Transcript WEEK 8
WEEK 8
SYSTEMS OF EQUATIONS
DETERMINANTS AND CRAMER’S RULE
OBJECTIVES
At the end of this session , you will be able to:
Evaluate a second order determinant.
Evaluate a third-order determinant.
Evaluate higher order determinants.
Solve a system of linear equations in two variables using Cramer’s rule.
Solve a system of linear equations in three variables using Cramer’s rule.
Use determinant’s to identify inconsistent systems and systems with dependent equations.
INDEX
1.
Determinants 2.
Solving a system of linear equations using determinants.
2.1 Solution of system of linear equations in two variables by Cramer’s rule 2.2 Solution of System of Linear Equations in three variables by Cramer’s rule 3. Summary
1. DETERMINANTS
A determinant is a real number associated with every square matrix.
Corresponding to each square matrix written as det
A A
|
A a a
...
a
| 11 21
n
1
a a a a a
11 21 ...
12 22 ...
n
2
a a
...
...
...
...
there is associated an expression, called the determinant of A, denoted by det A or |A|, 12 22 ...
a a a
...
...
...
1
n
2 ...
n nn
a a
1
n
2 ...
n a n
1
a n
2 ...
a nn
A matrix is an arrangement of numbers and so it has no fixed value, while each determinant has a fixed value.
A determinant having n rows and n columns is known as a determinant of order n.
NOTE: 1. Only square matrices have determinants. The determinants of non square matrices are not defined.
2. Determinants are useful in solving a system of linear equations and help in determining if the system is consistent or inconsistent.
Determinant of square matrix of order 1: If A = [a 11 ] is a square matrix of order 1, then the determinant of A is defined as |A| = a 11
1. DETERMINANTS(Cont…)
then
A
a a
21
a a
12 22 the expression a 11 a 22 – a 12 a 21 is defined as the determinant of A, that is, |
A
|
a a
11 21
a
12
a
22
a a
11 22
a a
12 21 We also say that the value of second order determinant
a
11
a
21
a
12
a
22
a a
11 22
a a
12 21 Thus, the determinant of a square matrix of order 2 is equal to the product of the diagonal elements minus the product of off-diagonal elements.
Example: Let us evaluate 5 2 4 3 From the above definition, we have 5 2 4 3 23
1. DETERMINANTS(Cont…)
NOTE:
The determinant is a real number, it is not a matrix. The determinant can be a negative number. It is not associated with absolute value at all except that they both use vertical lines. The determinant only exists for square matrices (2×2, 3×3, ... n×n). The determinant of a 1×1 matrix is that single value in the determinant.
Determinant of order 3 or more:
For finding the value of a determinant of order 3 or more, we need the following definitions: Minor : The minor of an element a
ij
in |A| is defined as the value of the determinant obtained by deleting the ith row and jth column of |A|, and is denoted by M
ij
.
Cofactor: The cofactor C
ij
of an element is defined as C
ij = (-1) i+j M ij. Now let us find the minors and cofactors of the elements of the determinant
|
A
|
a
11
a
21
a
31
a
12
a
22
a
32 Let M
ij
denote the minor of a
ij
in |A|.
a
13
a
23
a
33
1. DETERMINANTS(Cont…)
Now a
11
occurs in the first row and first column. In order to find the minor of a
11,
we delete the first row and first column of |A|.
|
A
|
a a a
1 1 21 3 1
a a
12
a
22 32
a a a
13 23 3 3 We have to find the minor of
a 11
, so we delete the first row and second column The minor M
11
of a
11 is given by,
M
11
a
22
a
32
a
23
a
33 (
a a
22 33
a a
23 32 ) Next to find the minor of a
12,
that is, the element in the first row, second column, we delete the first row and second column of |A|.
|
A
|
a
11
a
21
a
3 1
a a
1 2
a
22 32
a a
13 23
a
33 We have to find the minor of
a 12
, so we delete the first row and second column The minor M
12
of a
12
is given by,
M
12
a
21
a
31
a
23
a
33 (
a a
21 33
a a
23 31 )
1. DETERMINANTS(Cont…)
Similarly, we have
M
13
a
21
a
31
M
21
a
12
a
32
a a
22 32 ( 32 31
a
13
a
33 (
a a
12 33
a a
13 32 ) ) Similarly, we may obtain the minor of each of the remaining elements.
Now, if we denote the cofactor of a
ij
by C
ij
, then
C 11 is the cofactor of a 11 , that is, cofactor of the element in the first row, first column.
By definition of cofactor, C ij = (-1) i+j M ij. C 11 = (-1) 1+1 M 11 = (-1) = M
(
11
a a
22
2 M 11
33
a a
23 32 )
11
) Similarly,
C C
12 13
C
21 .
M
12
M
12 ( .
M
13 .
M
21
M
13
M
21 ( ( 32 33 33 31 ) 31 32 ) )
Similarly, the cofactor of each of the remaining elements of |A| can be determined.
1. DETERMINANTS(Cont…)
Example: Let us find the minor and cofactor of each element of 1 4 3 The minors of the elements of |A|are given by |
A
|
M
11
1 2 5 2
= (-1).(2) – (2).(5) = -2 – 10 = -12
M
21
3 5 2 2
= (-3).(2) – (5).(2) = -6 – 10 = -16
M
31
3 2 1 2
= (-3).(2) – (-1).(2) = -6 +2 = -4
M
12
4 3 2 2
= (4).(2) – (2).(3) = 8 – 6 = 2
M
22
1 3 2 2
= (1).(2) – (2).(3) = 2 – 6 = -4
M
32
1 2 4 2
= (1).(2) – (4).(2) = 2 – 8 = -6
M
13
3 1 5
4 3 1 5
2 2 2
M
23
= (4).(5) – (-1). (3) = 20 + 3 = 23
1 3 3 5
M
33 = (1).(5) – (-3).(3) = 5 + 9 = 14
1 4 3 1
= (1).(-1) – (-3).(4) = -1 + 12 = 11
1. DETERMINANTS(Cont…)
Now let us find the cofactors of the corresponding elements of |A|. From the definition of cofactors,
C ij = (-1) i+j M ij. C 11 = (-1) 1+1 . M 11 = (-1) 2 . M 11 = M 11 = -12 C 12 = (-1) 1+2 . M 12 = (-1) 3 . M 12 = - M 12 = -2
Substituting the
C 21 = (-1) 2+1 . M 21 = (-1) 3 . M 21 = - M 21 = 16
values of M 11, M 12, and
C 22 = (-1) 2+2 . M 22 = (-1) 4 . M 22 = M 22 = -4 C 31 = (-1) 3+1 . M 31 = (-1) 4 . M 31 = M 31 = -4 C 32 = (-1) 3+2 . M 32 = (-1) 5 . M 32 = - M 32 = 6 C 13 = (-1) 1+3 . M 13 = (-1) 4 . M 13 = M 13 = 23
M 13 respectively .
C 23 = (-1) 2+3 . M 23 = (-1) 5 . M 23 = M 23 = -14 C 33 = (-1) 3+3 . M 33 = (-1) 6 . M 33 = M 33 = 11
1. DETERMINANTS(Cont…)
Value of a Determinant: The value of a determinant is the sum of the products of the elements of a row (or a column) with their corresponding cofactors.
We find the determinant of a matrix of order three or more by expanding along any arbitrarily chosen row or column.
Expansion of a Determinant: Expanding the determinant of order three along first row, we have
a
11
a
21
a
31
a
12
a
22
a
32
a
13
a
23
a
33 = a
11
. (Its cofactor) + a
12
. (Its cofactor) + a
13
. (Its cofactor) = a
11
. C
11 = a 11
. M
11
+ a
12
. C
12
- a
12
. M
12
+ a
13
. C
13
+ a
13
. M
13 (As C 12 = -M 12 ) =
a
11 .
a
22
a
32 =
a
11 .(
a a a
23
a
33 22 33
a
12 .
a
21
a
31
a
23
a
33
a
13 .
a
21
a
31
a a
23 32 )
a
12 .(
a a
21 33
a a
31 23 )
a
22
a
32
a
13 .(
a a
21 32
a a
22 31 ) We may expand by any row or column.
1. DETERMINANTS(Cont…)
NOTE: 1. We can expand a determinant by minors about any row or column. We use alternating plus and minus signs to precede the numerical factors of the minors, keeping in view the following sign array:
2. If a row or column of a determinant consists of all zeros, the value of the determinant is zero.
Now let us understand how to evaluate a determinant of matrix of order three with help of an example: Let us evaluate the determinant of 3 0 2 2 1 5 0 5 1 Expanding the given determinant about first row, we get 3 2 2 0 1 5 0 1 3 1 5 5 1 0 2 2 5 1 0 2 2 1 5 = 3 .{(1).(-1) - (-5).(5)} – = 3.
{-1 + 25} – 0 + 0 = 3 . 24 = 72 0 {(2).(1) – (-5).(2)} + 0 .(2. 5 + 1. 2)
1. DETERMINANTS(Cont…)
NOTE: To expand a determinant choose the row or column with the most zeros in it.
Since each minor or cofactor is multiplied by the element in the matrix, picking a row or column with lots of zeros in it means that you will be multiplying by a lot of zeros. In fact, if the element is zero, you don't need to even find the minor or cofactor. Let us evaluate the determinant of 4X4 matrix
A
1 2 3 1 2 1 1 1 1 2 2 0 3 3 1 2 To evaluate the determinant of a square matrix of order 4 or more we follow the same procedure as discussed in evaluating the determinant of a square matrix of order 3.
|
A
|
1
2 3 1 1 1 1
1
2 2 0
3
3 1 2
= 1 .(its cofactor) + 2 .(its cofactor) + (-1) .(its cofactor) + 3 .(its cofactor)
1. DETERMINANTS(Cont…)
1
2 3 1
Solving further, we get,
2
1 1 1
1
2 2 0
3
3 1 2
= 1 . C 11 + 2 . C 12 + (-1) . C 13 + 3 . C 14 = 1 . M 11 + 2 .(-M 12 ) + (-1) . M 13 + 3 . (-M) 14 (C ij = (-1) i+j M ij ) . = 1 . M 11 2 .M
12 + (-1) . M 13 3 . M 14 1 1 . 1 1 2 2 0 3 1 2 2 2 3 1 2 2 0 3 1 2 ( 1) 2 3 1 1 1 3 1 1 2 3 2 3 1 1 1 1 Now evaluating each of the above determinants of order 3 as explained before, we get the following result: |A|= 1 .(16) – 2 .(12) + (-1) .(-11) – 3 .(14) = 16 – 24 + 11 – 42 = -39 2 2 0
2. SOLVING A SYSTEM OF LINEAR EQUATIONS USING DETERMINANTS
2.1 SOLUTION OF SYSTEM OF LINEAR EQUATIONS IN TWO VARIABLES BY CRAMER’S RULE:
We now intend to solve a system of simultaneous linear equations by Cramer’s rule named after the Swiss mathematician Gabriel Cramer. Cramer’s Rule: The solution of the system of simultaneous linear equations a 1 x + b 1 y = c 1 a 2 x + b 2 y = c 2 D x and D y are obtained by replacing the x coefficients and y-coefficients in D respectively with the constants c 1 and c 2 .
is given by
x
D x D
,
y
D y D D
a a
2
b b
2 ,
D
c b
,
D
a c
c
2
b
2
a
2
c
2 0.
D
a a
2
b b
2
a
1
a
2
b
1
b
2 NOTE: 1. Three different determinants are used to find x and y. The determinants in the denominators for x and y are identical. The determinants in the numerators for x and y differ.
2. SOLVING A SYSTEM OF LINEAR EQUATIONS USING DETERMINANTS (Cont…)
2. D x , the determinant in the numerator of x, is obtained by replacing the x-coefficients in D, a
1
and a 2, with the constants on the right side of the equations, c
1
and c 2.
D
a a
1 2
b
1
b
2 ,
D x
c c
1 2
b b
1 2 3. D y , the determinant in the numerator of y, is obtained by replacing the y-coefficients in D, b
1
and b 2, with the constants on the right side of the equations, c
1
and c 2.
D
a a
1 2
b b
1 2 ,
D y
a a
1 2
c
1
c
2
Conditions for Consistency:
For a system of two simultaneous linear equations with two unknowns we have the following conditions: If D 0 then the system is consistent and has a unique solution given by
x
D x
,
y
D y D D
If D = 0, D x = 0, and D If D = 0, and one of D x y = 0, the system is consistent and has infinite number of solutions. The equations in the system are dependent.
and D y is non-zero, then the system is inconsistent.
2. SOLVING A SYSTEM OF LINEAR EQUATIONS USING DETERMINANTS (Cont…)
Example: Solve the following system of equations with the help of determinants 2x – y = 17 3x + 5y = 6
D
a a
2
D
2 3
b b
2 1 5
D x
c
1
b
1
c
2
b
2
D x
17 6 1 5 Next,
D y
a
1
a
2
c
1
c
2 91
2. SOLVING A SYSTEM OF LINEAR EQUATIONS USING DETERMINANTS
Substituting the values from the given system of equations, we have,
D y
2 17 3 6 39 Now we have found the values of all the three determinants, using Cramer’s rule, we have,
D y x
D x
,
y
D D
Substituting the values, we get,
x
y
D x D D y D
91 13 39 13 7, 3 Hence, the solution set for the given system of equations is {(7, -3)}.
CHECK: We can always check the solution (7, -3) by substituting these values into the original system of equations.
2. SOLVING A SYSTEM OF LINEAR EQUATIONS USING DETERMINANTS (Cont…)
Additional Examples:
SYSTEM OF LINEAR EQUATIONS 2x – y = 17 3x + 5y = 6 VALUES OF D, D x AND D y D = 13 D D y x = 91 = -39 REMARKS As D 0, the given system is consistent and has a unique solution given by x = 7 and y = -3 x + 2y = 3 4x + 8y = 12 D = 0 D D y x = 0 = 0 As D = D x = D y = 0, the given system has infinite number of solutions 2x + y = 3 4x + 2y = 4 D = 0 D x = -1 As D = 0 and D x 0, so the given system is inconsistent.
2. SOLVING A SYSTEM OF LINEAR EQUATIONS USING DETERMINANTS (Cont…)
2.2 SOLUTION OF SYSTEM OF LINEAR EQUATIONS IN THREE VARIABLES BY CRAMER’S RULE:
The solution of the system of linear equations a 1 x + b 1 y + c 1 z = d 1 a 2 x + b 2 y + c 2 z = d 2 a 3 x + b 3 y + c 3 z = d 3 is given by
x
D x D
,
y
D y
,
z
D D z D
where
D a
1
a
2
a
3
b
1
b
2
b
3
c
1
c
2 ,
D x
d d
2 1
c
3
d
3
b
1
b
2
b
3
c
1
c
2 ,
D y
a a
2 1
c
3
a
3
d
1
d
2
d
3
c
1
c
2 ,
D z
a a
2 1
c
3
a
3
b
1
b
2
b
3
d
1
d
2
d
3 provided that D 0 REMARK: Here D is the determinant of the coefficient matrix.
2. SOLVING A SYSTEM OF LINEAR EQUATIONS USING DETERMINANTS (Cont…)
Conditions for Consistency: For a system of three simultaneous linear equations in three unknowns, If D 0, then the given system of equations is consistent and has a unique solution given by
x
D x D
,
y
D y
,
z
D D z D
If D = 0 and D x many solutions.
= D y = D z = 0, then the system of equations is consistent with infinitely If D = 0 and at least one of the determinants D x of equations is inconsistent.
6x – 8y – z = 15 , D y , D z is non-zero, then the given system Example: Let us solve the following system of equations using Cramer’s rule: 5x – 7y +z = 11 3x + 2y – 6z = 7 We know
a
1
b
1
c
1 5 7 1
D
a a
2 3
b b
2 3
c c
2 3 6 3 8 2 1 6 (Substituting the values from the given system of equations)
2. SOLVING A SYSTEM OF LINEAR EQUATIONS USING DETERMINANTS (Cont…)
Expanding about first row, we get,
D
5
6
7
8
1 5
8 2 3 2 6
D
5
D
5
(48 2)
7
36 3)
D
5
(50)
D
D
5 5
7
36 1 6
1
(12 6 3 24) 1 6
1
6 3 8 2
1
{(6).(2)
2. SOLVING A SYSTEM OF LINEAR EQUATIONS USING DETERMINANTS (Cont…)
Similarly,
D x
11 15 7
D x
550 7 8 2 1 6 11 (48 55 2) 7) 1 (30 56) 5
D y
6 3 11 15 7 1 6 5 ( 90 7) 11 3) 1 (42 45)
D y
415 363
D z
5 6 7 8 11 15 55
D z
3 2 7 430 2 1 3 96 55 30) 7 (42 45) 1 1 (12 24)
2. SOLVING A SYSTEM OF LINEAR EQUATIONS USING DETERMINANTS (Cont…)
So, by Cramer’s rule, we have,
x
D x
55 1
D
55
y
D y
55 1
D
55
z
D D z
55 55 1 Hence, the solution set for the given system of equations is {(1, -1, -1)}.
The solution (1, -1, -1) can be checked by substituting the values into the original system of equations.
2. SOLVING A SYSTEM OF LINEAR EQUATIONS USING DETERMINANTS (Cont…)
Additional Examples:
SYSTEM OF LINEAR EQUATIONS 5x – 7y + z = 11 6x - 8y - z = 15 3x + 2y – 6z = 7 x + y + z = 1 x + 2y + 3z = 4 X + 3y + 5z = 7 2x – y + z = 4 x + 3y + 2z = 12 3x + 2y + 3z = 10 VALUES OF D, D x D z , D y, AND D = 55 D D y D z x = 55 = -55 = -55 D = 0 D D y D z x = 0 = 0 = 0 REMARKS As D 0, the given system is consistent and has a unique solution given by x = 1, y = -1 and z = -1 As D = D x = D y = D number of solutions.
z = 0, the given system has infinite D = 0 D x = 30 0 As D = 0 and D x 0, so the given system is inconsistent.
3. SUMMARY
Let us recall what we have learnt so far: A determinant is a real number associated with every square matrix. The value of second order determinant
a a
11 21
a a
12 22
a a
11 22
a a
12 21 Minor : The minor of an element a
ij
in |A| is defined as the value of the determinant obtained by deleting the ith row and jth column of |A|, and is denoted by M
ij
.
Cofactor: The cofactor C
ij
of an element id defined as C
ij = (-1) i+j M ij.
Value of the third order determinant
a
11
a
21
a
31
a a a
12 22 32
a a a
13 23 33 =
a
11 .(
a a
22 33
a a
23 32 )
a
12 .(
a a
21 33
a a
31 23 )
a
13 .(
a a
21 32
a a
22 31 ) Cramer’s Rule: The solution of the system of simultaneous linear equations a 1 x + b 1 y = c 1 a 2 x + b 2 y = c 2 is given by
x
D D x
,
y
D D y
3. SUMMARY(Cont…)
Cramer’s Rule: The solution of the system of linear equations a 1 x + b 1 y + c 1 z = d 1 a 2 x + b 2 y + c 2 z = d 2 a 3 x + b 3 y + c 3 z = d 3 is given by
Conditions for Consistency:
x
D D x
,
y
D y D
,
z
D D z
For a system of two simultaneous linear equations with two unknowns we have the following conditions: If D 0 then the system is consistent and has a unique solution given by
x
D x
,
y
D y D D
If D = 0, D x = 0, and D y If D = 0, and one of D x = 0, the system is consistent and has infinite number of solutions.
and D y is non-zero, then the system is inconsistent.
For a system of three simultaneous linear equations in three unknowns, If D 0, then the given system of equations is consistent and has a unique solution given by
x
D x D
,
y
D y
,
z
D D z D
If D = 0 and D x = D y = D z = 0, then the system of equations is consistent with infinitely many solutions.
If D = 0 and at least one of the determinants D x , D y , D z is non-zero, then the given system of equations is inconsistent.