Download Determinants

Determinants
After completing this section, you should be able to:
Use determinants to solve simultaneous equations
Set up second-order determinants to solve two simultaneous equations
Set up third-order determinants to solve three simultaneous equations
Evaluate determinants using either the expansion method or the cofactor method
Determinants
A determinant is an array of numbers or symbols in rows and columns and is written between
vertical lines.
a
c
1.
2.
3.
4.
5.
6.
7.
b
d
or
1 3 5 or a11 a12 a13
2 1 4
a21 a22 a23
0 3 7
a31 a32 a33
The numbers or symbols are called elements
Any horizontal set of elements is called a row
Any vertical set of elements is called a column
A determinant is always square i.e. no of rows = number of columns
For subscripted elements aij i denotes the row and j the column
The elements shown arrowed form the principle diagonal
The first element in the first row (and column) is called the leasing element.
The order of Determinants
The order of a determinant is equal to the number of rows or column in the determinant. The
above determinants are 2nd, 3rd and 3rd order respectively.
Minors and Co-factors
Minors
Every element in a determinant has a minor. The minor of an element is the
determinant obtained by striking out the row and column containing that element. For
example consider:
a b
d e
g h
c
f
i
The minor of d is
The minor of I is
b
h
a
d
c
i
The minor of h is
a
d
c
f
b
e
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Co-Factors
The co-factor of an element is called the minor with the correct sign attached to it. The sign
depends upon the position of the element. The sign convention is:
+ - +
- + + - +
For example consider
a b c
d e f
g h i
The co-factor of d is –
b c
h i
d is negative according to the sign position
The co-factor of g is +
b c
e f
g is positive according to the sign position
The co-factor of f is –
a b
g h
f is negative according to the sign position
Expansion and Evaluation of a determinant
A determinant can be expanded about any row or column. For example row 1
a1 b1 c1 = a1 b2 c2 - b1 a2 c2 +c1 a1 b2
a2 b2 c2
b3 c3
a3 c3
b3 c3
a3 b3 c3
column 2 expanded gives:
- b1 a2 c2 + b2 a1 c1 - b3 a1 c1
a3 c3
a3 c3
a2 c2
Also
b2 c2 = b2 x c3 – c2 x b3
b3 c3
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Examples
1
Evaluate
2
Evaluate
3.
=
3
1
2 = 3 x 4 – 2 x 1= 10
4
4 1 = 4 x 1 – 1 x –2 = 6
-2 1
1 0 2
3 4 -1
0 3 5
Evaluate 
About row 1
=
1 4 -1 - 0 3 -1 + 2 3
3 5
0 5
0
4
3
 = 1(4 x 5 –(-1 x 3) – 0 + 2(3 x 3 – 4 x 0 ) = 23 + 18 = 41
Exercises
1.
Evaluate
3
7
2 =
1
2
Evaluate
-3 -2 =
-1 4
3.
=
1 2 1
0 -3 4
3 5 1
Evaluate 
Answers
1.
2.
3.
-11
–14
10
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Solving Two Simultaneous Equations for Two Unknowns
To illustrate the method of second-order determinants, we will assume two loop equations as
follows:
10I1 + 5I2 = 15
2I1 + 4I2 = 8
We want to find the value of I1 and I2. To do so, we form a determinant with the coefficients
of the unknown currents. A coefficient is the number associated with an unknown. For
example, 10 is the coefficient for I1 in the first equation.
The first column in the determinant consists of the coefficients of I1, and the second column
consists of the coefficients of I2. The resulting determinant appears as follows:
1st column
2nd column
10 5
2 4
This is called the characteristic determinant for the set of equations.
Next, we form another determinant and use it in conjunction with the characteristic
determinant to solve for I1 We form this determinant for our example by replacing the
coefficients of I1 in the characteristic determinant with the constants on the right side of the
equations. Doing this, we get the following determinant:
15 5
8 4
Replace coefficients of I1 with constants
from right sides of equations.
We can now solve for A by evaluating both determinants and then dividing by the
characteristic determinant. To evaluate the determinants, we cross-multiply and subtract the
resulting products. An evaluation of the characteristic determinant in this example is
illustrated in the following steps:
Step 1: Multiply the first number in the left column by the second number in the right
column.
10 5
2 4
=10 x 4 = 40
Step 2: Multiply the second number in the left column by the first number in the right column
and subtract from the product in Step 1. This result is the value of the determinant (30 in this
case).
10 5
2 4
= 40 - (2 x 5) = 40 - 10 = 30
Next, repeat the same procedure for the other determinant that was set up for I1.
50
15 5
8 4
= 15 x 4 = 60
15 5
8 4
= 60- (8 x 5) = 60-40 = 20
The value of this determinant is 20. Now we can solve for I1 by dividing the I1 determinant
by the characteristic determinant as follows:
15
8
10
2
5
4
5
4
= 20 = 0.667A
30
To find I2 we form another determinant by substituting the constants on the rig side of the
equations for the coefficients of I2.
Replace with coefficients of I2 with constants
10 15 from right sides of equations
8 2
We solve for I2 by dividing this determinant by the characteristic determinant already
evaluated.
I2 =
10 15
2 8 = (10 x 8) - (2 x 15)
30
30
80 - 30 = 50 = l.67A
30
30
Solve the following set of equations for the unknown currents:
2I1 - 5I2 = 10
6I1 + 10I2 = 20
Solution The characteristic determinant is evaluated as follows:
2
6
-5 = (2)(l0) - (-5)(6) = 20 - (-30) = 20 + 30 = 50
10
Solving for, I1 yields
10 -5
20 10
50
= (10)(l0) - (-5)(20) = 100 - (-100) = 200 = 4A
50
50
50
Solving for I2 yields
I2 = 2 10
51
6 20 = (2)(20) - (6)(l0) = 40 – 60 = 0.4A
50
50
50
In a circuit problem, a result with a negative sign indicates that the direction of actual current
is opposite to the assigned direction.
Exercise
Solve the following set of equations
5I1 + 3I2 = 10
I1 + 2I2 = -6
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Solving Three Simultaneous Equations For Three Unknowns
The Cofactor Method Unlike the expansion method, the cofactor method can be used to
evaluate determinants with higher orders than three and is therefore more versatile. We will
use a third-order determinant to illustrate the method, keeping in mind that fourth-, fifth-, and
higher-order determinants can be evaluated in a similar way. The following specific
determinant is used to demonstrate the cofactor method on a step-by-step basis.
1
4
5
3
0
1.5
-2
-l
6
Step 1: Select any one column or row in the determinant. Each number in the selected column
or row is used as a multiplying factor. For illustration, we will use the first column.
3
0
1.5
-2
-l
6
Step 2: Determine the cofactor for each number in the selected column (or row). The cofactor
for a given number is the determinant formed by all numbers that are not in the same column
or row as the given number. This is illustrated follows:
1 3 -2
4 0 -1
5 1.5 6
1 3 -2
4 0 -1
5 1.5 6
Cofactor for 1
Cofactor for 4
1 3 -2
4 0 -1
5 1.5 6
Cofactor for 5
Step 3: Assign the proper sign to each multiplying factor according to the following format
(note the alternating pattern).
+ - +
- + + - +
Step 4: Sum all of the products of each multiplying factor and its associated cofactor using
the appropriate sign.
1 x 0 -l -4 x 3 -2 + 5 x 3 -2
1.5 6
1.5 6
0 -l
=
1[(0)(6) - (1.5)(- 1)] - 4[(3)(6) + (l.5)(-2)] + 5[(3)(- 1) + (0)(-2)]
=
1(1.5) - 4(15) + 5(-3) = 1.5 - 60 - 15 = -73.5
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EXAMPLE
Use the cofactor method to find I2. The equations are repeated below.
2 I1 + 0.5 I2 + 1 I3 0
0.75I1 + 0 I2 + 2 I3 = 1.5
3I1 + 0.2 I2 + 0 I3 = -l
Solution Evaluate the characteristic determinant as follows:
2
0.75
3
0.5
0
0.2
1
2
0
= 2 0 2 -0.75 0.5 1 + 3 0.5
0.2 0
0.2 0
0
1
2
2[(0)(0) - (0.2)(2)]- 0.75[(0.5)(0) - (0.2)(l)] + 3[(0.5)(2) - (0)(l)]
=
2(-0.4) - 0.75(-0.2) + 3(1) = -0.8+0.15 + 3 = 2.35
The determinant for I2 is
2
0.75
3
0
1.5
-l
1 = 2 1.5 2 _ 0.75 0
2
-l
0
-1
0
1 + 3 0
0
1.5
1
2
=
2[(1.5)(0) - (- l)(2)] - 0.75[(0)(0) - (- l)(l)] + 3[(0)(2) - (l.5)(l)]
=
2(2) - 0.75(1) + 3(-1.5) = 4-0.75-4.5 = -1.25
Then,
I2 = -1.25 = -0.532 A = -532 mA
2.35
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