the elementary matrix corresponding to each row operation that you use, and hence in each case
express both the matrix and the inverse as products of elementary matrices.
(a)
1
3
4
3
(b)
1
0
0
0
0
1
1
2
3
(c)
0
0
1
1
1
1
0
1
2
Solution
(a)
1
3
1
0
4
3
0
1
1
3
1
0
E
1
=
1
0
0

9

4
1

4
1
1
3
1
0
E
2
=
1
0
0
1
4
/
9

1
/
9
0

1
/
9
1
0

1
/
3
1
/
3
E
3
=
1

3
0
1
4
/
9

1
/
9
0
1
So
A

1
=

1
/
3
1
/
3
4
/
9

1
/
9
.
Now
E
3
E
2
E
1
A
=
I
, so
A

1
=
E
3
E
2
E
1
=
1

3
0
1
1
0
0

1
/
9
1
0

4
1
, and
A
=
E

1
1
E

1
2
E

1
3
=
1
0
4
1
1
0
0

9
1
3
0
1
.
22
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(b)
1
0
0
1
0
0
0
0
1
0
1
0
1
2
3
0
0
1
1
0
0
1
0
0
E
1
=
1
0
0
0
0
1
0
1
0
0
1
0
0
2
3

1
0
1

1
0
1
1
0
0
1
0
0
E
2
=
1
0
0
0
2
3

1
0
1
0
0
1
0
0
1
0
1
0
0
1
0
1
0
0
1
0
0
E
3
=
1
0
0
0
1
3
/
2

1
/
2
0
1
/
2
0
1
/
2
0
0
0
1
0
1
0
0
0
1
1
0
0
1
0
0
E
4
=
1
0
0
0
1
0

1
/
2

3
/
2
1
/
2
0
1

3
/
2
0
0
1
0
1
0
0
0
1
so
A

1
=
1
0
0

1
/
2

3
/
2
1
/
2
0
1
0
.
A

1
=
E
4
E
3
E
2
E
1
=
1
0
0
0
1

3
/
2
0
0
1
1
0
0
0
1
/
2
0
0
0
1
1
0
0
0
0
1
0
1
0
1
0
0
0
1
0

1
0
1
and
A
=
E

1
1
E

1
2
E

1
3
E

1
4
=
1
0
0
0
1
0
1
0
1
1
0
0
0
0
1
0
1
0
1
0
0
0
2
0
0
0
1
1
0
0
0
1
3
/
2
0
0
1
.
(c)
0
0
1
1
0
0
1
1
1
0
1
0
0
1
2
0
0
1
1
1
1
0
1
0
E
1
=
0
1
0
0
0
1
1
0
0
1
0
0
0
1
2
0
0
1
0
0
1
1
1
1
0
1
0
E
2
=
1
0
0
0
1
2
0
0
1
0
0
1
0
0
1
1
0
0
0
1
0
1
0

1
0
1

1
E
3
=
1

1
0
0
1
2
0
0
1
0
1
0
0
0
1
1
0
0
0
0
1
1
0
0
1
1

1
E
4
=
1
0
1
0
1
2
0
0
1
0
1
0
0
0
1
1
0
0
0
0
1
1
0
0
1
1

1
E
5
=
1
0
0
0
1
0

2
0
1
0
1

2
0
0
1
1
0
0
0
0
1
so
A

1
=
1
1

1

2
0
1
1
0
0
.
23
Then we have
A

1
=
E
5
E
4
E
3
E
2
E
1
=
1
0
0
0
1

2
0
0
1
1
0
1
0
1
0
0
0
1
1

1
0
0
1
0
0
0
1
1
0
0
0
0
1
0
1
0
0
1
0
1
0
0
0
0
1
and
A
=
E

1
1
E

1
2
E

1
3
E

1
4
E

1
5
=
0
1
0
1
0
0
0
0
1
1
0
0
0
0
1
0
1
0
1
1
0
0
1
0
0
0
1
1
0

1
0
1
0
0
0
1
1
0
0
0
1
2
0
0
1
.
15. Find all values of
α
for which the following matrix is
not
invertible.
A
=
1
4

3

2

7
6

1
3
α
Solution
Row reducing
A
gives
1
0

3
0
1
0
0
0
α

3
. To further row reduce to the identity we need
α

3
6
= 0,
so
α
6
= 3. Thus
A
is not invertible if
α
= 3.
16. Let
A
and
B
be matrices of the same size and suppose that
B
is invertible.
(a) Prove that
A
is invertible if and only if
AB
is invertible.
(b) Hence prove that either
A
+
B
and
I
+
AB

1
are both invertible or else both not invertible.
Solution
(a) This is an ‘if and only if’ statement. Therefore, we need to prove two things
(with the assumption given, that
B
is invertible): (i) if
A
is invertible then
AB
is invertible, and
(ii) if
AB
is invertible, then
A
is invertible.
The key property of invertible matrices that we will use is the following fact: if
X
and
Y
are
invertible matrices of the same size then
XY
is also invertible (in fact we saw in class that
(
XY
)

1
=
Y

1
X

1
).
We prove (i). We are given that
A
is invertible and
B
is invertible. Therefore,
AB
is invertible
by the key property above.
We prove (ii).
Now we are given that
B
is invertible and
AB
is invertible.
Observe that
(
AB
)
B

1
=
A
(
BB

1
) =
AI
=
A
.
But this shows that
A
is the product of two invertible
matrices, namely
AB
and
B

1
, and hence is invertible by the key property above (recall that
we saw in lecture that
B

1
is invertible with (
B

1
)

1
=
B
).