36. By Exercise 35, the
i
th column of
A
is
Ae
i
, for
i
= 1
,
2
,
3. Therefore,
A
=
1
4
7
2
5
8
3
6
9
.
37. We have to solve the system
x
1
+
2
x
2
=
2
x
3
=
1
or
x
1
= 2

2
x
2
x
3
= 1
.
Let
x
2
=
t
. Then the solutions are of the form
x
1
x
2
x
3
=
2

2
t
t
1
, where
t
is an arbitrary
real number.
38. We will illustrate our reasoning with an example. We generate the “random” 3
×
3 matrix
A
=
0
.
141
0
.
592
0
.
653
0
.
589
0
.
793
0
.
238
0
.
462
0
.
643
0
.
383
.
Since the entries of this matrix are chosen from a large pool of numbers (in our case
1000, from 0.000 to 0.999), it is unlikely that any of the entries will be zero (and even
less likely that the whole first column will consist of zeros).
This means that we will
usually be able to apply Steps 2 and 3 of the GaussJordan algorithm to turn the first
41
Chapter 1
ISM:
Linear Algebra
column into
1
0
0
; this is indeed possible in our example:
0
.
141
0
.
592
0
.
653
0
.
589
0
.
793
0
.
238
0
.
462
0
.
643
0
.
383
→
1
4
.
199
4
.
631
0

1
.
680

2
.
490
0

1
.
297

1
.
757
.
Again, it is unlikely that any entries in the second column of the new matrix will be zero.
Therefore, we can turn the second column into
0
1
0
.
Likewise, we will be able to clear up the third column, so that rref(
A
) =
1
0
0
0
1
0
0
0
1
.
We summarize:
As we apply GaussJordan elimination to a random matrix
A
(of any size), it is unlikely
that we will ever encounter a zero on the diagonal. Therefore, rref(
A
) is likely to have all
ones along the diagonal.
39. We will usually get rref(
A
) =
1
0
0
a
0
1
0
b
0
0
1
c
, where
a, b
, and
c
are arbitrary.
40. We will usually have rref(
A
) =
1
0
0
0
1
0
0
0
1
0
0
0
.
(Compare with the summary to Exercise 38.)
41. If
Ax
=
b
is a “random” system, then rref(
A
) will usually be
1
0
0
0
1
0
0
0
1
, so that we will
have a unique solution.
42. If
Ax
=
b
is a “random” system of three equations with four unknowns, then rref(
A
) will
usually be
1
0
0
a
0
1
0
b
0
0
1
c
(by Exercise 39), so that the system will have infinitely many solutions
(
x
4
is a free variable).
42
ISM:
Linear Algebra
Section 1.3
43. If
Ax
=
b
is a “random” system of equations with three unknowns, then rref[
A
.
.
.
b
] will
usually be
1
0
0
.
.
.0
0
1
0
.
.
.0
0
0
1
.
.
.0
0
0
0
.
.
.1
, so that the system is inconsistent.
44. Let
E
= rref(
A
)
,
and note that all the entries in the last row of
E
must be zero, by
the definition of rref.
If
c
is any vector in
R
n
whose last component isn’t zero, then
the system
Ex
=
c
will be inconsistent.
Now consider the elementary row operations
that transform
A
into
E,
and apply the opposite operations, in reversed order, to the
augmented matrix
E
.
.
.
c
.
You end up with an augmented matrix
A
.
.
.
b
that
represents an inconsistent system
Ax
=
b,
as required.
45. Write
A
= [
v
1
v
2
. . . v
m
] and
x
=
x
1
. . .
x
m
.
Then
A
(
kx
) = [
v
1
. . . v
m
]
kx
1
. . .
kx
m
=
kx
1
v
1
+
· · ·
+
kx
m
v
m
and
k
(
Ax
) =
k
(
x
1
v
1
+
· · ·
+
x
m
v
m
) =
kx
1
v
1
+
· · ·
+
kx
m
v
m
. The
two results agree, as claimed.