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) (10) Let S = {A ∈ Rn×n | BA is symmetric}. Prove that S is a subspace of
Rn×n .
Proof : We first observe that the 0 matrix (the n × n matrix whose
entries are all zeros) is in S, since B(0) = 0 = (B(0))T . Next, we
suppose that C and D are in S. We show that C + D is in S. Since, C
and D are in S, we have that BC is symmetric (so that BC = (BC)T )
and BD is symmetric (so that BD = (BD)T ). Thus we have that
(B(C + D))T = (BC + BD)T = (BC)T + (BD)T = BC + BD =
B(C +D). So B(C +D) is symmetric and hence C +D is in S. Finally,
we let c be any scalar and D ∈ S. Then (B(cD))T = c(BD)T = cBD =
B(cD). So B(cD) is symmetric and hence cD ∈ S for all scalars c.
Hence S is a subspace of Rn×n .
4
Math 317: Linear Algebra
Exam 3 Solutions
Fall 2015
4. (a) (5) Suppose that A is a n × n matrix. Prove that det(cA) = cn det A for any
nonzero c ∈ R.
Proof : Suppose that A is an n × n matrix whose ith row is denoted
by Ai . Then we have




A1
cA1
 A2 
 cA2 
 


A =  ..  =⇒ cA =  ..  .
 . 
 . 
An
cAn
Using the linearity in row property for determinants we have that






cA1
A1
A1
 cA2 
 cA2 
 A2 






det cA = det  ..  = c det  ..  = c ∗ c det  ..  = . . .
 . 
 . 
 . 
cAn

cAn

cAn
A1
 A2 
 
= cn det  ..  = cn det A.
 . 
An
(b) (5) Suppose that A is a skew-symmetric matrix. (Recall that this means that
A = −AT .) Prove that A is a singular matrix if n is odd. (Hint: Recall that
det A = det AT .)
Proof : Following the hint, we use determinants to prove that A is
singular. That is, we show that det A = 0. Suppose that A is skewsymmetric and n is odd. From part (a) we have that det(AT ) =
det(−A) = (−1)n det A. Since n is odd, (−1)n = −1. So det AT =
− det A. However, we also have that det AT = det A. Thus det A = 0
and A is singular.
5
Math 317: Linear Algebra
Exam 3 Solutions
Fall 2015
 
 
1
1
3



5. (25, 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 T1 : R3 → R3 be the linear transformation which projects x ∈ R3 onto
V ⊥ . Find [T1 ]B , the matrix of T with respect to the basis B.
Noting that v1 , v2 ∈ V and v3 ∈ V ⊥ , we have that T1 (v1 ) = projV ⊥ (v1 ) =
0, T1 (v2 ) = projV ⊥ (v2 ) = 0, and T1 (v3 ) = projV ⊥ (v3 ) = v3 . We calculate the coordinates of each answer with respect to B as follows:
(T1 (v1 ))B = (0)B = 0v1 + 0v2 + 0v3
(T1 (v2 ))B = (0)B = 0v1 + 0v2 + 0v3
(T1 (v3 ))B = (v3 )B = 0v1 + 0v2 + 1v3 ,


0 0 0
and so [T1 ]B = 0 0 0.
0 0 1
(c) 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 ,
6
Math 317: Linear Algebra
Exam 3 Solutions
Fall 2015


1 0 0
so that [T2 ]B = 0 1 0 .
0 0 −1
(d) 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
(e) Calculate det [T2 ]stand .
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) If A = QR where Q is an orthogonal matrix and R is an upper triangular matrix
with positive diagonal entries, then the least squares solution x̄ associated with
Ax = b can be written as x̄ = R−1 QT b.
True. The least squares solution to Ax = b is given by x̄ = (AT A)−1 AT b.
Substituting A = QR and using the facts that QT Q = I and R−1 exist
(since its diagonal terms are nonzero) we have that
x̄ = (AT A)−1 AT b = ((QR)T (QR))−1 (QR)T b = (RT QT QR)−1 RT QT b
= (RT R)−1 RT QT b = R−1 (RT )−1 RT QT b = R−1 QT b
(b) There is a nonsingular matrix P such that
1 2
3 1
P =P
.
1 0
1 0
1 2
3 1
False. If such a matrix existed, then
would be similar to
.
1 0
1 0
However their traces are not the same and hence they cannot be similar
to each other.
1 2
1 2
(c) det
= det
.
3 4
4 6
True. The matrices differ by a Type III operation (adding one row to
another row) and we know that Type III operations do not change the
determinant.
8
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.
9