Math 317: Linear Algebra
Exam 1 Solutions
Spring 2016
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
2.2. Good luck!
2 1
1 2 1
1. Let A =
,B=
.
4 3
0 1 2
(a) (5 points) Is A a nonsingular matrix? Justify your claim.
To check and see if A is a nonsingular matrix, we show that rank(A) = 2.
Putting A in row echelon form, we see that
2 1 −2R1 +R2 →R2 2 1
−→
. We see that the row echelon form of A has two
4 3
0 1
nonzero rows and hence rank(A) = 2.
(b) (5 points) Is B a nonsingular matrix? Justify your claim.
B is not a square matrix (i.e. it is not an n × n matrix for a natural number
n) and hence B cannot be nonsingular.
(c) (10 points) Let C and D be n × n matrices such that CD = In where In is
the n × n identity matrix. Furthermore, suppose that C is nonsingular. Prove
that D is nonsingular. (Caution! You cannot use the fact that C is invertible.
(See directions.))
Proof: To prove that D is nonsingular, we show that Dx = 0 =⇒ x = 0.
Suppose that Dx = 0. Multiplying both sides by C, we obtain C(Dx) =
C(0) = 0 =⇒ (CD)x = 0 =⇒ In x = 0 =⇒ x = 0. Thus D is
nonsingular.
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Math 317: Linear Algebra
Exam 1 Solutions
Spring 2016
2. Let x and y be vectors in R3 .
(a) (10 points) Give an example to show that x · y = 0 does not necessarily imply
that x = 0 or y = 0.
Let x = (1, 2, 0) and y = (−2, 1, 0). Then x · y = (1, 2, 0) · (−2, 1, 0) =
1(−2) + 2(1) + 0(0) = 0, but x 6= 0 and y 6= 0.
(b) (10 points) Suppose that x · y = 0 for all x ∈ R3 . Prove that y = 0.
Proof : Let y = (y1 , y2 , y3 ). We show that y1 = y2 = y3 = 0. Let x1 =
(1, 0, 0). Then since x · y = 0 for every x ∈ R3 , then x1 · y = 0 =⇒
(1, 0, 0) · (y1 , y2 , y3 ) = 0 =⇒ 1(y1 ) + 0(y2 ) + 0(y3 ) = 0 =⇒ y1 = 0.
To show that y2 = y3 = 0, take x2 = (0, 1, 0) and x3 = (0, 0, 1). Then
0 = x2 · y = y2 and 0 = x3 · y = y3 . Thus y = 0.
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Math 317: Linear Algebra
Exam 1 Solutions
Spring 2016
3. 20 points, 5 points each.
(a) Find a parametric equation for the line 3x1 − x2 = 0. Write your answer as
span(a) for some vector a.
3x1 − x2 = 0 =⇒ x2 = 3x1 and so x = (x1 , x2 ) = (x1 , 3x1 ) = x1 (1, 3) +
(0, 0). Thus, the parametric line is given by x = {(0, 0) + t(1, 3); t ∈ R} =
span((1, 3)). So a = (1, 3).
(b) Let T1 : R2 → R2 be the linear transformation that projects x onto the line
3x1 − x2 = 0 in R2 . Give the standard matrix that represents T1 .
x·a
0
1
We calculate T1
and T1
where T1 (x) = projspan{a} x =
a.
0
1
kak2
(1, 0) · (1, 3) 1
1
T1
=
3
0
k(1, 3)k2
1
1
= 2
1 + 32 3
1/10
=
.
3/10
Similarly,
(0, 1) · (1, 3) 1
0
T1
=
3
1
k(1, 3)k2
2
1
= 2
2
1 +3 3
2/10
=
.
6/10
1/10 2/10
So the standard matrix for T1 is given by: A =
.
3/10 6/10
2
5
(c) Let T2 : R → R be the linear transformation such that T2
=
and
1
3
0
1
T2
=
. Give the standard matrix that represents T2 .
1
−3
2
2
1
0
0
Once again, we calculate T2
and T2
. We note that T2
=
0
1
1
3
Math 317: Linear Algebra
Exam 1 Solutions
Spring 2016
1
1
, and so we only need to find T2
. To do so, we note that
−3
0
2
1
0
T2
= 2T2
+ 1T2
=⇒
1
0
1
5
1
1
= 2T2
+
=⇒
3
0
−3
1
2
T2
=
.
0
3
2 1
So the standard matrix for T2 is given by: B =
.
3 −3
(d) Give the standard matrix that represents T2 ◦ T1 .
Recalling that the action of the transposition of linear transformations is
given by matrix multiplication,
we have that the standard matrix for T2 ◦ T1
2 1
1/10 2/10
is given by BA =
.
3 −3 3/10 6/10
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Math 317: Linear Algebra
Exam 1 Solutions
Spring 2016
4. (a) (10 points) Let b1 , b2 , b3 , b4 be constants in R. Find a row echelon form for
the following matrix.

1
0

1
2
1
1
1
1
2
1
1
2

b1
b2 
.
b3 
b4
We put the matrix in row echelon form as follows:


b1
1


b2  −2R1 +R4 →R4 0
−→
b3  −R1 +R3 →R3 0
b4
0

1 2
b1
 −R3 +R4 →R4
1 1
b2

−→
0 −1
b3 − b1 
0 −1 b4 − 2b1 − b2

1
0

1
2

1
0

0
0
1
1
1
1
2
1
1
2

2
b1
1
b2 
4 →R4
 R2 +R
−→
−1 b3 − b1 
−2 b4 − 2b1
1
1
0
−1

1
0

0
0

1 2
b1

1 1
b2


0 −1
b3 − b1
0 0 b4 − 2b1 − b2 − (b3 − b1 )
 
b1
b2 

(b) (5 points) Find constraint equations (if any) that b = 
b3  must satisfy in
b4
order for the following system of equations to be consistent:
x1 + x2 + 2x3
x2 + x3
x1 + x2 + x3
2x1 + x2 + 2x3
=
=
=
=
b1
b2
b3
b4
Hint: Use your work in (a) to help with this problem.
In matrix-vector form, the following system of equations can be expressed
as

1
0
[A|b] = 
1
2
1
1
1
1
2
1
1
2

b1
b2 
.
b3 
b4
This is precisely the matrix you worked with in part (a). As we have
already put this matrix in row echelon form, we see that the constraint
5
Math 317: Linear Algebra
Exam 1 Solutions
Spring 2016
equation for b = (b1 , b2 , b3 , b4 ) is given precisely by b4 − 2b1 − b2 − (b3 − b1 ) =
b4 − 3b1 − b2 − b3 = 0.
(c) (5 points) Determine if (1, 2, 1, 0) ∈ span {(1, 0, 1, 2), (1, 1, 1, 1), (2, 1, 1, 2)}.
Hint: Use your work in (b) to help with this problem.
To determine if (1, 2, 1, 0) ∈ span {(1, 0, 1, 2), (1, 1, 1, 1), (2, 1, 1, 2)} we need
to check and see that (1, 2, 1, 0) satisfy the constraint equations given in
part(b) noting that the columns in the matrix in part(b) correspond to the
vectors in our considered span set. Since b4 − 3b1 − b2 − b3 = 0 we have that
0 − 3(1) − 2 − 1 = 0. Thus, (1, 2, 1, 0) satisfies the constraint equation and
hence (1, 2, 1, 0) ∈ span {(1, 0, 1, 2), (1, 1, 1, 1), (2, 1, 1, 2)}.
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Math 317: Linear Algebra
Exam 1 Solutions
Spring 2016
5. (20 points, 5 points each) Determine if the following statements are true or false. If
the statement is true, give a brief justification (or an example if the problem asks
for one). If the statement is false, provide a counterexample or explain why the
statement is not true.
(a) Suppose that A is an m × n matrix such that Ax = b is consistent. If m < n,
then Ax = b has infinitely many solutions.
True. Since we know that Ax = b is consistent, we know that the linear
system either has one solution or infinitely many solutions. The system has
one solution if rank(A) = n and infinitely many solutions if rank(A) < n.
Recalling that rank(A) is the number of nonzero rows in any echelon form
of A, and noting that A has m rows, we see that rank(A) ≤ m. Since m < n
this immediately tells us that rank(A) < n hence giving us infinitely many
solutions.
(b) Suppose that A is an m × n matrix such that Ax = 0 has only the trivial
solution. Then Ax = b has a unique solution.
This statement is false as we do not know that Ax = b is consistent. For
instance

1
0
A=
0
0
0
1
0
0

0
0

1
0
has only the trivial solution to Ax = 0. However, the system Ax = b is
inconsistent for any b = (b1 , b2 , b3 , b4 ) whenever b4 6= 0.
(c) Suppose that A is a 2 × 2 matrix such that A 6= 02×2 . Then A2 6= 02×2 .
0 1
0 1 0 1
2
This is false. Consider the matrix A =
. Then A =
=
0 0
0 0 0 0
0 0
.
0 0
x1
x1 + 2x2
(d) Let T : R → R be defined by T
=
. T is a linear transx2
0
formation.
2
2
True. We verify that
T (x + y) =
T(x) + T (y) and cT (x) = T (cx) for any
x
y
c ∈ R where x = 1 and y = 1 .
x2
y2
x1
y1
x1 + y 1
(x1 + y1 ) + 2(x2 + y2 )
T
+
=T
=
x2
y2
x2 + y 2
0
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Math 317: Linear Algebra
Exam 1 Solutions
Spring 2016
x1 + 2x2
y1 + 2y2
=
+
= T (x) + T (y).
0
0
We also see that
x1
cx1
cx1 + 2(cx2 )
x1 + 2x2
T c
=T
=
=c
x2
cx2
0
0
x1
= cT
.
x2
8