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 − j 17.32) or 17.78∠−77
◦
Now try the following exercise
Exercise 94 Further problems on 2 by 2
determinants
1. Calculate the determinant of
3 −1
−4
7
[17]
2. Calculate the determinant of
−2
5
3 −6
[−3]
3. Calculate the determinant of
−1.3
7.4
2.5 −3.9
[−13.43]
4. Evaluate
j 2
− j 3
(1 + j )
j
[−5 + j 3]
5. 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 matrix A is A
−1 such that A × A
−1
= I ,
the unit matrix.
Let matrix A be
1 2
3 4
and let the inverse matrix, A −1
be
a b
c d
.
Then, since A × A −1 = I ,
1 2
3 4
×
a b
c d
=
1 0
0 1
Multiplying the matrices on the left hand side, gives
a + 2c b+ 2d
3a + 4c 3b + 4d
=
1 0
0 1
Equating corresponding elements gives:
b + 2d = 0, i.e. b = −2d
and 3a + 4c = 0, i.e. a = −
4
3
c
Substituting for a and b gives:
⎛
⎜
⎜
⎜
⎝
−
4
3
c + 2c
−2d + 2d
3
−
4
3
c
+ 4c 3(−2d) + 4d
⎞
⎟
⎟
⎟
⎠
=
1 0
0 1
i.e.
⎛
⎝
2
3
c 0
0 −2d
⎞
⎠ =
1 0
0 1
showing that
2
3
c = 1, i.e. c =
3
2
and −2d = 1, i.e. d =−
1
2
Since b =−2d, b = 1 and since a =−
4
3
c, a =−2.
Thus the inverse of matrix
1 2
3 4
is
a b
c d
that is,
⎛
⎝
−2
1
3
2
−
1
2
⎞
⎠
There is, however, a quicker method of obtaining the
inverse of a 2 by 2 matrix.
For any matrix
p q
r s
the inverse may be
obtained by:
(i) interchanging the positions of p and s,
(ii) changing the signs of q and r, and
(iii) multiplying this new matrix by the reciprocal of
the determinant of
p q
r s
Thus the inverse of matrix
1 2
3 4
is
1
4 − 6
4 −2
−3 1
=
⎛
⎝
−2
1
3
2
−
1
2
⎞
⎠
as obtained previously.
Problem 13. Determine the inverse of
3 −2
7
4