 + 2x3 - 2x4   = 0

into

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

The last equation is trivial. The 3rd equation gives us

x4 = 1/2 - 1/2x5, x5 arbitrary.

The 2nd equation gives us

x2 = 1/2 - 1/2x3 - 1/2x4 - 1/2x5 = 1/4 - 1/2x3 - 1/4x5, x3 arbitrary.

The 1st equation then gives us

x1 = - 3x2 - 2x3 - x5 = - 3/4 - 1/4x3 - 1/4x5.

The general solution is given by the expressions of x1, x2, x4 in terms of arbitrary x3, x5.