18.06 Problem Set 2
Due Wednesday, 20 February 2008 at 4 pm in 2106.
Problem 1:
a) Do problem 5 from section 2.4 (pg. 65) in the book.
b) Do problem 26 in section 2.4 (pg. 69).
Solution
(5+5 points)
a) We have
1
b
0
1
n
=
1
nb
0
1
and
2
2
0
0
n
=
2
n
2
n
0
0
If you insist on a rigorous proof, you can use induction.
b)
1
0
2
4
2
1
3
3
0
1
2
1
=
1
2
2
3
3
0 +
0
4
1
1
2
1
=
3
3
0
6
6
0
6
6
0
+
0
0
0
4
8
4
1
2
1
=
3
3
0
10
14
4
7
8
1
Problem 2:
Do problem 24 from section 2.4 (pg. 68).
Solution
(5+5 points)
In general, if we take an upper triangular matrix with 0s along the diagonal,
some power of it will be 0. The matrices I picked are all instances of this principle.
a)
A
=
0
1
0
0
will work.
b)
A
=
0
1
0
0
0
1
0
0
0
. Do you see the pattern?
1
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Problem 3:
Do problem 7 from section 2.5 (pg. 79).
Solution
(3+4+3 points)
a) Suppose we had a solution
x
. The equation
Ax
= (1
,
0
,
0) amounts to the three
equations row 1
·
x
= 1, row 2
·
x
= 0, row 3
·
x
= 0. But we know row 1 + row 2 =
row 3. The three dot products should sum correctly, but they don’t:
(row 1 + row 2)
·
x
= 1 + 0 = 0 = row 3
·
x
Thus, there can’t be a solution
x
. This shows that
A
is not invertible.
b) By the same reasoning as above, we must have
b
1
+
b
2
=
b
3
for any allowable
solution. It’s possible that even some of these won’t have solutions; we don’t have
enough information about
A
to say for sure. For example, the 0 matrix satisfies the
requirement to be
A
, but 0
x
=
b
certainly won’t have any solutions unless all the
b
i
are 0. We’ll learn a more precise way to discuss this when we talk about rank.
c) After elimination row 3 will become all 0.
How do we see this?
We know
from part a) that
A
is not invertible (if it were, every choice of
b
would have a a
solution). So
A
can’t have three (nonzero) pivots. Since
A
has three rows but at
most two pivots, at least one row must be all 0. (Remember that a pivot is the first
nonzero entry in a row; a row without a pivot must not have any nonzero entries
at all.) Because we always eliminate downwards, the row without a pivot will be
the bottom one (it’s possible that the earlier rows also will be all 0).
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 Spring '08
 CHANILLO
 Diagonal matrix, Triangular matrix, Row

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