Mathematics 9 (Complete Solutions)

Q3
Find the multiplicative inverse (if it exists) of each.

(i) A =
 
-1 3
2 0
 
(ii) B =
 
1 2
-3 -5
 
(iii) C =
 
-2 6
3 -9
 
(iv) D =
 
1
2
3
4
1 2
 
Solution:

(i) A =
 
-1 3
2 0
 
A-1 =
Adj A
∣ A ∣
So first we will find ∣ A ∣
∣ A ∣ =
-1 3
2 0
= (-1)(0) - (3)(2)
∣ A ∣ = 0 - 6
∣ A ∣ = -6
So now we will find Adj A
Adj A =
 
0 -3
-2 -1
 
A-1 =
 
0 -3
-2 -1
 
-6
A-1 =
-
1
6
 
0 -3
-2 -1
 
A-1 =
 
0 (-
1
6
)
-3 (-
1
6
)
-2 (-
1
6
)
-1 (-
1
6
)
 
A-1 =
 
0
6
3
6
2
6
1
6
 
A-1 =
 
0
1
2
1
3
1
6
 
         Answer

(ii) B =
 
1 2
-3 -5
 
B-1 =
Adj B
∣ B ∣
So first we will find ∣ B ∣
∣ B ∣ =
1 2
-3 -5
= (1)(-5) - (2)(-3)
∣ B ∣ = -5 + 6
∣ B ∣ = 1
So now we will find Adj B
Adj B =
 
-5 -2
3 1
 
B-1 =
 
-5 -2
3 1
 
1
B-1 =
 
-5 -2
3 1
 
         Answer
(iii) C =
 
-2 6
3 -9
 
C-1 =
Adj C
∣ C ∣
So first we will find ∣ C ∣
∣ C ∣ =
-2 6
3 -9
= (-2)(-9) - (6)(3)
∣ C ∣ = 18 - 18
∣ C ∣ = 0
It is a singular matrix. Its multiplication inverse does not exist.

(iv) D =
 
1
2
3
4
1 2
 
D-1 =
Adj D
∣ D ∣
So first we will find ∣ D ∣
∣ D ∣ =
1
2
3
4
1 2
∣ D ∣ = (
1
2
) (2) - (
3
4
) (1)
∣ D ∣ =
2
2
-
3
4
∣ D ∣ = 1 -
3
4
∣ D ∣ =
(4)1 - 3
4
∣ D ∣ =
4 - 3
4
∣ D ∣ =
1
4
So now we will find Adj D
Adj D =
 
2
-
3
4
-1
1
2
 
D-1 =
 
2
-
3
4
-1
1
2
 
1
4
D-1 = 4
 
2
-
3
4
-1
1
2
 
D-1 =
 
(4)2
(4) ( -
3
4
)
(4)(-1)
(4) (
1
2
)
 
D-1 =
 
8
-
12
4
-4
4
2
 
D-1 =
 
8 -3
-4 2
 
         Answer

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