Cramers Rule
Cramers Rule
Cramers Rule
Solution of simultaneous
equations by matrices
3x + 5y 7 = 0
(1)
4x 3y 19 = 0
(2)
a1 x + b1 y = c1
a2 x + b2 y = c2
4x 3y = 19
(ii) write the matrix equation corresponding to
these equations,
i.e.
a1
a2
x
c
b1
= 1
y
b2
c2
i.e.
1
a1 b2 b1 a2
b2
a2
b1
a1
a1
a2
b1
b2
=
4 3
y
19
3
5
(iii) The inverse of matrix
is
4 3
1
3 5
3
3 (3) 5 4 4
3
5
29 29
i.e.
4 3
29 29
(iv) Multiplying each side of (ii) by (iii) and remembering that A A1 = I, the unit matrix, gives:
3
5
1 0
x
7
= 29 29
0 1
y
19
4 3
29 29
i.e.
95
29
57
29
1
1
1
x
4
2 3
4 y = 33
3 2 2
z
2
(iii) The inverse matrix of
1
1
1
4
A = 2 3
3 2 2
y = 1
Checking:
equation (1),
3 4 + 5 (1) 7 = 0 = RHS
equation (2),
4 4 3 (1) 19 = 0 = RHS
is given by
A1 =
x
d1
a 1 b1 c1
a2 b2 c2 y = d2
z
a3 b3 c3
d3
(iii) determine the inverse matrix of
a1 b 1 c 1
a2 b2 c2 (see Chapter 61)
a3 b3 c3
adj A
|A|
14 16
5
0 5
5
7 2 5
and the transpose of this matrix gives
14
0
7
adj A = 16 5 2
5
5 5
The determinant of A, i.e. the sum of the products
of elements and their cofactors, using a first row
expansion is
1
3
2
4
2
1
2
3
4
2
+1
2
3
3
2
= (1 14) (1 ( 16)) + (1 5) = 35
(1)
2x 3y + 4z 33 = 0
(2)
3x 2y 2z 2 = 0
(3)
14
1
16
A1 =
35
5
0
5
5
7
2
5
(iv) Multiplying each side of (ii) by (iii), and remembering that A A1 = I, the unit matrix, gives
1 0 0
x
0 1 0 y
0 0 1
z
14
0
7
4
1
16 5 2 33
=
35
5
5 5
2
Section 10
Thus
21
+
x
= 29
y
28
29
x
4
=
y
1
simultaneous equations:
I1 + 2I2 + 4 = 0
5I1 + 3I2 1 = 0
(14 4) + (0 33) + (7 2)
1
(16 4) + (( 5) 33) + (( 2) 2)
35 (5 4) + (5 33) + (( 5) 2)
70
1
105
=
35
175
2
= 3
5
(v) By comparing corresponding elements, x = 2,
y = 3, z = 5, which can be checked in the
original equations.
[x = 4, y = 3]
2. 2p + 5q + 14.6 = 0
3.1p + 1.7q + 2.06 = 0
[ p = 1.2, q = 3.4]
Section 10
3. x + 2y + 3z = 5
2x 3y z = 3
3x + 4y + 5z = 3
62.2
Solution of simultaneous
equations by determinants
[x = 1, y = 1, z = 2]
4. 3a + 4b 3c = 2
2a + 2b + 2c = 15
7a 5b + 4c = 26
[a = 2.5, b = 3.5, c = 6.5]
5. p + 2q + 3r + 7.8 = 0
2p + 5q r 1.4 = 0
5p q + 7r 3.5 = 0
[ p = 4.1, q = 1.9, r = 2.7]
6. In two closed loops of an electrical circuit, the currents flowing are given by the
and then
(ii) the solution is given by
x
y
1
=
=
Dx
Dy
D
where
Dx =
b1
b2
c1
c2
a1
a2
c1
c2
a1
a2
b1
b2
(i) 3x 4y 12 = 0
7x + 5y 6.5 = 0
x
(4)(6.5) (12)(5)
y
(3)(6.5) (12)(7)
1
=
(3)(5) (4)(7)
x
y
1
=
=
26 + 60
19.5 + 84
15 + 28
x
y
1
=
=
86
64.5
43
1
86
x
=
then x =
=2
86
43
43
Similarly,
Since
and since
1
y
=
then
64.5
43
64.5
y=
= 1.5
43
Problem 4. The velocity of a car, accelerating at
uniform acceleration a between two points, is given
by v = u + at, where u is its velocity when passing
the first point and t is the time taken to pass
between the two points. If v = 21 m/s when
t = 3.5 s and v = 33 m/s when t = 6.1 s, use
determinants to find the values of u and a, each
correct to 4 significant figures.
3.5
6.1
21
33
= 12.6
1 21
Da =
1 33
= (1)(33) (21)(1)
= 12
i.e.
Du =
= (3.5)(33) (21)(6.1)
x
y
1
=
=
12
3 12
3 4
6.5
7 6.5
7
5
i.e.
(2)
u
a
1
=
=
Du
Da
D
i.e.
33 = u + 6.1a
7x + 5y = 6.5
4
5
(1)
3x 4y = 12
(ii)
21 = u + 3.5a
and
Thus
i.e.
and
D=
1
1
3.5
6.1
Section 10
and
I1
(6 + j8)
5
(8 + j3) (2 + j4)
=
I2
(9 + j12)
5
(6 + j8) (2 + j4)
I
=
(9 + j12) (6 + j8)
(6 + j8) (8 + j3)
I1
(20 + j40) + (40 + j15)
a1
D z = a2
a3
I2
(30 j60) (30 + j40)
1
=
(36 + j123) (28 + j96)
I1
I2
1
=
=
20 + j55 j100
64 + j27
20 + j55
Hence I1 =
64 + j27
a1
D = a2
a3
b1
b2
b3
c1
c2
c3
58.5270.02
= 0.8447.15 A
69.4622.87
10090
I2 =
= 1.4467.13 A
69.4622.87
=
and
d1
d2
d3
b1
b2
b3
I1 5I2 3I3 = 87
7I1 + 2I2 + 6I3 = 12
Use determinants to solve for I1 , I2 and I3
a1 x + b 1 y + c 1 z + d 1 = 0
I1 5I2 3I3 + 87 = 0
a2 x + b2 y + c2 z + d2 = 0
a3 x + b3 y + c3 z + d3 = 0
and then
(ii) the solution is given by
Section 10
y
z
1
x
=
=
=
Dx
Dy
Dz
D
where
b1
D = x b2
b3
c1
c2
c3
d1
d2
d3
c1
c2
c3
d1
d2
d3
4
3
6
3
6
26
87
12
87
5
(4)
12
2
87
12
+ (26)
= 3(486) + 4(114) 26(24)
= 1290
5
2
3
6
4
3
6
2
1
7
26
87
12
2
1
7
3
5
2
26
87
12
2
1
7
3
5
2
4
3
6
I1
I2
I3
1
=
=
=
1290
1806
1161
129
giving
1290
= 10 mA,
129
1806
I2 =
= 14 mA
129
1161
and I3 =
= 9 mA
129
I1 =
3. 3x + 4y + z = 10
2x 3y + 5z + 9 = 0
x + 2y z = 6
[x = 1, y = 2, z = 1]
3i1 2i2 + i3 = 5
2i1 3i2 + 2i3 = 6
Use determinants to solve for i1 , i2 and i3 .
[i1 = 5, i2 = 4, i3 = 2]
9. The forces in three members of a framework
are F1 , F2 and F3 . They are related by the
simultaneous equations shown below.
[x = 1.2, y = 2.8]
[m = 6.4, n = 4.9]
Section 10
I1 = 3.31722.57 A,
I2 = 1.96340.97A and
I3 = 1.010148.32 A
62.3
4
D = 33
2
a11
D = a21
a31
b1
Dx = b2
b3
a12
a22
a32
a12
a22
a32
a13
a23
a33
1
Dy = 2
3
Section 10
a11
Dy = a21
a31
b1
b2
b3
a13
a23
a33
a12
a22
a32
b1
b2
b3
4
33
2
1
4
2
1
3
2
4
33
2
a13
a23
a33
1
4
2
+ 1((66) (6)) = 56 + 74 60 = 70
where
1
3
2
1
4
2
Solution of simultaneous
equations using Cramers rule
1
3
2
Dy
Dx
70
105
=
= 2, y =
=
= 3
D
35
D
35
Dz
175
z=
=
=5
D
35
x=
and