Newberger Math 247 Spring 03
Solutions for Quiz 1.5 and 1.7
A=
1 2 −1
−2 8 −10
5−4 9
5 2 3
b=
0
−12
14
8
1. (a) (8 points) Solve the system Ax=bwhere Aand bare given above.
Put your answer in parametric vector form.
To solve, reduce the augmented matrix.
1 2 −1 0
−2 8 −10 −12
5−4 9 14
5 2 3 8
∼
1 2 −1 0
0 12 −12 −12
0−14 14 14
1−8 8 8
∼
1 2 −1 0
0 1 −1−1
0 0 0 0
0 0 0 0
∼
1 0 1 2
0 1 −1−1
0 0 0 0
0 0 0 0
Note that there are three columns, so the vector xis in R3.
x=
x1
x2
x3
=
2−x3
−1 + x3
x3
=
2
−1
0
+
−1
1
1
x3
(b) (4 points) Give a geometric description of the solution set that you
found in part (a).
The solution set to this matrix equation is a line in R3passing through
the vector
2
−1
0
parallel to the span of the vector
−1
1
1
.
1
