Higher Engineering Mathematics

THE THEORY OF MATRICES AND DETERMINANTS 275

F

Problem 17. Determine the inverse of the

matrix

⎛

⎝

34 − 1

207

1 − 3 − 2

⎞

⎠

The inverse of matrixA,A−^1 =

adjA

|A|

The adjoint ofAis found by:

(i) obtaining the matrix of the cofactors of the

elements, and

(ii) transposing this matrix.

The cofactor of element 3 is+

∣

∣

∣

∣

07

− 3 − 2

∣

∣

∣

∣=21.

The cofactor of element 4 is−

∣

∣

∣

∣

27

1 − 2

∣
∣
∣
∣=11, and
so on.

The matrix of cofactors is

(

21 11 − 6

11 − 513

28 − 23 − 8

)

The transpose of the matrix of cofactors, i.e. the
adjoint of the matrix, is obtained by writing the rows

as columns, and is

(

21 11 28

11 − 5 − 23

− 613 − 8

)

From Problem 14, the determinant of

∣
∣
∣
∣
∣

34 − 1

207

1 − 3 − 2

∣

∣

∣

∣

∣

is 113.

Hence the inverse of

(

34 − 1

207

1 − 3 − 2

)

is

(

21 11 28

11 − 5 − 23

− 613 − 8

)

113

or

1

113

(

21 11 28

11 − 5 − 23

− 613 − 8

)

Problem 18. Find the inverse of

(

15 − 2

3 − 14

− 36 − 7

)

Inverse=

adjoint

determinant

The matrix of cofactors is

(

− 17 9 15

23 − 13 − 21

18 − 10 − 16

)

The transpose of the matrix of cofactors (i.e. the

adjoint) is

(

− 17 23 18

9 − 13 − 10

15 − 21 − 16

)

The determinant of

(

15 − 2

3 − 14

− 36 − 7

)

=1(7−24)−5(− 21 +12)−2(18−3)

=− 17 + 45 − 30 =− 2

Hence the inverse of

(

15 − 2

3 − 14

− 36 − 7

)

=

(

− 17 23 18

9 − 13 − 10

15 − 21 − 16

)

− 2

=

(

8. 5 − 11. 5 − 9

− 4. 56. 55

− 7. 510. 58

)

Now try the following exercise.

Exercise 112 Further problems on the

inverse ofa3by3matrix

1. Write down the transpose of
   (
   4 − 76
   − 240
   57 − 4

)

[(

4 − 25

− 747

60 − 4

)]

2. Write down the transpose of
   ⎛

⎝

(^3612)
5 −^237
− (^1035)
⎞
⎠
⎡
⎣
⎛
⎝
35 − 1
6 −^230
1
2 7
3
5
⎞
⎠
⎤
⎦