Math 317: Linear Algebra
Exam 3 Solutions
Fall 2015
Name:
*No notes or electronic devices. You must show all your work to receive full credit. When
justifying your answers, use only those techniques that we learned in class up to Section
5.1. I am thinking of the number thirty seven. Unless otherwise stated, you do not have
to simplify your answers. Good luck!
1. Suppose that A is a 4 × 4 matrix with rows a, b, c and d such that det A = 5. That
is,
 
a
b

A=
c
d
(a) (10) Calculate det B if


3a + b + d


b

B=
 c+d 
d
Using the linearity in rows property for determinants and the fact that
if a matrix has a row of zero entries then its determinant is 0, we obtain

 
 
 

3a + b + d
3a
b
d


b
0
0
b

 
 
 
det B = det 
 c + d  = det  c  + det d + det  0 
d
d
0
0
 
a
b

= 3 det 
 c  = 3(5) = 15.
d
(b) (5) Calculate det A−1 or explain why it does not exist.
det A−1 =
1
1
=
det A
5
(c) (5) Calculate det AB (if possible). If it is not possible, explain why.
det AB = det A det B = (5)(15) = 75
1
Math 317: Linear Algebra
Exam 3 Solutions
Fall 2015
2. Let V = Span (v1 = (1, 0, 1, 0), v2 = (0, 1, 4, 2)) ⊂ R4 .
(a) (5) Find an orthogonal basis for V .
We apply the Gram-Schmidt
procedure on the basis for V . To begin,

1
0

we let w1 = 
1 . A vector w2 that is in V and is orthogonal to w1 is
0
given by
v2 · w1
w2 = v2 −
w1
kw1 k2
   
 
1
0
−2
1 4 0  1 
   

= 
4 − 2 1 =  2  ,
0
2
2
so that {v1 , v2 } is an orthogonal basis for V .
(b) (5) Find a basis for V ⊥ .
To find a basis for V ⊥ , we find a matrix
A so that
V = R(A). Then
1 0 1 0
V ⊥ = N (A). To begin, we write A =
. Since rank(A) = 2,
0 1 4 2
we have that x3 and x4 are the free variables. Thus, x1 = −x3 and
x2 = −4x3 − 2x4 , and a basis for V ⊥ = N (A) is given by
   
−1
0 


   

−4 −2

B =  ,  .
1
0 





0
1
(c) (5) Find the projection of (1, 1, 1, 1) on V ⊥ . Hint: Use the orthogonal basis
you obtained in (a) to your advantage.
Following the hint, we first find the projection of (1, 1, 1, 1) on V . Since
we have an orthogonal basis for V , we have that
projV (b) = projw1 (b) + projw2 (b) =
b · w1
b · w2
w1 +
w2
2
kw1 k
kw2 k2
2
Math 317: Linear Algebra
Exam 3 Solutions
Fall 2015
 
  

1
−2
7/13
  

2 0
 + 3  1  =  3/13  .
= 
2 1 13  2  19/13
0
2
6/13
Now, we use the fact that projV ⊥ (b) = b − projV (b) to obtain
  
 

1
7/13
6/13
1  3/13   10/13 
 
 

projV ⊥ (b) = 
1 − 19/13 = −6/13 .
1
6/13
7/13
3
Math 317: Linear Algebra
Exam 3 Solutions
Fall 2015
3. Let A be an n × n matrix and let Rn×n denote the space of all n × n matrices.
(a) (5) Define the trace of a matrix A.
trace(A) =
Pn
i=1
aii , where aii denotes the diagonal entries of A.
(b) Let < A, B >= trace(AB). Show that <A, B >
is not an inner product on
1 1
Rn×n . Hint: Consider < A, A > for A =
.
−2 1
Proof : We observe
the
that with
A given to
us in thehint, that <
−1 2
1 1
1 1
A, A >= trace
= trace
= −2 6≥ 0.
−2 1 −2 1
−4 −1
Hence, < A, B >= trace(AB) is not an inner product on Rn×n .
4
Math 317: Linear Algebra
Exam 3 Solutions
Fall 2015
4. (15 points) Find the QR-factorization of the following matrix:

1
1
A=
1
1

2
4
.
2
4
Solution: We begin by finding an orthonormal basis for C(A). To this
regard, we have that
 
1
1
 
1
1
 
1/2
1/2
q1 = √
=  .
12 + 12 + 12 + 12 1/2
1/2
q2 is then given by
v2 − (q1 · v2 )q1
kv2 − (q1 · v2 )q1 k
(2, 4, 2, 4) − ((1/2, 1/2, 1/2, 1/2) · (2, 4, 2, 4))(1/2, 1/2, 1/2, 1/2)
=
k(2, 4, 2, 4) − ((1/2, 1/2, 1/2, 1/2) · (2, 4, 2, 4))(1/2, 1/2, 1/2, 1/2)k
(2, 4, 2, 4) − 6(1/2, 1/2, 1/2, 1/2)
=
k(2, 4, 2, 4) − 6(1/2, 1/2, 1/2, 1/2)k


−1/2
 1/2 
(−1, 1, −1, 1)

= p
=
(−1)2 + 12 + (−1)2 + 12 −1/2
1/2
q2 =
From our definition, we have that r11 = 2 = r22 and r12 = 6. Thus the QR
factorization for A is given by

1
1
A=
1
1
 

2
1/2 −1/2 1/2 1/2  2 6
4
=

= QR
2 1/2 −1/2 0 2
4
1/2 1/2
5
Math 317: Linear Algebra
Exam 3 Solutions
Fall 2015
 
 
1
1
3



5. (20, 5 each) Let V ⊂ R be the plane spanned by v1 = 0 and v2 = 1.
2
0
(a) Find a vector v3 so that V ⊥ = span(v3 ). Note that B = {v1 , v2 , v3 } is a basis
for R3 .
We find
A so that V = R(A). Then V ⊥ = N (A). Let
a matrix
1 0 2
A =
. rank(A) = 2 since the vectors (that make up A)
1 1 0
are linearly independent and so we have one free variable x3 . Solving
the associated homogenous system of equations gives x1 =
 −2x
 3 and
−2
x1 = −x2 = −2x3 =⇒ x2 = 2x3 . Thus we can take v3 =  2  .
1
(b) Let T2 : R3 → R3 be the linear transformation which reflects x ∈ R3 over V .
Find [T2 ]B , the matrix of T2 with respect to the basis B.
Recalling that T2 (x) = projV (x) − projV ⊥ (x), we have that
(T2 (v1 ))B = (v1 − 0)B = 1v1 + 0v2 + 0v3
(T2 (v2 ))B = (v2 − 0)B = 0v1 + 1v2 + 0v3
(T2 (v3 ))B = (0 − v3 )B = 0v1 + 0v2 − 1v3 ,


1 0 0
so that [T2 ]B = 0 1 0 .
0 0 −1
(c) Use the change of basis formula to compute [T2 ]stand . That is, compute the
standard matrix of T2 . You do not need to carry out the matrix multiplication/inversion.
The change of basis formula tells us that [T2 ]stand = P [T2 ]B P −1 where
the columns of P form a basis for B. Thus, we have that
[T2 ]stand



−1
1 1 −2 1 0 0
1 1 −2
= 0 1 2  0 1 0  0 1 2  .
2 0 1
0 0 −1 2 0 1
(d) Calculate det [T2 ]stand .
6
Math 317: Linear Algebra
Exam 3 Solutions
Fall 2015
By the change of basis formula, [T2 ]stand is similar to [T2 ]B . Thus,
they have the same determinant (since the determinant is a similarity
invariant). [T2 ]B is a diagonal matrix and hence its determinant is
the product of its diagonal entries. Thus, det [T2 ]stand = det[T2 ]B =
(1)(1)(−1) = −1.
7
Math 317: Linear Algebra
Exam 3 Solutions
Fall 2015
6. (5 points each) Determine if the following statements are true or false. If the statement is true, give a brief justification. If the statement is false, provide a counterexample or explain why the statement is not true.
(a) Suppose that the columns of A contain a basis for a subspace V ⊂ Rm . If
A = QR where Q is an orthogonal matrix and R is an upper triangular matrix
with positive diagonal entries, then the projection matrix (which projects any
x ∈ Rn onto V ) is given by PV = QQT .
True. Recalling that the projection matrix PV is given by PV =
A(AT A)−1 AT (and that QT Q = I), we have that
PV
=
=
=
=
=
=
A(AT A)−1 AT
QR((QR)T QR)−1 (QR)T
QR(RT QT QR)−1 RT QT
QR(RT R)−1 RT QT
QR(R)−1 (RT )−1 RT QT
QQT
(b) det(A + B) = det A + det B for any two n × n matrices A and B.
1 0
0
False. As a counterexample, let A =
and B =
0 0
0
det(A + B) = det(I2×2 ) = 1, but det A = det B = 0 since A
singular.
0
. Then
1
and B are
(c) There is a nonsingular matrix P such that
2 2
1 1
P =P
.
0 4
0 5
2 2
1 1
False. If this were true, then
would be similar to
. How0 4
0 5
ever, the determinants (product of the diagonal entries since these
matrices are upper triangular) of the two matrices are not equal, and
hence cannot be similar to each other, since the determinant is a similarity invariant.
(d) Suppose that A and B are n × n matrices such that trace(A) = trace(B). Then
A is similar to B.
8
Math 317: Linear Algebra
Exam 3 Solutions
Fall 2015
False. The two matrices in part(c) work as a counterexample.
9
Math 317: Linear Algebra
Exam 3 Solutions
Fall 2015
7. (5 points) (Bonus): I am thinking of a number between 1-100. What is it?
37. See the directions.
10