Name
Math 311
Exam 1
Section 502
1.
25 points
Spring 2010
Solutions
P. Yasskin
1
/25
4
/10
2
/25
5
/20
3
/25 Total
/105
Consider the matrices
1 0 1 1
a p
0 1 1 0
A=
b q
X=
1 0 1 0
1
and
c r
0 2 3 3
1
1 −1
B=
1 −1
d s
1
1
a. Compute A −1 .
1 0 1 1
1 0 0 0
1 0 1
1
1
0
0 0
R1 − R3
0 1 1 0
0 1 0 0
0 1 1
0
0
1
0 0
R2 − R3
1 0 1 0
0 0 1 0
0 0 1
3
0
−2 0 1
0 2 3 3
0 0 0 1
0 0 0 −1
−1
0
R3 − R1
−R 4
1 0
1 0 1
1
1
0 0 0
1 0 0 −2
1
2
0
−1
R 1 + 2R 4
0 1 1
0
0
1 0 0
0 1 0 −3
0
3
0
−1
R 2 + 3R 4
0 0 1
3
0 −2
0
1
R 3 − 3R 4
0 0 0
1
1
−1
0
0 0 0 −1
−1 0 1 0
0 2 3
3
0
0 0 1
1 0 1
1
1
0
0 0
1 0 0 0
3
2
−2 −1
0 1 1
0
0
1
0 0
0 1 0 0
3
3
−3 −1
0 0 0 −1
−1
0
1 0
R4
0 0 1 0
−3 −2
3
1
0 0 1
0
−2 0 1
R3
0 0 0 1
1
−1
0
3
R 4 − 2R 2
A −1 =
2
−2 −1
3
3
−3 −1
−3 −2
3
1
1
−1
0
0
0
AX = B.
b. Solve the equation
X = A −1 B =
3
0
3
2
−2 −1
1
3
3
−3 −1
1 −1
−3 −2
3
1
1 −1
1
−1
0
1
0
1
1
=
2
2
2
2
−1 −3
0
2
1
2.
25 points
3x + 5y − 2z = 17
Consider the system of equations:
x+ y
=5
3y − 3z = p
a. Write out the augmented matrix and row reduce it to reduced row eschelon form. (Give reasons
for each step.)
3 5 −2
17
R2
1 1
0
5
1 1
0
5
R1
0 1 −1
1
0 3 −3
p
0 3 −3
p
1 1
0
5
1 0
3 5 −2
17
0 3 −3
p
1 1
0
5
0 2 −2
2
0 3 −3
p
R 2 − 3R 1
R 3 − 3R 2
1
4
R1 − R3
0 1 −1
1
R2 + R3
0 0
0
p−3
1 R2
2
b. For what value(s) of p are there
no solutions?
R1 − R2
(At least one answer is "No p".)
p≠3
exactly two solutions?
a unique solution?
No p
infinitely-many solutions?
No p
p=3
c. For those p’s for which there are solutions, what are the solutions?
For p = 3, the equations are:
x+z = 4
y−z = 1
.
x = 4−r
Since z is a free variable, the solutions are:
y = 1+r
z=r
2
3.
25 points
3x + 5y − 2z = 17
Consider the system of equations:
x+ y
=5
2y − z = p
a. Write out the augmented matrix and row reduce it to reduced row eschelon form. (Give reasons
for each step.)
3 5 −2
17
R2
1 1
0
5
1 1
0
5
R1
0 1 −1
1
0 2 −1
p
0 2 −1
p
1 1
0
5
1 0
3 5 −2
17
0 2 −1
p
1 1
5
0
0 2 −2
2
0 2 −1
p
R 2 − 3R 1
b. For what value(s) of p are there
no solutions?
No p
exactly two solutions?
R 3 − 2R 2
1
4
R1 − R3
0 1 −1
1
R2 + R3
0 0
1 R2
2
R1 − R2
1
p−2
1 0 0
6−p
0 1 0
p−1
0 0 1
p−2
(At least one answer is "No p".)
a unique solution?
No p
infinitely-many solutions?
All p
No p
c. For those p’s for which there are solutions, what are the solutions?
x = 6−p
y = p−1
z = p−2
3
4.
10 points Write out the equations which need to be solved to find a, b, c and d to balance the
chemical equation:
a Ca O 2 H 2 + b H 3 P O 4  c Ca 3 P 2 O 8 + d H 2 O
Then set up the augmented matrix for the equations. Do not solve the equations.
a = 3c
Ca:
5.
1 0 −3
a − 3c = 0
O:
2a + 4b = 8c + d
2a + 4b − 8c − d = 0
H:
2a + 3b = 2d
2a + 3b − 2d = 0
P:
b = 2c
b − 2c = 0
0
0
2 4 −8 −1
0
0
−2
0
0 1 −2
0
0
2 3
20 points Say whether each of the following statements is true or false for all scalars a, b and c,
all 3 × 3 matrices A, B and C and all 3 × 3 elementary matrices E 1 , E 2 and E 3 of types I, II and III
respectively where E 1 , E 2 and E 3 are NOT the unit matrix. Circle your answers.
a. (A + B)C = AC + BC.......................................................... True
False
b. (A + B) 2 = A 2 + 2AB + B 2 ................................................. True
False
c. (AB) 2 = A 2 B 2 .....................................................................True
False
d. If B = A − A T then B T = −B............................................. True
e. (AB) T = A T B T ....................................................................True
f. (AB) T = B T A T .................................................................... True
g. (AB) −1 = A −1 B −1 ................................................................ True
h. (AB) −1 = B −1 A −1 ................................................................ True
i. (aA + bB) T =
aA T + bB T ............................................... True
False
False
False
False
False
False
j. (aA + bB) −1 = (aA −1 + bB −1 )............................................. True
False
k. E 1 A = AE 1 ......................................................................... True
False
l. E 2 A = A T E 2 ....................................................................... True
False
T
T
m. E 3 A =
A T E T3
T
................................................................ True
False
n. det(cA) = c 3 det A............................................................... True
False
o. det(AB −1 ) = det A − det B...................................................True
p. det AB T
= det A det B...................................................... True
False
False
1 .............................................................. True
det A
r. det(E 2 A) = k det A with k ≠ 1............................................ True
False
s. det(E 3 A) = k det A with k ≠ 1............................................True
False
q. det(E 1 A) =
−1
t. det(E 1 E 2 E 3 AE −1
3 E 2 ) = − det A.......................................... True
False
False
4