Math 317: Linear Algebra
Homework 2 Solutions
Spring 2016
1. Section 1.4, Problem 5(c)
Give the general solution of the equation Ax = x in standard form.


0 −1 0
A =  0 0 −1 
1 0
0
We begin by noting that Ax = x implies that

   
0 −1 0
x1
x1
 0 0 −1  x2  = x2  ,
1 0
0
x3
x3
which in turns generates the following system of equations:
0x1 − x2 + 0x3 = x1
0x1 + 0x2 − x3 = x2
x1 + 0x2 + 0x3 = x3 ,
which we can rewrite as Bx = 0 by observing that the above system can be
rewritten to obtain
−x1 − x2 + 0x3 = 0
0x1 − x2 − x3 = 0
x1 + 0x2 − x3 = 0,
which gives the augmented matrix:

−1 −1 0 0
[B|0] =  0 −1 −1 0 
1
0 −1 0

which we can now but into echelon form so that we may solve the associated
system of equations.




−1 −1 0 0
−1 −1 0 0
R1 +R3 →R3
−R2 +R3 →R3
[B|0] =  0 −1 −1 0  −→  0 −1 −1 0 
−→
1
0 −1 0
0 −1 −1 0
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Math 317: Linear Algebra
Homework 2 Solutions
Spring 2016


−1 −1 0 0
 0 −1 −1 0 
0
0
0 0
From here we see that x1 , x2 are the pivot variables and x3 is the free variable
(since its column does not contain a pivot entry). Thus, we solve everything
in terms of x3 . Converting the augmented matrix back to a system of equations yields:
−x1 − x2 = 0
−x2 − x3 = 0
Solving this system of equations in terms of the free variable x3 yields:
x2 = −x3 and −x1 − x2 = 0 =⇒ −x1 − (−x3 ) = 0 =⇒ x1 = x3 . We write
the solution in standard form as follows:
  

 
x1
x3
1
x2  = −x3  = x3 −1 .
x3
x3
1
2. Section 1.4, Problem 11(b)
We wish to find all vectors x ∈ R4 which are orthogonal to a = (1, 1, 1, −1)
and b = (1, 2, −1, 1). This problem is equivalent to finding those vectors x
such that x · a = 0 and x · b = 0. Using the algebraic definition of the dot
product, this yields the following system of equations:
x · a = (x1 , x2 , x3 , x4 ) · (1, 1, 1, −1) = x1 + x2 + x3 − x4 = 0
x · b = (x1 , x2 , x3 , x4 ) · (1, 2, −1, 1) = x1 + 2x2 − x3 + x4 = 0,
which we can now solve using our typical Gaussian elimination method:
1 1 1 −1 0
1 2 −1 1 0
1 1 1 −1 0
−→
0 1 −2 2 0
1 0 3 −3 0
0 1 −2 2 0
−R1 +R2 →R2
−R2 +R1 →R1
−→
Now that our augmented matrix is in echelon form, we see that our free
variables are x3 and x4 . Rewriting our augmented matrix as a system of
linear equations yields:
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Math 317: Linear Algebra
Homework 2 Solutions
Spring 2016
x1 + 3x3 − 3x4 = 0
x2 − 2x3 + 2x4 = 0
Solving the above equations in terms of x3 and x4 yields x1 = −3x3 + 3x4
and x2 = −2x3 + 2x4 . In standard form, the solution becomes
  
 
 

 
 
x1
−3x3 + 3x4
−3x3
3x4
−3
3
x2   2x3 − 2x4   2x3  −2x4 
2
−2
 =
=
 

 
 
x3  
  x3  +  0  = x3  1  + x 4  0 
x3
x4
x4
0
x4
0
1
So all of vectors that are orthogonal to both a and b are precisely those
vectors that lie in span {(−3, 2, 1, 0), (3, −2, 0, 1)}.
3. Section 1.4, Problem 15
(a) Prove or give a counterexample: If A is an m × n matrix and x ∈ Rn
satisfies Ax = 0, then either every entry of A is zero or x = 0.


−1 −1 0
Counterexample: Consider the matrix A =  0 −1 −1 and the
1
0 −1
 
1
vector x = −1. You have shown in a previous problem that Ax = 0
1
but A is not the zero matrix nor is x the zero vector.
(b) If A is an m × n matrix and x ∈ Rn satisfies Ax = 0 for every x ∈ Rn
then every entry of A is zero.
Proof : Suppose that Ax = 0 for every x ∈ Rn . Recalling that Ax =
x1 a1 + x2 a2 + . . . + xn an , where ai is the ith column of A and x =
(x1 , x2 , . . . , xn ), our strategy will be to show that each column of A is
made up entirely of zeroes. To begin, we let x1 = (1, 0, 0, . . . , 0). By
assumption, we know that Ax1 = 0. This is equivalent to saying that
Ax1 = 1a1 + 0a2 + . . . + 0an = 0. This implies that a1 = 0 which
says that the first column of A is entirely full of zeroes. In general, to
show that the ith column of A is 0, we take xi = (0, 0, . . . , 1, 0, . . . , 0)
where the 1 is in the ith position of the vector. Once again, by our
hypothesis, we know that Axi = 0 which is equivalent to saying that
Axi = 0a1 + 0a2 + . . . + 1ai + 0ai+1 + . . . + 0an = 0 which implies that
ai = 0. Since we can do this for each i = 1, 2, . . . , n, then each column
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Math 317: Linear Algebra
Homework 2 Solutions
Spring 2016
of A is comprised entirely of zeroes which in turn says that A is entirely
comprised of zeros.
4. Suppose that [A|b] is the augmented matrix associated with a linear system. You
know that performing row operations on [A|b] does not change the solution of the
system. However, no mention of column operations was ever made because column
operations can alter the solution.
(a) Describe the effect on the solution of a linear system when the j th column of
A is interchanged with the k th column of A.
We illiustrate how column operations affect the solution to a system
of linear equations by consider the 2 × 2 case. That is, consider the
following system of 2 equations with 2 unknowns:
a11 x1 + a12 x2 = b1
a21 x1 + a22 x2 = b2 .
The resulting augmented system is given by
a11 a12 b1
.
a21 a22 b2
If we switch the first column with the second column, we obtain
a12 a11 b1
a22 a21 b2
.
which corresponds to the following system of equations:
a12 x1 + a11 x2 = b1
a22 x1 + a21 x2 = b2 .
We see that this column operation has the effect of interchanging the
order of the unknowns–x1 and x2 have been permuted.
(b) Describe the effect when column A∗j (i.e. the j th column of A) is replaced by
cA∗j for some c ∈ R.
Once again, consider the following system of 2 equations with 2 unknowns:
a11 x1 + a12 x2 = b1
a21 x1 + a22 x2 = b2 .
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Math 317: Linear Algebra
Homework 2 Solutions
Spring 2016
The resulting augmented system is given by
a11 a12 b1
.
a21 a22 b2
If we multiply the 2nd column by c 6= 0 , we obtain
a11 ca12 b1
a21 ca22 b2
.
which corresponds to the following system of equations:
a11 x1 + ca12 x2 = b1
a21 x1 + ca22 x2 = b2 .
The solution to the new system is the same as the solution to the old
system except that the solution for the 2nd unknown of the new system
is x̂2 = 1c x2 .
(c) By experimenting with a 2 × 2 system, describe the effect when A∗j is replaced
by A∗j + cA∗k where A∗k denotes the k th column of A.
Once again, consider the following system of 2 equations with 2 unknowns:
a11 x1 + a12 x2 = b1
a21 x1 + a22 x2 = b2 .
The resulting augmented system is given by
a11 a12 b1
.
a21 a22 b2
If we multiply the 2nd column by c 6= 0 and add it to the first column
, we obtain
a11 + ca12 a12 b1
a21 + ca22 a22 b2
.
which corresponds to the following system of equations:
(ca12 + a11 )x1 + a12 x2 = b1
(ca22 + a21 )x1 + a22 x2 = b2 .
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Math 317: Linear Algebra
Homework 2 Solutions
Spring 2016
The solution to the new system is the same as the solution for the old
system except that the solution for the 1st unknown in the new system
is x̂1 = x1 − cx2 .
5. Section 1.5, Problem 4(b)
We seek to find constraint equations for b so that b is an element of V =
span {(1, 0, 1, 1), (0, 1, 1, 2), (1, 1, 1, 0)}. For b to be in this set it must be of
the form:
 
 
 
 
b1
1
0
1
b2 
0
1
1
  = c1   + c2   + c3   ,
b3 
1
1
1
b4
1
2
0
where c1 , c2 and c3 are unknown.This corresponds to the following system
of equations:
c1 + c3
c2 + c3
c1 + c2 + c3
c1 + 2c2
=
=
=
=
b1
b2
b3
b4
.
We may solve this system of equations via Gaussian elimination. The associated augmented matrix is given by


b1
1
−R1 +R3 →R3  0
b2 


−→
b3  −R1 +R4 →R4  0
b4
0

b1
0 1
 −3R3 →R4
1 1
b2
 −→
0 −1 b3 − b1 − b2 
0 −3 b4 − b1 − 2b2

1
 0

 1
1

1
 0

 0
0
0
1
1
2
1
1
1
0

0 1
b1
+R4 →R4
1 1
b2 
 −2R2−→

1 0 b3 − b 1
−R2 +R3 →R3
2 −1 b4 − b1

1 0 1
b1
 0 1 0
b2

 0 0 −1
b4 − b1 − 2b2
0 0 0 b2 − 3b3 − 4b1 + b4


.

For the above system to be consistent, we need b2 − 3b3 − 4b1 + b4 = 0. This
is precisely our constraint equation.
6. Section 1.5, Problem 7(a)
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Math 317: Linear Algebra
Homework 2 Solutions
Spring 2016
If possible, we wish to find a matrix A such that
• The rows of A are orthogonal to (1, 0, 1, 0),
• For some nonzero vector b ∈ R2 , x = (1, 0, 1, 0) and y = (1, 1, 1, 1) are
solutions to Ax = b.
Since b ∈ R2 , we know that we have two equations in our system with four
unknowns (since the solution x has four components). Thus the associated
matrix equation has the form:
 
x1

a b c d 
 x 2  = b1
e f g h  x3 
b2
x4
From here, we know the matrix A must satisfy a + c = b1 and e + g = b2
(the solution associated with (1, 0, 1, 0)), where either b1 or b2 is nonzero
(since we are told that b 6= 0). However, we are also told that the rows of
A are orthogonal to (1, 0, 1, 0). This is equivalent to saying that (a, b, c, d) ·
(1, 0, 1, 0) = 0 =⇒ a + c = 0 and (e, f, g, h) · (1, 0, 1, 0) = 0 =⇒ e + g = 0.
This tells us that b1 = b2 = 0 which is not possible since we need at least
one component of b to be nonzero. Thus, it is not possible to construct such
an A.
7. Section 1.5, Problem 13
Suppose A is an m × n matrix with rank m and v1 , v2 , . . . , vk ∈ Rn are vectors such that Span(v1 , v2 , . . . , vk ) = Rn . Prove that Span(Av1 , Av2 , . . . , Avk ) =
Rm .
Proof : To show that two sets are equal, say A and B, we must show that A ⊂
B and B ⊂ A. Let us begin by showing that Span(Av1 , Av2 , . . . , Avk ) ⊂
Rm . Suppose that b ∈ Span(Av1 , Av2 , . . . , Avk ). We want to show that
b ∈ Rm . Since b ∈ Span(Av1 , Av2 , . . . , Avk ), we know that
b = c1 (Av1 ) + c2 (Av2 ) + . . . ck (Avk ),
for some c1 , c2 , . . . , ck ∈ R. If we think of b1 = Av1 as a matrix equation
which represents a system of linear equations, then the fact that A is an m×n
matrix tells us that the system of equations has m equations. This means
that the system of equations has m right-hand sides. This is equivalent to
saying that b has m components which of course is the same thing as saying
that Av1 has m components. The argument is similar for Avi , i = 1, 2, . . . k,
and so b must also contain m elements as it is just the addition of vectors,
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Math 317: Linear Algebra
Homework 2 Solutions
Spring 2016
all with m components. Thus b ∈ Rm . The more challenging portion of
this problem is showing that Rm ⊂ Span(Av1 , Av2 , . . . , Avk ). We proceed
as follows. Let b ∈ Rm . Since rank(A) = m, we know that the reduced row
echelon of A has no zero rows (since A is an m × n matrix). Thus for any
b ∈ Rm , we can always find an x ∈ Rn such that Ax = b. That is, Ax = b is
consistent for every b ∈ Rm . Now since x ∈ Rn and Rn = span(v1 , . . . , vk ),
we know that x ∈ span(v1 , . . . , vk ) and hence
x = c1 v 1 + c2 v 2 . . . + ck v k ,
for c1 , c2 , . . . , ck ∈ Rn . Thus we have that
b = Ax
= A (c1 v1 + c2 v2 . . . + ck vk )
= c1 Av1 + c2 Av2 + . . . ck Avk ,
where we have used the two properties in Problem 13, 1.4, A(x + y) =
Ax+Ay and A(cx) = cAx. Thus, b ∈ Span(Av1 , Av2 , . . . , Avk ). Therefore,
Span(Av1 , Av2 , . . . , Avk ) = Rm .
8. Section 1.5, Problem 15
(a) Suppose that A is an m×n matrix with column vectors a1 , . . . , an ∈ Rm
such that a1 + . . . an = 0. We want to prove that rank(A) < n.
Proof: We want to show that rank(A) < n. Following the hint, we
look at solutions to Ax = 0. Recall that if Ax = 0 has infinitely many
solutions, then rank(A) < n. Since Ax = 0 either has one solution
(the trivial solution x = 0) or infinitely many solutions, it suffices to
find just one nontrivial solution to Ax = 0. (You should try and prove
that a system Ax = 0 cannot have exactly two solutions. That would
make a great test question!) To try and find a solution, we rewrite the
system of equations Ax = 0 as
Ax = 0 =⇒ x1 a1 + x2 a2 + . . . xn an = 0,
where ai is the ith column of A. Recall that this representation comes
from regarding the right hand side (0 in this case) as an element of
the span of the columns of A. For an explicit example of how this
representation is used, look at Problem 5 on Worksheet 1. Using this
representation, we see that a solution to Ax = 0 is precisely x =
(1, 1, 1, . . . 1), since this will force a1 + . . . an = 0 which is true by
assumption. Thus, Ax = 0 has a nontrivial solution which immediately
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Math 317: Linear Algebra
Homework 2 Solutions
Spring 2016
implies that Ax = 0 has infinitely many solutions. Thus, rank(A) < n.
(b) Suppose that A is an m×n matrix with column vectors a1 , . . . , an ∈ Rm
such that c1 a1 + . . . cn an = 0 where c1 6= 0. We want to prove that
rank(A) < n.
Proof: The proof here works the exact same way as (a) except that at
this point:
Ax = 0 =⇒ x1 a1 + x2 a2 + . . . xn an = 0,
we pick x1 = c1 , x2 = c2 , . . . xn = cn , so that c1 a1 + . . . cn an = 0. Thus,
Ax = 0 has a nontrivial solution (since ci 6= 0) which immediately
implies that Ax = 0 has infinitely many solutions. Thus, rank(A) < n.
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