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 =
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

 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

(ii) B =

 1 2 -3 -5

B-1 =
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

 -5 -2 3 1

B-1 =

 -5 -2 3 1

1
B-1 =

 -5 -2 3 1

(iii) C =

 -2 6 3 -9

C-1 =
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 =
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

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