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
- Calculate the determinant of
(
3 − 1
− 47
)
[17]
- Calculate the determinant of(
− 25
3 − 6
)
[−3]
- Calculate the determinant of(
− 1. 37. 4
2. 5 − 3. 9
)
[−13.43]
- Evaluate
∣
∣
∣
∣
j 2 −j 3
( 1 +j) j
∣
∣
∣
∣ [−^5 +j3]
- 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
)