Name
ID
Math 311
Exam 1
Spring 2002
Section 503
P. Yasskin
Solutions
1.
10 points
1
/10
2
/10
3
/30
4
/25
5
/25
A matrix A satisfies E 3 E 2 E 1 A = U where
1
0 0
0 1 0
2
0 0 1
0 1 0
E1 =
E2 =
1 0 0
0 0 1
1 0 0
E3 =
0 1 0
0 2 1
∗
0 −3 ∗
0 0 −1
2
U=
5
and the ∗’s represent unknown non-zero numbers. Find det A.
det E 1 = −1
det A =
2.
det E 2 = 1
2
det E 3 = 1
det U
=
det E 1 det E 2 det E 3
6
1
2
 −1 
1 
det U = 2 −3 −1  = 6
= −12
10 points If c is a scalar, A is a 50 × 60 matrix and B is a 60 × 80 matrix, prove AcB  = cAB .
HINT: Write out the ij-component of each side.
∑A
60
AcB  ij
=
cAB  ij
= cAB  ij = c
k=1
ik cB  kj
=
∑ A cB
60
k=1
∑A B
ik
60
k=1
ik
kj
These are equal. So AcB  = cAB 
kj
=c
∑A B
60
k=1
ik
kj
3.
30 points
Consider the triangle with vertices
A = 2, 4, 0 
B = 4, 2, 1 
C = 2, 7, 4 
a. Find cos θ where θ is the angle at vertex A.
AB = B − A = 2, −2, 1 
cos θ =
AB ⋅ AC
AB AC
=
AC = C − A = 0, 3, 4 
−6 + 4
= −2
15
4 + 4 + 1 9 + 16
b. Find the area of the triangle ΔABC.
i
AB × AC =
j
k
2 −2 1
0
3
= i−8 − 3  − j8  + k6  = −11, −8, 6 
4
Area = 1 AB × AC = 1 121 + 64 + 36 = 1 221
2
2
2
c. Find a set of parametric equations for the line containing A and C.
x
X = A + t AC
2
=
y
+t
4
z
0
0
x=2
3
y = 4 + 3t
4
z = 4t
d. Find a set of parametric equations for the plane containing A, B and C.
x
X = A + s AB + t AC
y
z
2
=
4
2
+s
0
−2
+t
1
0
x = 2 + 2s
3
y = 4 − 2s + 3t
4
z = s + 4t
e. Find a non-parametric equation for the plane containing A, B and C.
N = AB × AC = −11, −8, 6 
N⋅X = N⋅A
−11x − 8y + 6z = −11 ⋅ 2 − 8 ⋅ 4 + 6 ⋅ 0 = −54
11x + 8y − 6z = 54
4.
25 points
Consider the system of equations:
2
AX = B
A=
where
0
1
1 −2
1
0
−1
3
x p
X=
y q
z r
Compute A −1 . (Give reasons for each step.)
2
0
1
1 0 0
R2
1 −2
1
0 1 0
R1
0
3
−1
0 0 1
1 −2
1
0 1 0
2
0
1
1 0 0
0
3
−1
0 0 1
1 −2
1
0
1
R 2 − 2R 1
0
−1
0 3 −1
0
0
1
1 −2
1
0
1
0
0
1
0
1 −2 −1
0
3
−1
0
0
1 −2 0
4
1 0
1
0 1
0
0 1 0
0 0 1
1
−3 −2
1 −2 −1
−3 6 4
2
0 0 −1
1 0 0
0
R2 − R3
R 1 + 2R 2
R 3 − 3R 2
R1 + R3
−R 3
−1 3 2
1 −2 −1
3 −6 −4
A −1
=
−1 3 2
1 −2 −1
3 −6 −4
Solve AX = B.
X=
A −1 B
=
−1 3 2
1 −2 −1
3 −6 −4
1
0
0
1
1 −1
1
=
1
−1
−1 −2
0
B=
1
0
0
1
1 −1
5.
25 points
Consider the system of equations:
3w + 6x + y
=
y − 3z =
w + 2x + y − 2z =
−2w − 4x + y − 5z =
5
2
3
b
Find the value(s) of b for which there exist solutions. (Give reasons for each step.)
3
6
1
0
5
0
0
1 −3
2
1
2
1 −2
3
−2 −4 1 −5
b
R3
R1
1
2
1 −2
3
0
0
1 −3
2
3
6
1
0
5
R 3 − 3R 1
−2 −4 1 −5
b
R 4 + 2R 1
−2
0 0 1 −3
0 0 −2 6
0 0 3 −9
3
1 2
1
1 2 0
R1 − R2
2
−4
b+6
1
1
0 0 1 −3
2
0 0 0
0
0
0 0 0
0
b
R 3 + 2R 2
R 4 − 3R 2
To have solutions we must have b = 0.
For that value (those values) of b what is the solution set?
w = 1 − 2s − t
x=s
y = 2 + 3t
z=t
Give a geometrical description of the solution set.
Plane in R 4 .