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 =
Adj AB
| AB |
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
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 =
Adj B
| B |
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
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 =
Adj A
| A |
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
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 =
Adj DA
| DA |
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
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 =
Adj A
| A |
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
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 =
Adj D
| D |
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
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

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