MATH 4305, Fall 2012
Midterm II, solution
Problem 1(20 pt) Let a = 1, b = 2 and c = 3. Find all values of x such
that the columns of the following matrices are linearly dependent. Explain.
A=
1
1
1
1
a
b
c
x
a2
b2
c2
x2
a3
b3
c3
x3
Solution: Columns of A
Math 4305, Fall 2012,
Midterm 1, Practice
Show all your work. You may use one side of a 3 5 index card for formulars
in this exam. Please give yourself 50 minutes.
Problem 1 Suppose that the matrix below is the augmented matix of a
system of linear equati
Math 4305, Fall 2012,
Midterm 1, Practice solutions
Problem 1 Suppose that the matrix below is the augmented matix of a
system of linear equations
12
00
0 2
00
0
1
0
1
0
3
1
k
1
2
3
h
a)(6 points) For what values of h and k , this system has no solution.
MATH 4305, Fall 2012,
Midterm 2, Practice
Show all your work. You may use one side of a 3 5 index card for formulars
in this exam. Calculator is not allowed. Please give yourself 50 minutes.
1
Problm 1 Let y = 2
3
spancfw_u, v.
1
, u = 1
0
1
, v = 0
1
,
a
MATH 4305, Fall 2012,
Midterm 2, Practice: Solution
Show all your work. You may use one side of a 3 5 index card for formulars
in this exam. Calculator is allowed. Please give yourself 50 minutes.
1
Problm 1 Let y = 2
3
spancfw_u, v.
1
, v = 0
1
1
, u = 1
MATH 4305, Fall 2012
Practice Final: Solutions
Problem 1 Please complete the on-line course survey. Thank you in advance.
Solutions: If you still did not do it, please do it right now.
4 2
1
Problem 2 Let A = 2 1 , b = 2
00
3
.
a) Find a basis for N ul(A)
MATH 4305, Fall 2012
Midterm I, Solutions
Problem 1(25 points) Suppose that the matrix below is the augmented
matix of a system of linear equations
1
0
0
0
2
0
0
0
0
1
0
1
0
3
1
h
1
2
3
k
a)(10 pt) Row reduce the matrix into REF and identify all values of
Math 4305, Fall 2012,
Homework 1, solutions
Section 1.1, 20: Set demand of A is a and the demand of B is b, we solve
the system
a = 1000 + 0.1b
b = 780 + 0.2a,
and obtain the unique solution
a = 1100, b = 1000.
Section 1.1, 30: We solve the system
f (1) =
Math 4305, Exam 1: 1.1-3.1 (practice)
Name/Section:
1. (25 points) (1.2.6) Find the general solution of the system of equations
x1 7x2 + x5 = 3
x3 2x5 = 2
x4 + x5 = 1.
Solution: We apply the row reduction algorithm to the augmented coecient matrix. We se
Math 4305, Exam 2: 3.2-5.2.1 (practice)
Name/Section:
1. (25 points) (3.7.6) Find the rank of the matrix
1 1 3
2 1 4
.
Solution: We apply the row reduction algorithm to obtain
1
1
3
0 1 2
.
From this we see there are two pivots and the rank is 2.
Name and
Math 4305, Fall 2012,
Homework 9, solutions
Section 6.1, 56
a) Expand across rst column, one has
dn = (1)n+1 dn1 = (1)n1 dn1 , f or n > 2.
Therefore, d2 = 1, d3 = 1, d4 = 1, d5 = 1, d6 = 1, d7 = 1, d8 = 1.
It repeated every four terms.
b) Actually, from a
MATH 4305, Summer 2009
Practice Final: Solutions
Problem 1 Please complete the on-line course survey. Thank you in advance.
Solutions: If you still did not do it, please do it right now.
4 2
1
Problem 2 Let A = 2 1 , b = 2
00
3
.
a) Find a basis for N ul(
Math 4305, Fall 2012,
Homework 2, solutions
Section 2.1, 44: Let v = (a1 , a2 , a3 )T , it is easy to show that T (x) = v x
is linear. This can be conrmed by
0 a3 a2
0 a1 x.
T (x) = [T (e1 ), T (e2 ), T (e3 )]x = a3
a2 a1
0
Section 2.1, 46: Now we know it
Math 4305, Fall 2012,
Homework 4, solutions
Section 3.1, 10 Solve the system Ax = 0, we see
ker(A) = spancfw_(1, 2, 1, 0)T .
Section 3.1, 44: (a) Yes, the system Ax = 0 has the same solution set as
B x = 0. The elementary row operations does not change th
Math 4305, Fall 2012,
Homework 5, solutions
Section 3.3, 24
rref =
1
0
0
0
2
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
,
cfw_(2, 1, 0, 0, 0)T is a basis of the kernel of A, the 1st, 3rd, 4th and 5th
columns of A form a basis of the image of A.
Section 3.3, 28: Thes
Math 4305, Fall 2012,
Homework 6, solutions
Section 4.1, 44 Let S = [u1 , u2 , u3 ], it reads that Au1 = u1 , Au2 = u2 and
Au3 = 0. Therefore, if n is a normal vector of V , while v1 , v2 is a basis of V ,
we see that
u1 = av1 + bv2 , u2 = cv1 + dv2 , u3
Math 4305, Fall 2012,
Homework 8, solutions
Section 5.4, 8
a) Since ker(A) = cfw_0, columns of A are linear independent. We thus
know that
L+ (y) = (AT A)1 AT y.
Thus L+ is linear.
b) If L is invertible, A is invertible, and (AT A)1 AT = A. Thus L = L+ .