MATH 1105, PARTIAL EXAM 1 SOLUTIONS TO PRACTICE ...

MATH 1105, PARTIAL EXAM 1
SOLUTIONS TO PRACTICE PROBLEMS
JOHN GOODRICK
Problem 1. For each of the matrices below, determine whether or not it is
in row echelon form. If it is not, put it into row echelon form using a single
elementary row operation.
A =
⎡
⎣ 1 2 3 −1
0 1 7 7
0 1 3 0
⎤
⎦ B =
⎡
⎣ 0 0 1
0 1 3
1 −1 2
⎤
⎦ C =
⎡
⎣ 1 3 1
0 1 7
0 0 −4
SOLUTION: The matrix A is not in row echelon form – the leading entry of
row 3 is not to the right of the first nonzero entry of row 2. (See the definition
on p. 57 of the textbook.) However, by subtracting row 2 from row 3, we get
the following matrix, which is in row echelon form:
⎡
⎣ 1 2 3 −1
0 1 7 7
0 0 −4 −7
The matrix B is also not in row echelon form, for the same reason as A is
not. By swapping row 1 and row 3, we do get a matrix in row echelon form:
B ∼
⎡
⎣ 1 −1 2
0 1 3
0 0 1
The matrix C is in row echelon form. (It is not in reduced row echelon
form, since the pivot −4 in row 3 is not equal to 1, but this wasn’t what I was
asking about.) If this is unclear, you should review the definition on p. 57 of
the textbook.
Problem 2.
(a) Find the general solution to the linear system
⎤
⎦
⎤
⎦
x 1 + 2x 2 − x 3 + 2x 4 = 0
3x 1 + 6x 2 − 3x 3 + 9x 4 = 0.
SOLUTION: Put the system into an augmented matrix, then row reduce,
just as described in section 1.4 of the textbook:
1
⎤
⎦

2 JOHN GOODRICK
[ ] [ ] [ ]
1 2 −1 2 | 0 1 2 −1 2 | 0 1 2 −1 2 | 0
∼
∼
3 6 −3 9 | 0 0 0 0 3 | 0 0 0 0 1 | 0
[ ]
1 2 −1 0 | 0
∼
.
0 0 0 1 | 0
From here, we can read off the general solution: x 1 = −2x 2 + x 3 − 2x 4 ,
x 4 = 0, and since there are no pivots in columns 2 and 3, the variables x 2 and
x 3 are free.
(Don’t forget to say that x 2 and x 3 are free variables!)
(b) Is there any pair of numbers b 1 and b 2 in R such that the system
x 1 + 2x 2 − x 3 + 2x 4 = b 1
3x 1 + 6x 2 − 3x 3 + 9x 4 = b 2
has no solution for x 1 , x 2 , x 3 , and x 4 ? Explain why or why not.
If you write out the augmented matrix for the system with b 1 and b 2 and
row-reduce it like in part (a), you will get:
[ ] [ ]
1 2 −1 2 | b1 1 2 −1 2 | b1
∼
3 6 −3 9 | b 2 0 0 0 3 | b 2 − 3b 1
[ ] [ ]
1 2 −1 2 | b1 1 2 −1 0 | 3b1 − 2b 2
∼
b
0 0 0 1 | 23
∼
3
b
− b 1 0 0 0 1 | 23
.
− b 1
Notice that no matter what the numbers b 1 and b 2 are, the final matrix has
pivots in columns 1 and 4, but no pivot in the rightmost column – therefore
the solution is always consistent, and there is no pair of numbers b 1 and b 2
for which the system has no solution.
Problem 3. (a) Let
⎧⎡
⎨
S = ⎣ 1 2
⎩
3
⎤
⎦ ,
⎡
⎣ 4 5
6
⎤
⎦ ,
⎡
⎣ 7 8
9
⎤⎫
⎬
⎦
⎭ .
Is S linearly independent? Explain why or why not.
SOLUTION: The best general way to do a problem like this is to put the
vectors into a matrix A (as columns), then row-reduce:
⎡
A = ⎣ 1 4 7
⎤ ⎡
2 5 8 ⎦ ∼ ⎣ 1 4 7
⎤ ⎡
0 −3 −6 ⎦ ∼ ⎣ 1 4 7
⎤
0 −3 −6 ⎦ .
3 6 9 0 −6 −12 0 0 0
This last matrix is in row echelon form, but there is no pivot in the third
column – therefore S is linearly dependent.

MATH 1105, PARTIAL EXAM 1 SOLUTIONS TO PRACTICE PROBLEMS 3
(b) Is S a basis for R 3 ?
SOLUTION: No, S is not a basis for R 3 . Any basis must be linearly
independent (see Theorem 2.1 in the textbook), and by part (a), S is not
linearly independent.
(c) Find a basis for span(S).
SOLUTION: To find the basis, we put the vectors into a matrix A as
columns and row reduce to row echelon form, as in part (a). Since there are
pivots in columns 1 and 2 (but not in column 3), a basis for the span of S is
the first two vectors in S:
⎧⎡
⎤ ⎡ ⎤⎫
⎨
⎣ 1 2 ⎦ ,
⎩
3
⎣ 4 5
6
⎬
⎦
⎭ .
Problem 4. For each of the following matrices, determine whether or not
it is invertible. If it is invertible, find its inverse.
⎡
[ ]
2 1
A =
B = ⎣ 1 0 1
⎤
[ ]
0 1 0 ⎦ 1 0 1
C =
1 2
0 1 0
1 0 1
Solution:
D =
⎡
⎢
⎣
0 1 0 0
1 0 0 0
2 0 1 0
2
3 0 0 −1
⎤
⎥
⎦ .
The matrix A is invertible:
[ ] [ ] [
2 1 | 1 0 2 1 | 1 0 1
1 1
| 0
∼
1 2 | 0 1 0 3 | − 1 ∼ 2 2
1 0 3 | − 1 1
2 2 2 2
[ ] [ ]
1
1 1
| 0
∼ 2 2 1 0 |
2
−
0 1 | − 1 2 ∼
1 3 3
0 1 | − 1 2 .
3 3
3 3
]
The matrix B is not invertible. We can see this is by counting its pivots:
by subtracting Row 1 from Row 3, we get the reduced row echelon matrix
⎡
⎣ 1 0 1
⎤
0 1 0 ⎦ ,
0 0 0

4 JOHN GOODRICK
which has pivots in only 2 of the 3 columns, and so B cannot be row reduced
to I 3 . Therefore B is not invertible.
The matrix C is not invertible, simply because it is not a square matrix.
Recall that in the definition of an invertible matrix (on page 75 of the textbook),
we require that the matrix be n × n for some n.
For the matrix D, we augment the matrix by I 4 on the right and then rowreduce
the left half (as in section 1.5):
∼
⎡
⎢
⎣
⎡
⎢
⎣
0 1 0 0 | 1 0 0 0
1 0 0 0 | 0 1 0 0
2 0 1 0 | 0 0 1 0
2
3 0 0 −1 | 0 0 0 1
1 0 0 0 | 0 1 0 0
0 1 0 0 | 1 0 0 0
0 0 1 0 | 0 −2 1 0
2
0 0 0 −1 | 0 −3 0 1
⎤
⎡
⎥
⎦ ∼ ⎢
⎣
⎤ ⎡
⎥
⎦ ∼ ⎢
⎣
1 0 0 0 | 0 1 0 0
0 1 0 0 | 1 0 0 0
2 0 1 0 | 0 0 1 0
2
3 0 0 −1 | 0 0 0 1
⎤
⎥
⎦
1 0 0 0 | 0 1 0 0
0 1 0 0 | 1 0 0 0
0 0 1 0 | 0 −4 2 0
0 0 0 1 | 0 3 0 −1
Since we were able to row-reduce the left half to the identity matrix I 4 , the
matrix D is invertible, and the row reduction above shows that
⎡
⎤
0 1 0 0
D −1 = ⎢ 1 0 0 0
⎥
⎣ 0 −4 2 0 ⎦ .
0 3 0 −1
Problem 5. (a) Write the solution set of
3x 1 + 2x 2 + 7x 3 + x 4 = 2
3x 1 + 3x 2 + 6x 3 + x 4 = 17
as ⃗p + span(⃗v 1 , ⃗v 2 ) for some vectors ⃗p, ⃗v 1 , and ⃗v 2 .
SOLUTION: First, you should find the general solution by the method in
section 1.4:
[
3 2 7 1 | 2
3 3 6 1 | 17
⎤
⎥
⎦ .
] [ ] [ ]
3 2 7 1 | 2 1
2 7 1 2
|
∼
∼ 3 3 3 3
0 1 −1 0 | 15 0 1 −1 0 | 15
[ ]
1 0 3
1
| −
∼
28 3 3 .
0 1 −1 0 | 15
This is in reduced row echelon form, so we can read of the general soution:
x 1 = −3x 3 − 1 3 x 4 − 28 3 , x 2 = x 3 + 15, and x 3 and x 4 are free.
From here, we can proceed as in section 1.6: the solution set consists of all
(x 1 , x 2 , x 3 , x 4 ) in R 4 such that

MATH 1105, PARTIAL EXAM 1 SOLUTIONS TO PRACTICE PROBLEMS 5
⎡ ⎤ ⎡
x 1
⎢ x 2
⎥
⎣ x 3
⎦ = ⎢
⎣
x 4
−3x 3 − 1 3 x 4 − 28 3
x 3 + 15
x 3
x 4
⎤
⎥
⎦ = x 3
⎡
⎢
⎣
−3
1
1
0
⎤
⎥
⎦ + x 4
⎡
⎢
⎣
− 1 3
0
0
1
⎤ ⎡
⎥
⎦ + ⎢
⎣
Therefore we can let ⃗p = [ − 28 3 , 15, 0, 0] , ⃗v 1 = [−3, 1, 1, 0], and ⃗v 2 = [ − 1 3 , 0, 0, 1] .
(This is not the only possible solution to this problem.)
(b) Write the solution set of
as the span of two vectors.
3x 1 + 2x 2 + 7x 3 + x 4 = 0
3x 1 + 3x 2 + 6x 3 + x 4 = 0
SOLUTION: Note that this homogeneous linear system is the same as the
system in part (a) except that the numbers on the right sides of the equations
are 0. So using the same row reduction as in part (a), we get:
[
3 2 7 1 | 0
3 3 6 1 | 0
]
∼
[
1 0 3
1
3
| 0
0 1 −1 0 | 0
Now using the same technique as in part (a), we can see that the general
solution is
⎡ ⎤ ⎡
x 1 −3x 3 − 1 ⎢ x 2
⎥
⎣ x 3
⎦ = x ⎤ ⎡
3 4
⎢ x 3
⎥
⎣ x 3
⎦ = x ⎢
3 ⎣
x 4 x 4
with x 3 and x 4 free. So the solution set is
⎛⎡
⎤ ⎡
−3 − 1 3
span ⎜⎢
1
⎥
⎝⎣
1 ⎦ , ⎢ 0
⎣ 0
0 1
−3
1
1
0
⎤
⎥
⎦ + x 4
⎤⎞
⎥⎟
⎦⎠ .
Problem 6. Let S be the set of all vectors ⃗v in R 3 such that
⃗v · [1, 3, 11] = 0.
Is S a subspace of R 3 ? Why or why not?
SOLUTION: Yes, the set S is a subspace of R 3 . One way to see this is to
note that S is the set of all ⃗v = [v 1 , v 2 , v 3 ] such that
v 1 + 3v 2 + 11v 3 = 0,
⎡
⎢
⎣
]
.
− 1 3
0
0
1
⎤
⎥
⎦ ,
− 28 3
15
0
0
⎤
⎥
⎦ .

6 JOHN GOODRICK
which is a homogeneous linear system (with only one equation), and as explained
in section 1.6, the solution set to a homogeneous linear system is always
a subspace of the appropriate R n .
Alternatively, you could have checked that the zero vector [0, 0, 0] is in S
and that S is closed under vector addition and scalar multiplication – see
Definition 1.16 on page 89 of the textbook.
Problem 7. For how many different vectors ⃗v in R 3 is it true that
⎧⎡
⎨
Span ⎣ 1 ⎤ ⎡
0 ⎦ , ⎣ 0 ⎤ ⎫
⎬
1 ⎦ , ⃗v
⎩
0 0
⎭ = R3 ?
Explain.
SOLUTION: There are infinitely many such vectors ⃗v.
Here one way to prove this: if we write
⎡
⃗v = ⎣ v ⎤
1
v 2
⎦ ,
v 3
then by Theorem 1.16 of section 1.6, the span of the set
⎧⎡
⎨
⎣ 1 ⎤ ⎡
0 ⎦ , ⎣ 0 ⎤ ⎫
⎬
1 ⎦ , ⃗v
⎩
0 0
⎭
is all of R 3 if and only if the matrix
⎡
A =
⎣ 1 0 v ⎤
1
0 1 v 2
⎦
0 0 v 3
is row equivalent to I 3 .
The matrix A is already in row echelon form. It has either two or three
pivots, depending on whether or not v 3 = 0: the two 1’s are pivots in any case,
the v 3 in the lower right is a pivot if it is nonzero, and otherwise v 3 is not a
pivot. As long as v 3 ≠ 0, we can row reduce the matrix A to I 3 by adding
appropriate multiples of row 3 to rows 1 and 2. (The values of v 1 and v 2 do
not matter.)
Since there are infinitely many such choices for v 3 (it can be any nonzero
number in R), there are infinitely many such vectors ⃗v.
Problem 8. Is the vector (0, 3, 1) in
⎧⎡
⎨
Span ⎣ 1 ⎤ ⎡
2 ⎦ , ⎣ 2 ⎤⎫
⎬
1 ⎦
⎩
1 1
⎭ ?
Explain why or why not.

MATH 1105, PARTIAL EXAM 1 SOLUTIONS TO PRACTICE PROBLEMS 7
SOLUTION: Yes, the vector (0, 3, 1) is in this span.
The vector (0, 3, 1) is in this span if and only if the system
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
x 1 ·
⎣ 1 2
1
⎦ + x 2 ·
⎣ 2 1
1
⎦ =
⎣ 0 3
1
has a solution for x 1 and x 2 . But this is equivalent to the linear system
x 1 + 2x 2 = 0
2x 1 + x 2 = 3
x 1 + x 2 = 1.
We convert this into an augmented matrix, then row reduce:
⎡
⎣ 1 2 | 0
2 1 | 3
1 1 | 1
⎤
⎦ ∼
⎡
⎣ 1 2 | 0
0 −3 | 3
0 −1 | 1
⎤
⎦ ∼
⎡
⎦
⎣ 1 2 | 0
0 −3 | 3
0 0 | 0
This last matrix is in row echelon form, with pivots in columns 1 and 2, but
no pivot in the rightmost column. Therefore the system is consistent (it has a
solution).
⎤
⎦ .