Math 152 Test #1 version A
B2: For each of the systems described below as augmented matrices that
have been put into reduced row echelon form, determine whether the
systems have
• no solutions
• a single solution (in this case, find the solution)
• an infinite number of solutions (in this case, find a parametric
description of all solutions).
Each part is worth 1 mark.

1 2 3 0
(a)
0 0 0 1
2
3
1 0
21
3 5
(b) 4 0 1
0 0
0

1 2 0 9 3
(c)
0 0 1 8 2

1 0 5
(d)
0 1 9
2
3
1 0 3 2
(e) 4 0 1 0 5 5
0 0 0 0
6
Test #1, Version A, MATH

2 3
0
0 0
1
solutions:
the
last
2
3 line means 0x + 0y + 0z = 1, which is not true for any real values of x, y, and z.
1 0
21
3 5
(b) 40 1
0 0
0

21
Precisely one solution:
3

1 2 0 9
3
(c)
0 0 1 8
2
Infinitely many solutions.
The easiest way to parameterize this is to let x2 = s and x4 = t. Then the first line says x1 = 3 2s 9t,
and the second line says x3 = 2 8t. So:
2 3 2
3 2 3
2 3
2 3
x1
3 2s 9t
3
2
9
6x2 7 6
7 607
617
607
s
6 7=6
7 6 7
6 7
6 7
4x3 5 4 2 8t 5 = 425 + s 4 0 5 + t 4 85
x4
t
0
0
1
(a)
1
0
No
Solution
will look di↵erent if di↵erent variables are chosen for parameters.

1 0
5
(d)
0 1
9

5
Precisely one solution:
9
2
3
1 0 3
2
55
(e) 40 1 0
0 0 0
0
Infinitely many solutions. The easiest way to parameterize this is to let x3 = s. Then:
2 3 2
3 2 3
2 3
x1
2 3s
2
3
4 x 2 5 = 4 5 5 = 4 55 + s 4 0 5
x3
s
0
1
Solutions might look di↵erent, but they will all have a direction vector that is a scalar multiple of [ 3, 0, 1].
2
MWF Version 2: One corner of a parallelepiped in R3 is at the point A = (1, 1, 1).
The three corners adjacent to that corner lie on the points B = (1, 1, 1), C = ( 1, 2, 4),
and D = (2, 0, 3).
(a) [2] What is the volume of the parallelepiped?
(b) [1] What is the area of the triangle ABC?
(c) [2] What are the other four corners of the parallelepiped?
Solution: Let
!
b = AB = (1, 1, 1)
!
c = AC = ( 1, 2, 4)
!
d = AC = (2, 0, 3)
(1, 1, 1) = (0, 2, 2),
(1, 1, 1) = ( 2, 1, 3),
(1, 1, 1) = (1, 1, 2).
(a) The volume is the absolute value of
0
0
@
2
det
1
2
1
1
1
2
3 A.
2
Expanding this determinant in the first row, we obtain
0 · (2 + 3)
( 2) · ( 4
3) + ( 2) · (2
1) =
16 .
Answer: The volume in of the parallelepiped is 16.
(b) The area of ABC is
1
||b ⇥ c|| =
2
0
i
1
||b ⇥ c|| = det @0
2
1
1
j
k
1
2
2A || = ||( 2 2)i
2
1 2
1
= || 4i + 4j 4k|| = 2||
2
(0
i+j
4)j(0
4)k||
p
k|| = 2 3.
(c) Denote the parallelepiped in the problem by P . Let us translate this parallelepiped
!
by the vector OA, so that the vertex A gets translated to the origin, O = (0, 0, 0), and
the vetrices B, C and D, adjacent to A get translated to (0, 2, 2), ( 2, 1, 3), (1, 1, 2),
respectively. I will denote this new (translated) parallelepiped by P0 and its vertices O =
(0, 0, 0), (0, 2, 2), ( 2, 1, 3), (1, 1, 2), by B0 , C0 and D0 , respectively.
Note that P0 is the parallelepiped spanned by b, c and d. Its vertices A0 , B0 , C0 , D0 , E0 , F0 , G0 , H0
are given as follows:
!
OA0 = 0 = (0, 0, 0),
!
OB0 = b = (0, 2, 2),
!
OC0 = c = ( 2, 1, 3),
!
OD0 = d = (1, 1, 2),
!
OE0 = b + c = ( 2, 1, 1),
!
OF0 = b + d = (1, 3, 0),
!
OG0 = c + d = ( 1, 0, 5).
!
OH0 = b + c + d = ( 1, 2, 3).
!
The vertices of P are obtained by adding OA = (1, 1, 1) back to each of these. The first
four,
!
!
OA = OA0 + (1, 1, 1) = (1, 1, 1),
!
!
OB = OB0 + (1, 1, 1) = (1, 1, 1),
!
!
OC = OC0 + (1, 1, 1) = ( 1, 2, 4),
!
!
OD = OD0 + (1, 1, 1) = (2, 0, 3),
yield back the four corners A, B, C, D in the problem. The other four corners are given by
!
!
OE = OE0 + (1, 1, 1) = ( 1, 0, 2),
!
!
OF = OF0 + (1, 1, 1) = (2, 2, 1),
!
!
OG = OG0 + (1, 1, 1) = (0, 1, 6),
!
!
OH = OH0 + (1, 1, 1) = (0, 1, 4).