Mathematics 9 (Complete Solutions)

# Q6 verify that

If
A =

 4 0 -1 2

,
B =

 -4 -2 1 -1

,
D =

 3 1 -2 2

, then verify that

 (i) (AB)-1 = B-1A-1 (ii) (DA)-1 = A-1D-1

Solution:

(i) (AB)-1 = B-1A-1
First we will solve L.H.S
So first we will find AB
So,
AB =

 4 0 -1 2

 -4 -2 1 -1

AB =

 4(-4) + 0(1) 4(-2) + 0(-1) -1(-4) + 2(1) -1(-2) + 2(-1)

AB =

 -16 + 0 -8 + 0 4 + 2 2 - 2

AB =

 -16 -8 6 0

(AB)-1 =
So now we will find | AB |
| AB | =
 -16 -8 6 0
| AB | = (-16)(0) - (-8)(6)
| AB | = 0 + 48
| AB | = 48
So now we will find Adj AB

 0 8 -6 -16

(AB)-1 =

 0 8 -6 -16

48
(AB)-1 =
 1 48

 0 8 -6 -16

(AB)-1 =

0 (
 1 48
)
8 (
 1 48
)
-6 (
 1 48
)
-16 (
 1 48
)

(AB)-1 =

 0 48
 8 48
-
 6 48
-
 16 48

(AB)-1 =

0
 1 6
-
 1 8
-
 1 3

Now we will solve R.H.S
So first we will find B-1
So,
B-1 =
So now we will find | B |
| B | =
 -4 -2 1 -1
| B | = (-4)(-1) - (-2)(1)
| B | = 4 + 2
| B | = 6
So now we will find Adj B

 -1 2 -1 -4

B-1 =

 -1 2 -1 -4

6
B-1 =
 1 6

 -1 2 -1 -4

B-1 =

-1 (
 1 6
)
2 (
 1 6
)
-1 (
 1 6
)
-4 (
 1 6
)

B-1 =

-
 1 6
 2 6
-
 1 6
-
 4 6

B-1 =

-
 1 6
 1 3
-
 1 6
-
 2 3

and now we will find A-1
So,
A-1 =
So now we will find | A |
| A | =
 4 0 -1 2
| A | = (4)(2) - 0(-1)
| A | = 8 - 0
| A | = 8
So now we will find Adj A

 2 0 1 4

A-1 =

 2 0 1 4

8
A-1 =
 1 8

 2 0 1 4

A-1 =

2 (
 1 8
)
0 (
 1 8
)
1 (
 1 8
)
4 (
 1 8
)

A-1 =

 2 8
 0 8
 1 8
 4 8

A-1 =

 1 4
0
 1 8
 1 2

So now we will find B-1A-1
So,
B-1A-1 =

-
 1 6
 1 3
-
 1 6
-
 2 3

 1 4
0
 1 8
 1 2

B-1A-1 =

-
 1 6
(
 1 4
)
+
 1 3
(
 1 8
)
-
 1 6
(0) +
 1 3
(
 1 2
)
-
 1 6
(
 1 4
)
+
( -
 2 3
)
(
 1 8
)
-
 1 6
(0) +
( -
 2 3
)
(
 1 2
)

B-1A-1 =

-
 1 24
+
 1 24
0 +
 1 6
-
 1 24
-
 2 24
0 -
 2 6

B-1A-1 =

 -1 + 1 24
 1 6
 -1 - 2 24
-
 2 6

B-1A-1 =

 0 24
 1 6
-
 3 24
-
 1 3

B-1A-1 =

0
 1 6
-
 1 8
-
 1 3

Hence it is verified that (AB)-1 = B-1A-1

(ii) (DA)-1 = A-1D-1
First we will solve L.H.S
So first we will find DA
So,
DA =

 3 1 -2 2

 4 0 -1 2

DA =

 3(4) + 1(-1) 3(0) + 1(2) -2(4) + 2(-1) -2(0) + 2(2)

DA =

 12 - 1 0 + 2 -8 - 2 0 + 4

DA =

 11 2 -10 4

(DA)-1 =
So now we will find | DA |
| DA | =
 11 2 -10 4
| DA | = (11)(4) - (2)(-10)
| DA | = 44 + 20
| DA | = 64
So now we will find Adj DA

 4 -2 10 11

(DA)-1 =

 4 -2 10 11

64
(DA)-1 =
 1 64

 4 -2 10 11

(DA)-1 =

4 (
 1 64
)
-2 (
 1 64
)
10 (
 1 64
)
11 (
 1 64
)

(DA)-1 =

 4 64
-
 2 64
 10 64
 11 64

(DA)-1 =

 1 16
-
 1 32
 5 32
 11 64

Now we will solve R.H.S
So first we will find A-1
So,
A-1 =
So now we will find | A |
| A | =
 4 0 -1 2
| A | = (4)(2) - 0(-1)
| A | = 8 - 0
| A | = 8
So now we will find Adj A

 2 0 1 4

A-1 =

 2 0 1 4

8
A-1 =
 1 8

 2 0 1 4

A-1 =

2 (
 1 8
)
0 (
 1 8
)
1 (
 1 8
)
4 (
 1 8
)

A-1 =

 2 8
 0 8
 1 8
 4 8

A-1 =

 1 4
0
 1 8
 1 2

and now we will find D-1
So,
D-1 =
So now we will find | D |
| D | =
 3 1 -2 2
| D | = (3)(2) - (1)(-2)
| D | = 6 + 2
| D | = 8
So now we will find Adj D

 2 -1 2 3

D-1 =

 2 -1 2 3

8
D-1 =
 1 8

 2 -1 2 3

D-1 =

2 (
 1 8
)
-1 (
 1 8
)
2 (
 1 8
)
3 (
 1 8
)

D-1 =

 2 8
-
 1 8
 2 8
 3 8

D-1 =

 1 4
-
 1 8
 1 4
 3 8

So now we will find A-1D-1
So,
A-1D-1 =

 1 4
0
 1 8
 1 2

 1 4
-
 1 8
 1 4
 3 8

A-1D-1 =

 1 4
(
 1 4
)
+ 0
(
 1 4
)
 1 4
( -
 1 8
)
+ 0
(
 3 8
)
 1 8
(
 1 4
)
+
 1 2
(
 1 4
)
 1 8
( -
 1 8
)
+
 1 2
(
 3 8
)

A-1D-1 =

 1 16
+ 0
-
 1 32
+ 0
 1 32
+
 1 8
-
 1 64
+
 3 16

A-1D-1 =

 1 16
-
 1 32
 1 + 4 32
 -1 + 12 64

A-1D-1 =

 1 16
-
 1 32
 5 32
 11 64

Hence it is verified that (DA)-1 = A-1D-1