Higher Engineering Mathematics, Sixth Edition

(Nancy Kaufman) #1

236 Higher Engineering Mathematics






5 ∠ 30 ◦ 2 ∠− 60 ◦
3 ∠ 60 ◦ 4 ∠− 90 ◦




∣=(^5 ∠^30

◦)( 4 ∠− 90 ◦)

−( 2 ∠− 60 ◦)( 3 ∠ 60 ◦)

=( 20 ∠− 60 ◦)−( 6 ∠ 0 ◦)

=( 10 −j 17. 32 )−( 6 +j 0 )

=(4−j17.32)or17.78∠− 77 ◦

Now try the following exercise

Exercise 94 Further problems on 2 by 2
determinants


  1. Calculate the determinant of


(
3 − 1
− 47

)

[17]


  1. Calculate the determinant of(
    − 25
    3 − 6


)
[−3]


  1. Calculate the determinant of(
    − 1. 37. 4
    2. 5 − 3. 9


)
[−13.43]


  1. Evaluate






j 2 −j 3
( 1 +j) j




∣ [−^5 +j3]


  1. Evaluate







2 ∠ 40 ◦ 5 ∠− 20 ◦
7 ∠− 32 ◦ 4 ∠− 117 ◦

∣ ∣ ∣ ∣ ∣ [

(− 19. 75 +j 19. 79 )
or 27. 96 ∠ 134. 94 ◦

]

22.5 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 3a+ 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
inverseof a 2 by 2 matrix.
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

)
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