Higher Engineering Mathematics

THE THEORY OF MATRICES AND DETERMINANTS 273

F

3. Determine the inverse of

(

− 1. 37. 4

2. 5 − 3. 9

)

⎡

⎣

(

0 .290 0. 551

0 .186 0. 097

)

correct to 3 dec. places

⎤

⎦

25.6 The determinant ofa3by3

matrix

(i) Theminorof an element ofa3by3matrix is

the value of the 2 by 2 determinant obtained by

covering up the row and column containing that

element.

Thus for the matrix

(

123

456

789

)

the minor of

element 4 is obtained by covering the row

(4 5 6) and the column

(

1

4

7

)

, leaving the 2 by

2 determinant

∣

∣

∣

∣

23

89

∣

∣

∣

∣, i.e. the minor of element

4is(2×9)−(3×8)=−6.

(ii) The sign of a minor depends on its posi-

tion within the matrix, the sign pattern

being

(

+−+

−+−

+−+

)

. Thus the signed-minor

of element 4 in the matrix

(

123

456

789

)

is

−

∣

∣

∣

∣

23

89

∣

∣

∣

∣=−(−6)=6.

The signed-minor of an element is called the

cofactorof the element.

(iii)The value ofa3by3determinant is the
sum of the products of the elements and their
cofactors of any row or any column of the
corresponding 3 by 3 matrix.
There are thus six different ways of evaluating a 3× 3
determinant—and all should give the same value.

Problem 14. Find the value of

∣

∣

∣

∣

∣

34 − 1

207

1 − 3 − 2

∣

∣

∣

∣

∣

The value of this determinant is the sum of the prod-

ucts of the elements and their cofactors, of any row

or of any column. If the second row or second col-

umn is selected, the element 0 will make the product

of the element and its cofactor zero and reduce the

amount of arithmetic to be done to a minimum.

Supposing a second row expansion is selected.

The minor of 2 is the value of the determinant

remaining when the row and column containing the

2 (i.e. the second row and the first column), is cov-

ered up. Thus the cofactor of element 2 is

∣

∣

∣

∣

4 − 1

− 3 − 2

∣

∣

∣

∣

i.e.−11. The sign of element 2 is minus, (see (ii)

above), hence the cofactor of element 2, (the signed-

minor) is∣ +11. Similarly the minor of element 7 is

∣

∣

∣

34

1 − 3

∣

∣

∣

∣i.e.−13, and its cofactor is+13. Hence the

value of the sum of the products of the elements and

their cofactors is 2× 11 + 7 ×13, i.e.,

∣

∣

∣

∣

∣

34 − 1

207

1 − 3 − 2

∣

∣

∣

∣

∣

=2(11)+ 0 +7(13)= 113

The same result will be obtained whichever row or

column is selected. For example, the third column

expansion is

(−1)

∣

∣

∣

∣

20

1 − 3

∣

∣

∣

∣−^7

∣

∣

∣

∣

34

1 − 3

∣

∣

∣

∣+(−2)

∣

∣

∣

∣

34

20

∣

∣

∣

∣

= 6 + 91 + 16 = 113 , as obtained previously.

Problem 15. Evaluate

∣

∣

∣

∣

∣

14 − 3

− 526

− 1 − 42

∣

∣

∣

∣

∣

Using the first row:

∣

∣

∣

∣

∣

14 − 3

− 526

− 1 − 42

∣

∣

∣

∣

∣

= 1

∣

∣

∣

∣

26

− 42

∣

∣

∣

∣−^4

∣

∣

∣

∣

− 56

− 12

∣

∣

∣

∣+(−3)

∣

∣

∣

∣

− 52

− 1 − 4

∣

∣

∣

∣

=(4+24)−4(− 10 +6)−3(20+2)

= 28 + 16 − 66 =− 22

Using the second column:

∣

∣

∣

∣

∣

14 − 3

− 526

− 1 − 42

∣

∣

∣

∣

∣

=− 4

∣

∣

∣

∣

− 56

− 12

∣

∣

∣

∣+^2

∣

∣

∣

∣

1 − 3

− 12

∣

∣

∣

∣−(−4)

∣

∣

∣

∣

1 − 3

− 56

∣

∣

∣

∣

=−4(− 10 +6)+2(2−3)+4(6−15)

= 16 − 2 − 36 =− 22