MATH 51 FINAL EXAM SOLUTIONS
Professors Clingher, Munson, and White
March 15, 2004
1
Consider the matrices
A
=
1
1
0
0
2
2
2
0
1
6
0
1

1 1
3

1

2
1
1

1
and
R
=
1 0
1
0 1
0 1

1 0 1
0 0
0
1 2
0 0
0
0 0
.
The matrix
R
is the row reduced echelon form of
A
. (You do not need to check
this.)
1(a).
Find a basis for the column space of
A
. Solution:
R
has pivots in columns
1, 2 and 4, so the corresponding columns of
A
form the basis:
1
2
0

1
,
1
2
1

2
,
0
1
1
1
1(b).
Find a basis for the column space of
R
. Solution:
R
is its own rref, so again
the pivot columns of
R
(that is, the ﬁrst, second, and fourth columns) form the
basis:
e
1
,
e
2
, and
e
3
.
1(c).
Find a basis for the nullspace of
A
. The nullspace of
A
consists of all solutions
of
A
x
= 0, or, equivalently, of
R
x
= 0. From
R
x
= 0, we see that the free variables
are
x
3
and
x
5
and that
x
=

x
3

x
5
x
3

x
5
x
3

2
x
5
x
5
=
x
3

1
1
1
0
0
+
x
5

1

1
0

2
1
.
Thus

1
1
1
0
0
and

1

1
0

2
1
form a basis for
N
(
A
).
2.
Find all solutions of
x
1
+ 2
x
2
+
x
3
+
x
4
=
7
x
1
+ 2
x
2
+ 2
x
3

x
4
= 12
2
x
1
+ 4
x
2
+ 6
x
4
=
4
.
Solution: see example 6.4 on page 44 of the Levandosky text. The rref is
1 2 0
3
.
.
. 2
0 0 1

2
.
.
. 5
0 0 0
0
.
.
. 0
1
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View Full Documentfrom which we get the solution:
x
=
2
0
5
0
+
x
2

2
1
0
0
+
x
4

3
0
2
1
.
3(a).
Find all eigenvalues of the matrix
A
=
5 0 0
1 2 1
1 1 2
. Solution:
0 = det(
λI

A
) =
ﬂ
ﬂ
ﬂ
ﬂ
ﬂ
ﬂ
λ

5
0
0

1
λ

2

1

1

1
λ

2
ﬂ
ﬂ
ﬂ
ﬂ
ﬂ
ﬂ
= (
λ

5)
ﬂ
ﬂ
ﬂ
ﬂ
λ

2

1

1
λ

2
ﬂ
ﬂ
ﬂ
ﬂ
= (
λ

5)((
λ

2)
2

(

1)
2
) = (
λ

5)(
λ

3)(
λ

1)
so the eigenvalues are 5, 3, and 1.
3(b).
The matrix
M
=
5

6

6

1
4
2
3

6

4
has
λ
= 2 as one of its eigenvalues. (You
need not check this.) Let
V
be the eigenspace corresponding to this eigenvalue.
(In other words,
V
consists of all eigenvectors with eigenvalue 2 together with the
origin.) Find a basis for
V
.
Solution: 2
I

M
=

3
6
6
1

2

2

3
6
6
. Note that
V
is the nullspace of 2
I

M
. We
ﬁnd this nullspace by solving (2
I

M
)
v
= 0. Solving by Gaussian elimination, we
get the single equation
x

2
y

2
z
= 0
so
y
and
z
are free and
v
=
x
y
z
=
2
y
+ 2
z
y
z
=
y
2
1
0
+
z
2
0
1
.
Thus
2
1
0
and
2
0
1
form a basis for
V
.
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 '07
 Staff
 Linear Algebra, Algebra, Differential Calculus, Matrices, basis, ax

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