248 Higher Engineering Mathematics
Hence
x =
D x
D
=
70
35
= 2, y =
D y
D
=
−105
35
= −3
and z =
D z
D
=
175
35
= 5
Now try the following exercise
Exercise 100 Further problems on solving
simultaneous equations using Cramers rule
1. Repeat problems 3, 4, 5, 7 and 8 of Exercise
98 on page 241, using Cramers rule.
2. Repeat problems 3, 4, 8 and 9 of Exercise 99
on page 244, using Cramers rule.
23.4 Solution of simultaneous
equations using the Gaussian
elimination method
Consider the following simultaneous equations:
x + y + z = 4
( 1 )
2x − 3y + 4z = 33
(2)
3x − 2y − 2z = 2
( 3 )
Leaving equation (1) as it is gives:
x + y + z = 4
( 1 )
Equation (2) − 2 × equation (1) gives:
0 − 5y + 2z = 25
(2 )
and equation (3) − 3 × equation (1) gives:
0 − 5y − 5z = −10
(3 )
Leaving equations (1) and (2 ) as they are gives:
x + y + z = 4
( 1 )
0 − 5y + 2z = 25
(2 )
Equation (3 ) − equation (2 ) gives:
0 + 0 − 7z = −35
(3
)
By appropriately manipulating the three original equations we have deliberately obtained zeros in the positions shown in equations (2 ) and (3 ).
Working backwards, from equation (3
),
z =
−35
−7
= 5,
from equation (2 ),
−5y + 2(5) = 25,
from which,
y =
25 − 10
−5
= −3
and from equation (1),
x + (−3) + 5 = 4,
from which,
x = 4 + 3 − 5 = 2
(This is the same example as Problems 2 and 7, and
a comparison of methods can be made). The above
method is known as the Gaussian elimination method.
We conclude from the above example 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
the three-step procedure to solve simultaneous equations in three unknowns using the Gaussian elimination method is:
1. Equation (2) −
a 21
a 11
× equation (1) to form equation (2 ) and equation (3) −
a 31
a 11
× equation (1) to
form equation (3 ).
2. Equation (3 ) −
a 32
a 22
× equation (2 ) to form equation (3 ).
3. Determine z from equation (3
), then y from
equation (2 ) and finally, x from equation (1).
Problem 8. A d.c. circuit comprises three closed
loops. Applying Kirchhoff’s laws to the closed
loops gives the following equations for current flow
in milliamperes:
2I 1 + 3I 2 − 4I 3 = 26
(1)
I 1 − 5I 2 − 3I 3 = −87
(2)
−7I 1 + 2I 2 + 6I 3 = 12
(3)
Hence
x =
D x
D
=
70
35
= 2, y =
D y
D
=
−105
35
= −3
and z =
D z
D
=
175
35
= 5
Now try the following exercise
Exercise 100 Further problems on solving
simultaneous equations using Cramers rule
1. Repeat problems 3, 4, 5, 7 and 8 of Exercise
98 on page 241, using Cramers rule.
2. Repeat problems 3, 4, 8 and 9 of Exercise 99
on page 244, using Cramers rule.
23.4 Solution of simultaneous
equations using the Gaussian
elimination method
Consider the following simultaneous equations:
x + y + z = 4
( 1 )
2x − 3y + 4z = 33
(2)
3x − 2y − 2z = 2
( 3 )
Leaving equation (1) as it is gives:
x + y + z = 4
( 1 )
Equation (2) − 2 × equation (1) gives:
0 − 5y + 2z = 25
(2 )
and equation (3) − 3 × equation (1) gives:
0 − 5y − 5z = −10
(3 )
Leaving equations (1) and (2 ) as they are gives:
x + y + z = 4
( 1 )
0 − 5y + 2z = 25
(2 )
Equation (3 ) − equation (2 ) gives:
0 + 0 − 7z = −35
(3
)
By appropriately manipulating the three original equations we have deliberately obtained zeros in the positions shown in equations (2 ) and (3 ).
Working backwards, from equation (3
),
z =
−35
−7
= 5,
from equation (2 ),
−5y + 2(5) = 25,
from which,
y =
25 − 10
−5
= −3
and from equation (1),
x + (−3) + 5 = 4,
from which,
x = 4 + 3 − 5 = 2
(This is the same example as Problems 2 and 7, and
a comparison of methods can be made). The above
method is known as the Gaussian elimination method.
We conclude from the above example 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
the three-step procedure to solve simultaneous equations in three unknowns using the Gaussian elimination method is:
1. Equation (2) −
a 21
a 11
× equation (1) to form equation (2 ) and equation (3) −
a 31
a 11
× equation (1) to
form equation (3 ).
2. Equation (3 ) −
a 32
a 22
× equation (2 ) to form equation (3 ).
3. Determine z from equation (3
), then y from
equation (2 ) and finally, x from equation (1).
Problem 8. A d.c. circuit comprises three closed
loops. Applying Kirchhoff’s laws to the closed
loops gives the following equations for current flow
in milliamperes:
2I 1 + 3I 2 − 4I 3 = 26
(1)
I 1 − 5I 2 − 3I 3 = −87
(2)
−7I 1 + 2I 2 + 6I 3 = 12
(3)