Systems of Equations with Upper Triangular Coefficient Matrices
Let A1 and A2 be m × n matrices and let b1 and b2 be n × 1 matrices. The systems of
equations A1x = b1 and A2x = b2 are said to be equivalent if the systems have the same
solution sets, i.e.,
{x | A1x = b1} = {x | A2x = b2}.
We know that every system of equations A1x = b1 is equivalent to a system of equations
A2x = b2 with A2 in upper triangular form. (Of course, what the system A2x = b2 is
depends on what the original system A1x = b1 is.) So it is critical for us to be able to find
the solution of a system of equations whose coefficient matrix is in upper triangular form.
We practice reading off the solutions of a system of equations whose coefficient matrix is
in upper triangular form in the following examples.
Example 1
3x1 + 2x2 + x3 -
x4 =
0
x2 + 2x3 + x4 =
0
x3 + x4 =
1
-
x4 = -2
This system has the coefficient matrix
3 2 1 -1
0 -1 2 1
0 0 1 1
0 0 0 -1
and the augmented matrix
3
0
0
0
2
-1
0
0
1
2
1
0
-1
1
1
-1
0
0
1
-2
and the solution can be read off as
x4 = 2, x3 = 1 - x4 = -1, x2 = 2x3 + x4 = 0, and x1 = (1/3)(x4 - x3 - 2x2) = 1.
Example 2.
3x1 + 2x2 + x3 - x4 = 0
2x3 + x4 = 1.
This system has the augmented matrix
3
2
1 -1
0
0
0
2
1
1
.
The solutions are read off as follows:
Systems of Equations with Upper Triangular Coefficient Matrices page 1
The variable x4 can be an arbitrary number. In fact, we can read off a solution no matter
what value we assign for the x4. For this reason, let x4 = t. (The reason we choose t will be
apparent later in the course). Then
2x3 + t = 1 ! x 3 = (1 - t)/2.
Now the variable x2 can be an arbitrary number. In fact, we can read off a solution no matter
what value we assign to x2 (as well as to x4 which we do not need to consider at this stage).
For this reason let x2 = s. Then
3x1 + 2x2 + x3 - x4 = 0
! x1 = -(2s + x3 - t)/3 = (-2s - ((1 - t )/2) + t)/3
= -(2/3)s + (1/2)t - (1/6).
So we may write the solution as (the sum of 4 × 1 matrices)
x1
x2
x3
(-2/3)s + (1/2)t - (1/6)
=
x4
s
(1 - t)/2
-2/3
=s
t
1
0
1/6
+t
0
0
-1/2
-1/6
+
1
0
1/2
0
where s and t are arbitrary.
Example 3. The system with augmented matrix
1 2 3 -1 0
0 0 1 2 -1
0 0 0 0 9
has no solution since we would need 0x4 = 9, which is unobtainable.
Exercise. Find the solutions of the system 3x1 + 2x2 - x3 + 6x4 = 4,
x3 - 2x4 = 4 and put the solution in the form of Example 2.
Systems of Equations with Upper Triangular Coefficient Matrices page 2