The solution of simultaneous equations by matrices and determinants 247
8. Kirchhoff’s laws are used to determine the
current equations in an electrical network and
show that
i 1 + 8i 2 + 3i 3 = −31
3i 1 − 2i 2 + i 3 = −5
2i 1 − 3i 2 + 2i 3 = 6
Use determinants to find the values of i 1 , i 2
and i 3 .
[i 1 = −5, i 2 = −4, i 3 = 2]
9. The forces in three members of a framework
are F 1 , F 2 and F 3 . They are related by the
simultaneous equations shown below.
1.4F 1 + 2.8F 2 + 2.8F 3 = 5.6
4.2F 1 − 1.4F 2 + 5.6F 3 = 35.0
4.2F 1 + 2.8F 2 − 1.4F 3 = −5.6
Find the values of F 1 , F 2 and F 3 using
determinants.
[F 1 = 2, F 2 = −3, F 3 = 4]
10. Mesh-current analysis produces the following
three equations:
20∠0 ◦ = (5 + 3 − j 4)I 1 − (3 − j 4)I 2
10∠90 ◦ = (3 − j 4 + 2)I 2 − (3 − j 4)I 1 − 2I 3
−15∠0 ◦ − 10∠90 ◦ = (12 + 2)I 3 − 2I 2
Solve the equations for the loop currents I 1 , I 2
and I 3 .
⎡
⎣
I 1 = 3.317∠22.57 ◦ A
I 2 = 1.963∠40.97 ◦ A
I 3 = 1.010∠−148.32 ◦ A
⎤
⎦
23.3 Solution of simultaneous
equations using Cramers rule
Cramers rule states that if
a 11 x + a 12 y + a 13 z = b 1
a 21 x + a 22 y + a 23 z = b 2
a 31 x + a 32 y + a 33 z = b 3
then x =
D x
D
, y =
D y
D
and z =
D z
D
where D =
a 11 a 12 a 13
a 21 a 22 a 23
a 31 a 32 a 33
D x =
b 1 a 12 a 13
b 2 a 22 a 23
b 3 a 32 a 33
i.e. the x-column has been replaced by the R.H.S.
b column,
D y =
a 11 b 1 a 13
a 21 b 2 a 23
a 31 b 3 a 33
i.e. the y-column has been replaced by the R.H.S.
b column,
D z =
a 11 a 12 b 1
a 21 a 22 b 2
a 31 a 32 b 3
i.e. the z-column has been replaced by the R.H.S.
b column.
Problem 7. Solve the following simultaneous
equations using Cramers rule.
x + y + z = 4
2x − 3y + 4z = 33
3x − 2y − 2z = 2
(This is the same as Problem 2 and a comparison of
methods may be made). Following the above method:
D =
1
1
1
2 −3
4
3 −2 −2
= 1(6 − (−8)) − 1((−4) − 12)
+ 1((−4) − (−9)) = 14 + 16 + 5 = 35
D x =
4
1
1
33 −3
4
2 −2 −2
= 4(6 − (−8)) − 1((−66) − 8)
+ 1((−66) − (−6)) = 56 + 74 − 60 = 70
D y =
1 4
1
2 33
4
3 2 −2
= 1((−66) − 8) − 4((−4) − 12) + 1(4 − 99)
= −74 + 64 − 95 = −105
D z =
1
1 4
2 −3 33
3 −2 2
= 1((−6) − (−66)) − 1(4 − 99)
+ 4((−4) − (−9)) = 60 + 95 + 20 = 175
8. Kirchhoff’s laws are used to determine the
current equations in an electrical network and
show that
i 1 + 8i 2 + 3i 3 = −31
3i 1 − 2i 2 + i 3 = −5
2i 1 − 3i 2 + 2i 3 = 6
Use determinants to find the values of i 1 , i 2
and i 3 .
[i 1 = −5, i 2 = −4, i 3 = 2]
9. The forces in three members of a framework
are F 1 , F 2 and F 3 . They are related by the
simultaneous equations shown below.
1.4F 1 + 2.8F 2 + 2.8F 3 = 5.6
4.2F 1 − 1.4F 2 + 5.6F 3 = 35.0
4.2F 1 + 2.8F 2 − 1.4F 3 = −5.6
Find the values of F 1 , F 2 and F 3 using
determinants.
[F 1 = 2, F 2 = −3, F 3 = 4]
10. Mesh-current analysis produces the following
three equations:
20∠0 ◦ = (5 + 3 − j 4)I 1 − (3 − j 4)I 2
10∠90 ◦ = (3 − j 4 + 2)I 2 − (3 − j 4)I 1 − 2I 3
−15∠0 ◦ − 10∠90 ◦ = (12 + 2)I 3 − 2I 2
Solve the equations for the loop currents I 1 , I 2
and I 3 .
⎡
⎣
I 1 = 3.317∠22.57 ◦ A
I 2 = 1.963∠40.97 ◦ A
I 3 = 1.010∠−148.32 ◦ A
⎤
⎦
23.3 Solution of simultaneous
equations using Cramers rule
Cramers rule states that if
a 11 x + a 12 y + a 13 z = b 1
a 21 x + a 22 y + a 23 z = b 2
a 31 x + a 32 y + a 33 z = b 3
then x =
D x
D
, y =
D y
D
and z =
D z
D
where D =
a 11 a 12 a 13
a 21 a 22 a 23
a 31 a 32 a 33
D x =
b 1 a 12 a 13
b 2 a 22 a 23
b 3 a 32 a 33
i.e. the x-column has been replaced by the R.H.S.
b column,
D y =
a 11 b 1 a 13
a 21 b 2 a 23
a 31 b 3 a 33
i.e. the y-column has been replaced by the R.H.S.
b column,
D z =
a 11 a 12 b 1
a 21 a 22 b 2
a 31 a 32 b 3
i.e. the z-column has been replaced by the R.H.S.
b column.
Problem 7. Solve the following simultaneous
equations using Cramers rule.
x + y + z = 4
2x − 3y + 4z = 33
3x − 2y − 2z = 2
(This is the same as Problem 2 and a comparison of
methods may be made). Following the above method:
D =
1
1
1
2 −3
4
3 −2 −2
= 1(6 − (−8)) − 1((−4) − 12)
+ 1((−4) − (−9)) = 14 + 16 + 5 = 35
D x =
4
1
1
33 −3
4
2 −2 −2
= 4(6 − (−8)) − 1((−66) − 8)
+ 1((−66) − (−6)) = 56 + 74 − 60 = 70
D y =
1 4
1
2 33
4
3 2 −2
= 1((−66) − 8) − 4((−4) − 12) + 1(4 − 99)
= −74 + 64 − 95 = −105
D z =
1
1 4
2 −3 33
3 −2 2
= 1((−6) − (−66)) − 1(4 − 99)
+ 4((−4) − (−9)) = 60 + 95 + 20 = 175