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
2
1 0
4
0 1
(a)
0 0

1 2
(b)
0 0

1 0
(c)
0 1

1 2
(d)
0 0
2
1 0
(e) 4 0 1
0 0
worth 1 mark.
3
3 2
0 5 5
0 0
3
0
0
1
5
9
0 9
1 8
3
2
3
21
3 5
0
6
Test #1, Version A, Math
2
3
1 0 3
2
55
(a) 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].

1 2 3
0
(b)
0 0 0
1
No solutions: the last line means 0x + 0y + 0z = 1, which is not true for any real values of x, y, and z.
1 0
5
(c)
0 1
9

5
Precisely one solution:
9

1 2 0 9
3
(d)
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
Solution
will look di↵erent
if di↵erent variables are chosen for parameters.
2
3
1 0
21
3 5
(e) 40 1
0 0
0

21
Precisely one solution:
3
1
Mathematics 152, Midterm 1, Solutions to Problem B3
MWF Version 1: 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 = (3, 0, 2).
(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)
(1, 1, 1) = ( 2, 2, 0),
!
c = AC = (1, 2, 4)
(1, 1, 1) = (0, 1, 3),
!
d = AC = (3, 0, 2)
(1, 1, 1) = (2, 1, 1).
(a) The volume is the absolute value of
0
2
det @ 0
2
1
2 0
1 3A .
1 1
Expanding this determinant in the first row, we obtain
( 2) · (1 + 3)
( 2) · (0
6) + 0 · (0
2) =
8
12 =
20 .
Answer: The volume in of the parallelepiped is 20.
(b) The area of ABC is
0
i
1
1
||b ⇥ c|| = || det @ 2
2
2
0
1
j k
1
2 0 A || = ||( 6 0)i ( 6 0)j + ( 2 0)k||
2
1 3
p
1
= || 6i + 6j 2k|| = || 3i + 3j k|| = 19.
2
(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 ( 2, 2, 0), (0, 1, 3), (2, 1, 1),
respectively. I will denote this new (translated) parallelepiped by P0 and its vertices O =
(0, 0, 0), ( 2, 2, 0), (0, 1, 3), (2, 1, 1), 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 = ( 2, 2, 0),
!
OC0 = c = (0, 1, 3),
!
OD0 = d = (2, 1, 1),
!
OE0 = b + c = ( 2, 1, 3),
!
OF0 = b + d = (0, 3, 1),
!
OG0 = c + d = (2, 0, 4).
!
OH0 = b + c + d = (0, 2, 4).
!
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) = (3, 0, 2),
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, 4),
!
!
OF = OF0 + (1, 1, 1) = (1, 2, 2),
!
!
OG = OG0 + (1, 1, 1) = (3, 1, 5),
!
!
OH = OH0 + (1, 1, 1) = (1, 1, 5).