244 Higher Engineering Mathematics
D y =
a 1 c 1
a 2 c 2
i.e. the determinant of the coefficients left
when the y-column is covered up,
and
D =
a 1 b 1
a 2 b 2
i.e. the determinant of the coefficients left
when the constants-column is covered up.
Problem 3. Solve the following simultaneous
equations using determinants:
3x − 4y = 12
7x + 5y = 6.5
Following the above procedure:
(i) 3x − 4y − 12 = 0
7x + 5y − 6.5 = 0
(ii)
x
−4 −12
5 −6.5
=
−y
3 −12
7 −6.5
=
1
3 −4
7
5
i.e.
x
(−4)(−6.5) − (−12)(5)
=
−y
(3)(−6.5) − (−12)(7)
=
1
(3)(5) − (−4)(7)
i.e.
x
26 + 60
=
−y
−19.5 + 84
=
1
15 + 28
i.e.
x
86
=
−y
64.5
=
1
43
Since
x
86
=
1
43
then x =
86
43
= 2
and since
−y
64.5
=
1
43
then y = −
64.5
43
= −1.5
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.
Substituting the given values in v = u +at gives:
21 = u + 3.5a
(1)
33 = u + 6.1a
(2)
(i) The equations are written in the form
a 1 x + b 1 y + c 1 = 0,
i.e.
u + 3.5a − 21 = 0
and
u + 6.1a − 33 = 0
(ii) The solution is given by
u
D u
=
−a
D a
=
1
D
where D u is the determinant of coefficients left
when the u column is covered up,
i.e.
D u =
3.5 −21
6.1 −33
= (3.5)(−33) − (−21)(6.1)
= 12.6
Similarly, D a =
1 −21
1 −33
= (1)(−33) − (−21)(1)
= −12
and
D =
1 3.5
1 6.1
= (1)(6.1) − (3.5)(1) = 2.6
Thus
u
12.6
=
−a
−12
=
1
26
i.e.
u =
12.6
2.6
= 4.846 m/s
and
a =
12
2.6
= 4.615 m/s 2 ,
each correct to 4 significant
figures.
Problem 5. Applying Kirchhoff’s laws to an
electric circuit results in the following equations:
(9 + j 12)I 1 − (6 + j 8)I 2 = 5
−(6 + j 8)I 1 + (8 + j 3)I 2 = (2 + j 4)
Solve the equations for I 1 and I 2
D y =
a 1 c 1
a 2 c 2
i.e. the determinant of the coefficients left
when the y-column is covered up,
and
D =
a 1 b 1
a 2 b 2
i.e. the determinant of the coefficients left
when the constants-column is covered up.
Problem 3. Solve the following simultaneous
equations using determinants:
3x − 4y = 12
7x + 5y = 6.5
Following the above procedure:
(i) 3x − 4y − 12 = 0
7x + 5y − 6.5 = 0
(ii)
x
−4 −12
5 −6.5
=
−y
3 −12
7 −6.5
=
1
3 −4
7
5
i.e.
x
(−4)(−6.5) − (−12)(5)
=
−y
(3)(−6.5) − (−12)(7)
=
1
(3)(5) − (−4)(7)
i.e.
x
26 + 60
=
−y
−19.5 + 84
=
1
15 + 28
i.e.
x
86
=
−y
64.5
=
1
43
Since
x
86
=
1
43
then x =
86
43
= 2
and since
−y
64.5
=
1
43
then y = −
64.5
43
= −1.5
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.
Substituting the given values in v = u +at gives:
21 = u + 3.5a
(1)
33 = u + 6.1a
(2)
(i) The equations are written in the form
a 1 x + b 1 y + c 1 = 0,
i.e.
u + 3.5a − 21 = 0
and
u + 6.1a − 33 = 0
(ii) The solution is given by
u
D u
=
−a
D a
=
1
D
where D u is the determinant of coefficients left
when the u column is covered up,
i.e.
D u =
3.5 −21
6.1 −33
= (3.5)(−33) − (−21)(6.1)
= 12.6
Similarly, D a =
1 −21
1 −33
= (1)(−33) − (−21)(1)
= −12
and
D =
1 3.5
1 6.1
= (1)(6.1) − (3.5)(1) = 2.6
Thus
u
12.6
=
−a
−12
=
1
26
i.e.
u =
12.6
2.6
= 4.846 m/s
and
a =
12
2.6
= 4.615 m/s 2 ,
each correct to 4 significant
figures.
Problem 5. Applying Kirchhoff’s laws to an
electric circuit results in the following equations:
(9 + j 12)I 1 − (6 + j 8)I 2 = 5
−(6 + j 8)I 1 + (8 + j 3)I 2 = (2 + j 4)
Solve the equations for I 1 and I 2