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WebAssign - Linear Algebra Tutorial Bank 1/e (Homework)

James Finch

Math - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 11 / 12

Due : Sunday, January 27, 2030 00:00 EST

Last Saved : n/a Saving...  ()

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Points
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1/1 4/4 1/1 3/3 1/1 1/1 0/1
Total
11/12 (91.7%)

  • Instructions

    Enrich your course with the WebAssign Linear Algebra Tutorial Bank collection. Authored by the WebAssign Community of Teachers, this collection features more than 100 multi-step tutorial questions that guide students to a deeper understanding of the skills and concepts. Every question features an algorithmically stepped-out solution for further student support. Additionally, video explanations available at the topic level present a variety of learning avenues for tackling tough concepts. Ideal as extra problems for your assignments, or as extra practice for students, this Additional Resource can be added to any WebAssign course at no cost.

    All questions contain a tutorial and a solution.

    Question 1 uses special systems of equations grading that lets the student enter any form of the dependent solution.

    Question 2 utilizes the expandable matrix tool. A student must add or subtract rows and columns to create the correct matrix size prior to entering their answer.

    Question 3 lets a student enter any form of the general solution in the matrix tool.

    Question 4 asks for the student to perform the given operations on two vectors. Grading allows component form as well as i, j, k form of the vector. This grading also enforces simplification so that the trivial answer of 2 is not accepted.

    Question 5 has a tutorial that teaches the student about vector spaces. The tutorial then guides the student through further explanation that the set of all integers is not a real vector space.

    Question 6 uses special grading within the matrix tool to allow any correct version of the vector set.

    Question 7 verifies that the matrix entered is in fact orthonormal.

    View the complete list of WebAssign questions available for this textbook. This demo assignment allows many submissions and allows you to try another version of the same question for practice wherever the problem has randomized values.

    The answer key and solutions will display after the first submission for demonstration purposes. Instructors can configure these to display after the due date or after a specified number of submissions.

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1. 1/1 points  |  Previous Answers WALinAlgTutBank1 1.1.001.Tut. My Notes
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1/1
 
Determine whether the given system of linear equations is consistent. If so, solve. (Let α represent an arbitrary number. If the system is inconsistent, enter INCONSISTENT.)
− x1 + 2x2 − x3 + 4x4 = 2
x1 − 3x2 + 4x3 + x4 = 0
x1 + x3 − 2x4 = 1
(x1, x2, x3, x4) = 
4α−16​,32​−α,76​−2α,α
Correct: Your answer is correct. webMathematica generated answer key


Solution or Explanation
We are asked to determine whether or not a given system of linear equations is consistent and, if so, solve. The given system is included below.
−x1 + 2x2 − x3 + 4x4 = 2
x1 − 3x2 + 4x3 + x4 = 0
x1 + x3 − 2x4 = 1
A system of linear equations is consistent if and only if there is at least one solution. An overdetermined or underdetermined system of linear equations does not necessarily imply the system is inconsistent or consistent, respectively.
Further, any combination of switching two equations, multiplying any equation by a nonzero constant, or adding any multiple of one equation to another equation results in an equivalent system of linear equations, hence, one with the same solution set.
The given system has three equations in four unknowns and is underdetermined. Note that this does not necessarily imply the system is consistent.
To determine whether or not the system is consistent we apply a sequence of operations resulting in a simpler, but equivalent, system. First, we isolate an
x1
 term with coefficient 1 in the first equation by applying appropriate operations. Then we eliminate the
x1
 term in equations two and three.
Switching equations one and two results in a system with an
x1
 term in the first equation with coefficient 1.
x1 − 3x2 + 4x3 + x4 = 0
−x1 + 2x2 − x3 + 4x4 = 2
x1 + x3 − 2x4 = 1
Adding equation one to equation two results in the system
x1 − 3x2 + 4x3 + x4 = 0
−x2 + 3x3 + 5x4 = 2
x1 + x3 − 2x4 = 1.
Further, subtracting equation one from equation three results in the system
x1 − 3x2 + 4x3 + x4 = 0
−x2 + 3x3 + 5x4 = 2
3x2 − 3x3 − 3x4 = 1.
Next, eliminate
x2
 from equation three by adding three times equation two to equation three.
x1 − 3x2 + 4x3 + x4 = 0
−x2 + 3x3 + 5x4 = 2
6x3 + 12x4 = 7
Note that
x4
 can be treated as a free variable and moved to the right-hand side of each equation.
x1 − 3x2 + 4x3 = −x4
−x2 + 3x3 = 2 − 5x4
6x3 = 7 − 12x4
Using α as a parameter that represents values of
x4,
 each of
x1,
x2,
 and
x3
 can be solved in terms of α.
x1 − 3x2 + 4x3 = −α
−x2 + 3x3 = 2 − 5α
6x3 = 7 − 12α
First,
x3 = 
7
6
 − 2α.
 Plugging the value of
x3
 into the second equation
x2
 can be solved in terms of α.
−x2 + 3x3 = 2 − 5α
−x2 + 3
7
6
 − 2α
 = 2 − 5α
x2 = 
7
2
 − 6α − 2 + 5α
x2 = 
3
2
 − α
Lastly, using the values of
x2
 and
x3
 in terms of α in equation one, the value of
x1
 can be expressed in terms of α.
x1 − 3x2 + 4x3 = −α
x1 − 3
3
2
 − α
 + 4
7
6
 − 2α
 = −α
x1 −
9
2
 + 3α +
14
3
 − 8α		 = −α
x1 +
1
6
 − 5α		 = −α
x1 = −
1
6
 + 4α
Therefore,
−x1 + 2x2 − x3 + 4x4 = 2
x1 − 3x2 + 4x3 + x4 = 0
x1 + x3 − 2x4 = 1
is consistent with infinitely many solutions, which can be expressed in terms of a parameter α,
(x1, x2, x3, x4) = 
−
1
6
 + 4α,
3
2
 − α, 
7
6
 − 2α, α
.

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2. 4/4 points  |  Previous Answers WALinAlgTutBank1 1.2.002.Tut. My Notes
Question Part
Points
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2/100 1/100 3/100 2/100
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4/4
 
Given A and B, compute AB, BA,
BTAT,
 and
ATBT.
 (If the answer does not exist, enter DNE in any cell of the matrix.)
A =
−81
69
7−1
,
B =
10−7
−760
AB =

Correct: Your answer is correct. seenKey

[-15, 6, 56; -57, 54, -42; 14, -6, -49]
BA =

Correct: Your answer is correct. seenKey

[-57, 8; 92, 47]
BTAT =

Correct: Your answer is correct. seenKey

[-15, -57, 14; 6, 54, -6; 56, -42, -49]
ATBT =

Correct: Your answer is correct. seenKey

[-57, 92; 8, 47]


Solution or Explanation
Given A and B, we are asked, if possible, to compute AB, BA,
BTAT,
 and
ATBT.
A =
−81
69
7−1
,
B =
10−7
−760
Two matrices M and N can be multiplied as MN if and only if the number of columns of M is equal to the number of rows of N. The same is not true of NM because matrix multiplication is not commutative. For NM to be defined the number of columns of N must equal the number of rows of M.
To compute AB the number of columns of A must equal the number of rows of B. To compute BA the number of columns of B must equal the number of rows of A.
A has 3 rows and 2 columns, whereas B has 2 rows and 3 columns. Therefore, AB and BA are defined.
Beginning with AB, the
(i, j)th
 element is computed by taking the (inner) product of the
ith
 row of A and the
jth
 column of B. In symbols this is
2aikbkj,
k = 1
 where a is an element of A and b is an element of B.
AB = 
−81
69
7−1
10−7
−760
 = 
−8(1) + 1(−7)−8(0) + 1(6)−8(−7) + 1(0)
6(1) + 9(−7)6(0) + 9(6)6(−7) + 9(0)
7(1) + (−1)(−7)7(0) + (−1)(6)7(−7) + (−1)(0)
 = 
−15656
−5754−42
14−6−49
The
(i, j)th
 element of BA is computed by taking the (inner) product of the
ith
 row of B and the
jth
 column of A. In symbols this is
3bikakj,
k = 1
 where a is an element of A and b is an element of B.
BA = 
10−7
−760
−81
69
7−1
 = 
1(−8) + 0(6) + (−7)(7)1(1) + 0(9) + (−7)(−1)
−7(−8) + 6(6) + 0(7)(−7)(1) + 6(9) + 0(−1)
 = 
−578
9247
The rows of a matrix M become the columns of its transpose,
MT.
 In other words, the
(i, j)th
 element of M becomes the
(j, i)th
 element of
MT.
Further if M is an r ✕ s matrix then
MT
 is an s ✕ r matrix.
In this case A is a 3 ✕ 2 matrix and B is a 2 ✕ 3 matrix. Therefore,
AT
 is a 2 ✕ 3 matrix and
BT
 is a 3 ✕ 2 matrix.
AT =
−867
19−1
BT =
1−7
06
−70
Based on the sizes of
AT
 and
BT,
 the products
ATBT
 and
BTAT
 are defined.
BTAT = 
1−7
06
−70
−867
19−1
 = 
1(−8) + (−7)(1)1(6) + (−7)(9)1(7) + (−7)(−1)
0(−8) + 6(1)0(6) + 6(9)0(7) + 6(−1)
−7(−8) + 0(1)−7(6) + 0(9)−7(7) + 0(−1)
 = 
−15−5714
654−6
56−42−49
ATBT = 
−867
19−1
1−7
06
−70
 = 
−8(1) + 6(0) + 7(−7)−8(−7) + 6(6) + 7(0)
1(1) + 9(0) + (−1)(−7)1(−7) + 9(6) + (−1)(0)
 = 
−5792
847
Therefore, given
A =
−81
69
7−1
 and
B =
10−7
−760
,
AB =
−15656
−5754−42
14−6−49
,
BA =
−578
9247
,
BTAT =
−15−5714
654−6
56−42−49
,
 and
ATBT =
−5792
847
.

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1/1
 
Determine the solution set of
Ax = 0
 given the matrix
A.
 (If the system has infinitely many solutions, enter a general solution in terms of a and b.)
A =
2913
14−1−7
Correct: Your answer is correct. seenKey

[13*a + 75*b;-3*a - 17*b;a;b]


Solution or Explanation
We are asked to determine the solution set of
Ax = 0
 given the matrix
A.
A =
2913
14−1−7
A linear system is homogeneous if it can be written in the form
Ax = 0,
 hence, we are asked to determine solutions to a homogeneous linear system.
It is not possible for a homogeneous linear system of the form
Ax = 0
 to have no solutions. The zero vector is always a solution.
To find solutions, Gauss-Jordan elimination can be applied to the matrix
A
 or to
A
 augmented by the zero vector so that reduced row echelon form results.
The system
Ax = 0
 with
A =
2913
14−1−7
is equivalent to
2913
14−1−7
x1
x2
x3
x4
=
0
0
,
or the following.
2x1 + 9x2 + x3 + 3x4 = 0
x1 + 4x2 − x3 − 7x4 = 0
Therefore, applying Gauss-Jordan elimination to either
A
 or
A
 augmented by the column vector
0
0
will result in reduced row echelon form and the solution set can be determined.
Gauss-Jordan elimination applied to
29130
14−1−70
results in the following reduced row echelon form.
10−13−750
013170
Linear systems and their associated augmented matrices that are row equivalent have the same solution set. In other words the linear system
Ax = 0
 has the same solution set as
10−13−750
013170
x = 0,
or
10−13−75
01317
x1
x2
x3
x4
=
0
0
,
or the following.
x1 = 13x3 + 75x4
x2 = −3x3 − 17x4
x3 = x3
x4 = x4
Therefore, the system
Ax = 0
 for
A =
2913
14−1−7
has infinitely many solutions consisting of vectors of the form
13a + 75b
−3a − 17b
a
b
,
for any a and any b.

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4. 3/3 points  |  Previous Answers WALinAlgTutBank1 3.1.001.Tut. My Notes
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Suppose
u = 
1, 7
 and
v =
−1, 4
 are vectors in
the set of real numbers2.
 Determine
2u,
−v,
 and
u + 2v.
 (Simplify your answers completely.)
2u =
(2,14)
Correct: Your answer is correct. webMathematica generated answer key −v =
(1,−4)
Correct: Your answer is correct. webMathematica generated answer key u + 2v =
(−1,15)
Correct: Your answer is correct. webMathematica generated answer key


Solution or Explanation
A vector
x ∈ the set of real numbersn
 is an ordered list of n real numbers
(x1, x2,   , xn)
 such that
x = 
x1, x2,   , xn
.
 When a vector x is multiplied by a scalar α, each component of the vector is also multiplied by the scalar. Thus
αx = 
αx1, αx2,   , αxn
.
For
u = 
1, 7
,
2u = 2
1, 7
 = 
(2)(1), (2)(7)
 = 
2, 14
.
For
v = 
−1, 4
,
−v = −1
−1, 4
 = 
(−1)(−1), (−1)(4)
 = 
1, −4
.
When a vector x is added to a vector y, the individual components of the vectors are added together. Thus for
x = 
x1, x2,   , xn
 and
y = 
y1, y2,   , yn
,
x + y = 
x1 + y1, x2 + y2,   , xn + yn
.
Then for
u = 
1, 7
 and
v = 
−1, 4
,
u + 2v = 
1, 7
 + 2
−1, 4
 = 
1 − 2, 7 + 8
 = 
−1, 15
.

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Determine whether the set of all integers is a real vector space.
     Correct: Your answer is correct.


Solution or Explanation
A nonempty set V is a vector space provided the following conditions apply.
  1. V is closed under vector addition.
  2. V is closed under scalar multiplication.
  3. Vector addition in V is commutative.
  4. Vector addition in V is associative.
  5. There exists an additive identity in V.
  6. There exists an additive inverse in V.
  7. There exists a multiplicative identity in V.
  8. Scalar multiplication in V is associative.
  9. Scalar multiplication in V is distributive across vector sums.
  10. Scalar multiplication in V is distributive across scalar sums.
If any one of the above conditions is not met, the set V is not a vector space.
Let V be the set of all integers. Then
V = {  , −2, −1, 0, 1, 2,   }.
 For V to be a real vector space, the scalars referenced in the conditions above are taken from the set of real numbers.
Consider condition 1. Let x, y ∈ V. Then
x + y ∈ V.
 Thus condition 1 is satisfied.
Consider condition 2. If
x = 2 ∈ V
 and
α = 
1
3
 ∈ the set of real numbers,
 then
αx = 
2
3
 ∉ V.
 Thus condition 2 is not satisfied.
Since condition 2 is not satisfied, V is not a real vector space.

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Find vectors that span the solution set of the homogeneous linear system.
x1 + 3x2 + 20x3 − x4 = 0
x1 − 3x2 − 22x3 + 5x4 = 0
x1 + 4x2 + 27x3 − 2x4 = 0

Correct: Your answer is correct. seenKey

[1,-2; -7, 1; 1, 0; 0, 1]


Solution or Explanation
Finding vectors that span the solution set amounts to solving for solutions of the matrix equation
Ax = 0.
 One way to find such solutions is by row reducing the corresponding augmented matrix into reduced row-echelon form.
1320−10
1−3−2250
1427−20
 → 
10−120
017−10
00000
Since
10−120
017−10
00000
is the reduced row-echelon form of
[A 0],
 it is possible to solve
Ax = 0.
 If
xi
 corresponds to the
ith
 column, then in terms of
x3, x4
 only,
x1 = x3 − 2x4
 and
x2 = −7x3 + x4.
Since
x1 = x3 − 2x4
 and
x2 = −7x3 + x4,
x3
 and
x4
 are free variables. Using the fact that
x1 = x3 − 2x4
 and
x2 = −7x3 + x4,
x1
x2
x3
x4
= x3
1
−7
1
0
+ x4
−2
1
0
1
,
which yields
1
−7
1
0
,
−2
1
0
1
as a set of vectors that spans the solution set of the system.

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Use the Gram-Schmidt procedure to find an orthonormal basis for the vector space spanned by the given vectors. (Enter sqrt(n) for
n
.)
x1 =
6
3
2
,
x2 =
2
−6
3

Incorrect: Your answer is incorrect. seenKey

[]


Solution or Explanation
We are asked to use the Gram-Schmidt procedure to find an orthonormal basis for the vector space spanned by
x1 =
6
3
2
,
and x2.
x2 =
2
−6
3
The Gram-Schmidt procedure finds an orthonormal set of vectors
{u1, u2,   , un}
 from a linearly independent set of vectors
{x1, x2,   , xn}.
 The given vectors are linearly independent if
c1x1 + c2x2 = 0
 implies that
c1 = c2 = 0.
Since
0 = c1x1 + c2x2 = c1 
6
3
2
 + c2
2
−6
3
,
the unknown coefficients can be solved using a system of linear equations. Thus
Ac = 0
 yields the following.
62
3−6
23
c1
c2
 = 
0
0
0
Row reducing
[A 0]
 into reduced row echelon form will allow the unknown coefficients to be found. Thus the following results.
620
3−60
230
 → 
100
010
000
Since
100
010
000
is the reduced row echelon form of
[A 0],
c1 = 0
 and
c2 = 0.
 Thus the set of vectors is linearly independent.
A set of vectors is orthonormal if the set is composed of orthogonal unit vectors. The set
{x1, x2,   , xn}
 will need to be transformed into a set of orthogonal unit vectors that span the same space.
To do so, an iterative process is used. First, u1 is the following.
u1 = 
x1
x1
 = 
1
(6)2 + (3)2 + (2)2
6
3
2
Simplifying u1 yields the following.
u1 = 
x1
x1
 = 
1
(6)2 + (3)2 + (2)2
6
3
2
 = 
1
7
6
3
2
The second step of the process is to find
u2
 where
u2 = 
w2
w2
 and
w2 = x2 − 
x2, u1
u1.
 Thus w2 is the following.
w2 =
2
−6
3
 − 
2
−6
3
,
6/7
3/7
2/7
6/7
3/7
2/7
Simplifying w2 yields the following.
w2 =
2
−6
3
 − 
2
−6
3
,
6/7
3/7
2/7
6/7
3/7
2/7
 =
2
−6
3
Thus
u2
 is the following.
u2 = 
w2
w2
 = 
1
(2)2 + (−6)2 + (3)2
2
−6
3
Simplifying u2 yields the following.
u2 = 
w2
w2
 = 
1
(2)2 + (−6)2 + (3)2
2
−6
3
 = 
1
$s2
2
−6
3
Since
u1 =
6/7
3/7
2/7
and
u2 =
,
an orthonormal basis for the vector space spanned by
x1 =
6
3
2
,
and
x2 =
2
−6
3
is the following.
6/7
3/7
2/7
,
The Gram-Schmidt procedure finds an orthonormal set of vectors
{u1, u2,   , un}
 from a linearly independent set of vectors
{x1, x2,   , xn}
 using an iterative process. Are the given vectors linearly independent? What does it mean for a set of vectors to be orthonormal?

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