Higher Engineering Mathematics

(Greg DeLong) #1
272 MATRICES AND DETERMINANTS

25.5 The inverse or reciprocal of a


2 by 2 matrix


The inverse of matrixAisA−^1 such thatA×A−^1 =I,
the unit matrix.
Let matrixAbe


(
12
34

)
and let the inverse matrix,

A−^1 be

(
ab
cd

)
.

Then, sinceA×A−^1 =I,
(
12
34


)
×

(
ab
cd

)
=

(
10
01

)

Multiplying the matrices on the left hand side, gives
(
a+ 2 cb+ 2 d
3 a+ 4 c 3 b+ 4 d

)
=

(
10
01

)

Equating corresponding elements gives:

b+ 2 d=0, i.e.b=− 2 d

and 3 a+ 4 c=0, i.e.a=−

4
3

c

Substituting foraandbgives:




4
3

c+ 2 c − 2 d+ 2 d

3

(

4
3

c

)
+ 4 c 3(− 2 d)+ 4 d



⎠=

(
10
01

)

i.e.

(
2
3

c 0
0 − 2 d

)

=

(
10
01

)

showing that

2
3

c=1, i.e.c=

3
2

and− 2 d=1, i.e.

d=−

1
2
Sinceb=− 2 d,b=1 and sincea=−

4
3

c,a=−2.

Thus the inverse of matrix


(
12
34

)
is

(
ab
cd

)
that

is,

(
− 21
3
2


1
2

)

There is, however,a quicker method of obtaining
the inverseofa2by2matrix.


For any matrix

(
pq
rs

)
the inverse may be

obtained by:
(i) interchanging the positions ofpands,
(ii) changing the signs ofqandr, and

(iii) multiplying this new matrix by the reciprocal of

the determinant of

(
pq
rs

)

Thus the inverse of matrix

(
12
34

)
is

1
4 − 6

(
4 − 2
− 31

)
=

(
− 21
3
2


1
2

)

as obtained previously.

Problem 13. Determine the inverse of
(
3 − 2
74

)

The inverse of matrix

(
pq
rs

)
is obtained by inter-

changing the positions ofpands, changing the signs
ofqandrand multiplying by the reciprocal of the

determinant





pq
rs




∣. Thus, the inverse of
(
3 − 2
74

)
=

1
(3×4)−(− 2 ×7)

(
42
− 73

)

=

1
26

(
42
− 73

)
=





2
13

1
13
− 7
26

3
26





Now try the following exercise.

Exercise 110 Further problems on the
inverse of 2 by 2 matrices


  1. Determine the inverse of


(
3 − 1
− 47

)









7
17

1
17
4
17

3
17










  1. Determine the inverse of






1
2

2
3


1
3


3
5

⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣





7

5
7

8

4
7

− 4

2
7

− 6

3
7








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