Math 51 Midterm I Solutions
(These ’solutions’ are written more in the form of a guide to correct think
ing, and a guide to the level of understanding we were trying to test. This
is for students who want to learn from their mistakes and improve their
understanding.)
PROBLEM 1(a).
This is a routine row reduced echelon form problem.
It
tests a computational skill only. Here is the answer.
1
0
1
0
1
0
1
2
0
3
0
0
0
1
1
0
0
0
0
0
If you missed it, just check your arithmetic and see where you slipped up.
PROBLEM 1(b).
The point here is that even though we don’t know the
entries a and b yet, we can start the rref process and get a lot of information.
Remember, the final rref form will tell us the dimension of the null space,
which will be the number of free variables.
1
1
1
1
2
a
1
3
b
→
1
1
1
0
1
a

1
0
2
b

1
→
1
1
1
0
1
a

1
0
0
1 +
b

2
a
At this point you see that rref(
A
) will definitely have pivots in the first two
rows and columns. One of two things happens after that. If 1+
b

2
a
= 0 then
the rref will have three pivots in the end and
N
(
A
) =
{
0
}
. If 1 +
b

2
a
= 0
then the last row of rref(
A
) will be all 0’s. There is then one free variable
and the null space is a line in
R
3
.
PROBLEM 2(i). There are two basic views of a matrixvector product
A
v
.
The first view is tested here. Multiply the respective columns of
A
by the
entries of
v
and add. So here
A
v
= 1
a
1
+ 2
a
2
+ 3
a
3
+ 4
a
4
(See the bottom of page 47 of the text.)
1
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Notice this view makes it crystal clear that vectors of the form
A
v
are exactly
the vectors in the column space of
A
, that is, in the span of the columns of
A
PROBLEM 2(ii). This is the second important view of a matrixvector prod
uct. The entries of
A
v
are obtained as dot products of the rows of
A
and
the vector
v
. Let’s ignore the business with transposes and just treat a row
as a vector, even though it is written sideways, not up and down. Then
A
v
=
r
1
v
r
2
v
r
3
v
.
(See the bottom of page 48 of the text.)
Notice this view makes it crystal clear that vectors
v
in the null space of
A
are exactly the vectors that are orthogonal to all rows of
A
.
PROBLEM 2(iii). The answer is ‘no’. Any four vectors in
R
3
are linearly
dependent because if you put those vectors in the columns of a 3 by 4 matrix,
like
A
here, then rref(
A
) has at most three pivots, therefore there are free
variables and hence nonzero vectors in
N
(
A
). But any nonzero vector in
N
(
A
) gives rise to a linear dependence relation between the columns of
A
,
using the formula for
A
v
as a linear combination of columns of
A
.
PROBLEM 2(iv).
Infinitely many.
The explanation just above in 2(iii)
explains why the null space of
A
contains at least a line in
R
4
. There is at
least one free variable, maybe more.
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