# Math 1090 Quiz 3 Solutions February 23rd, 2015 ```Math 1090
Quiz 3 Solutions
February 23rd, 2015
Answer the following question in the space provided. The value of every
question is indicated at the beginning. Time: 12 minutes.
Name:
UID:
1. (20 points) Consider the following system of equations.
3x − 2y − 7z = 0
x−y−z =1
−x + 2y − 3z = a
Solve this system for a = −4 and a = 2 and write a one-sentence geometric interpretation
of the solution.
Solution. The augmented matrix for the system

3 −2 −7
 1 −1 −1
−1 2 −3
is

0
1 
a
Our goal is to apply row operations until the 3 &times; 3 matrix on the left becomes uppertriangular (namely, until we have zeros below the diagonal.


3 −2 −7 0
 1 −1 −1 1 
−1 2 −3 a


3 −2 −7
0
 ↔ 3 −   0 −1 4
3 
 ↔  + 
0 1 −4 a + 1


3 −2 −7
0
 0 −1 4
3 
 ↔  + 
0 0
0 a+4
Case a = −4 For a = −4, our augmented matrix becomes

3 −2 −7 0
 0 −1 4 3 
0 0
0 0

The last equation is 0 = 0, so we are left with only two equations for 3 variables
3x − 2y − 7z = 0
−y + 4z = 3
This means that we will be getting infinitely many solutions. To find them, fix one of
the variables, say z, and use the equations o write x and y in terms of z. From the
second equation we get
y = 4z − 3
and plugging that into the first equation yields
y
z }| {
3x − 2 (4z − 3) −7z = 0
3x − 8z + 6 − 7z = 0
3x − 15z + 6 = 0
3x = 15z − 6
x = 5z − 2
so our solutions are
y
x
z }| { z }| {
(5z − 2, 4z − 3, z)
and for every value of z we get one valid solution. For instance, for z = 0, we get
the point (−2, −3, 0), and you can check that it satisfies all 3 equations of the original
system.
Geometric interpretation. Remember that a linear equation in three variables represents a plane. Normally, when you intersect 3 planes you get a point. In this case you
are getting infinitely many points, so what this means is that your 3 planes intersect in
a line.
Case a = 2 For a = 2, our augmented

3
 0
0
matrix becomes

−2 −7 0
−1 4 3 
0
0 6