so that the matrix is not invertible in this case. On the other hand, when
a
=
0
,
after one rowreduction step we get
⎡
⎣

b

c0
00
c
b
⎤
⎦
,
so the rref can’t have a pivot in the second column, and again we see that the matrix
is not invertible. So the matrix in question is never invertible.
*34
The matrix
A
is invertible if and only if none of
a
,
b
,
c
is zero, and then the
inverse is
⎡
⎣
1
a
0
1
b
0
1
c
⎤
⎦
.
The general case is the same: an
n
×
n
diagonal matrix is invertible if and only if
none of the diagonal entries
a
11
,
a
22
,...,
a
nn
is zero, and then the inverse will be
a diagonal matrix whose entries are the reciprocals
1/a
ii
.
*43
We basically did this in class: the transformation that sends
~
x
to
A
(
B
~
x
)
is
the one that corresponds to the matrix
AB
. So the problem is asking us to Fnd
(
AB
)

1
(under the assumption that both
A
and
B
are invertible). We know that
(
AB
)

1
=
B

1
A

1
, and that is the answer to the problem, i.e.,
~
x
=
B

1
(
A

1
~
y
)
.
b. Section
2
.
4
31
The problem has inFnitely many solutions, as we can see by rowreducing the
appropriate matrix. The solutions are:
⎡
⎣
2
+
t

1
+
s

1

2t
1

s
ts
⎤
⎦
,
where
t
and
s
can be arbitrary.
*32
The equation
B
~
x
=
~
0
unpacks into a homogeneous linear system with three
unknowns and two equations. We showed, early in the game, that such systems
always have inFnitely many solutions, soinpa
r
t
icu
la
rtheremus
tbeanon
zero
solution
~
x
0
.
Now, if
AB
=
I
3
,wewou
ldhave
~
x
0
=
I
3
~
x
0
=
AB
~
x
0
=
A
(
B
~
x
0
)=
A
~
0
=
~
0,
which contradicts the fact that
~
x
0
is not zero. Hence, it is impossible for
AB
to be
equal to
I
3
.
When a true genius appears in the world, you may know him by this sign, that the dunces are all in confederacy against him. – Jonathan Swift