week 8 systems of equations determinants and cramer’s rule

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.

INDEX1. Determinants2. Solving a system of linear equations using determinants.

2.1 Solution of system of linear equations in two variables by Cramer’s rule2.2 Solution of System of Linear Equations in three variables by Cramer’s rule

3. Summary

1. DETERMINANTSA determinant is a real number associated with every square matrix.Corresponding to each square matrix

there is associated an expression, called the determinant of A, denoted by det A or |A|, written as

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 = [a11] is a square matrix of order 1, then the determinant of A is defined as |A| = a11

11 12 1

21 22 2

1 2

...

...... ... ... ...

...

n

n

n n nn

a a aa a aA

a a a

11 12 1

21 22 2

1 2

...

...det | | ... ... ... ...

...

n

n

n n nn

a a aa a aA A

a a a

1. DETERMINANTS(Cont…)Determinant of square matrix of order 2: If is a square matrix of order 2, then

the expression a11a22 – a12a21 is defined as the determinant of A, that is,

We also say that the value of second order determinant

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

From the above definition, we have

5 42 3

11 12

21 22

a aA a a

11 1211 22 12 21

21 22| | a aA a a a aa a

11 1211 22 12 21

21 22

a a a a a aa a

5 4 5.3 4.( 2) 15 8 232 3

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 aij 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 Mij .

Cofactor: The cofactor Cij of an element is defined as Cij = (-1)i+j Mij.

Now let us find the minors and cofactors of the elements of the determinant

Let Mij denote the minor of aij in |A|.

11 12 13

21 22 23

31 32 33

| |a a a

A a a aa a a

1. DETERMINANTS(Cont…)Now a11 occurs in the first row and first column. In order to find the minor of a11, we delete the first row and first column of |A|.

The minor M11 of a11 is given by,

Next to find the minor of a12, that is, the element in the first row, second column, we delete the first row and second column of |A|.

The minor M12 of a12 is given by,

Next let us find M12 , that is, we have to find the minor of a12

So, M12 = -3

Similarly, M21 = Minor of a21 = -7

We have to find the minor of a11, so we delete the first row and

second column

We have to find the minor of a12, so we delete the first

row and second column

12 13

21 22 23

32 3

1

33

1

1

| |A a aa

a aaa

a

a

22 2311 22 33 23 32

32 33( )a aM a a a aa a

21

11 13

22

32

23

3 33

2

1

1| |

a aa a

a aA a

a

a

21 2312 21 33 23 31

31 33( )a aM a a a aa a

1. DETERMINANTS(Cont…) Similarly, we have

Similarly, we may obtain the minor of each of the remaining elements. Now, if we denote the cofactor of aij by Cij , then

C11 is the cofactor of a11, that is, cofactor of the element in the first row, first column.

By definition of cofactor, Cij = (-1)i+j Mij.

C11 = (-1)1+1M11

= (-1)2 M11

= M11

= (Substituting the value of M11)

Similarly,

Similarly, the cofactor of each of the remaining elements of |A| can be determined.

21 2213 21 32 22 31

31 32

12 1321 12 33 13 32

32 33

( )

( )

a aM a a a aa a

a aM a a a aa a

1 212 12 12 21 33 23 31

1 313 13 13 21 32 22 31

2 121 21 21 12 33 13 32

( 1) . ( )

( 1) . ( )

( 1) . ( )

C M M a a a a

C M M a a a a

C M M a a a a

22 33 23 32( )a a a a

1. DETERMINANTS(Cont…)Example: Let us find the minor and cofactor of each element of

The minors of the elements of |A|are given by

1 3 2| | 4 1 2

3 5 2A

= (-1).(2) – (2).(5) = -2 – 10 = -12

= (4).(2) – (2).(3) = 8 – 6 = 2

= (4).(5) – (-1). (3) = 20 + 3 = 23

= (-3).(2) – (5).(2) = -6 – 10 = -16

= (1).(2) – (2).(3) = 2 – 6 = -4

= (1).(5) – (-3).(3) = 5 + 9 = 14

= (-3).(2) – (-1).(2) = -6 +2 = -4

= (1).(2) – (4).(2) = 2 – 8 = -6

= (1).(-1) – (-3).(4) = -1 + 12 = 11

221 23 2

M

111 25 2

M 124 23 2

M 134 13 5

M

213 25 2

M 231 33 5

M

313 21 2

M

321 24 2

M 331 34 1

M

1. DETERMINANTS(Cont…)Now let us find the cofactors of the corresponding elements of |A|.

From the definition of cofactors, Cij = (-1)i+j Mij.

C11 = (-1)1+1 . M11

= (-1)2 . M11

= M11 = -12

Substituting the

C12 = (-1)1+2 . M12

= (-1)3 . M12

= - M12 = -2

values of M11, M12,and

C13 = (-1)1+3 . M13

= (-1)4 . M13

= M13 = 23

M13 respectively.

C21 = (-1)2+1 . M21

= (-1)3 . M21

= - M21 = 16

C22 = (-1)2+2 . M22

= (-1)4 . M22

= M22 = -4

C23 = (-1)2+3 . M23

= (-1)5 . M23

= M23 = -14

C31 = (-1)3+1 . M31

= (-1)4 . M31

= M31 = -4

C32 = (-1)3+2 . M32

= (-1)5 . M32

= - M32 = 6

C33 = (-1)3+3 . M33

= (-1)6 . M33

= M33 = 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

= a11. (Its cofactor) + a12. (Its cofactor) + a13. (Its cofactor)

= a11. C11 + a12. C12 + a13. C13

= a11. M11 - a12. M12 + a13. M13 (As C12 = -M12)

= =

We may expand by any row or column.

11 12 13

21 22 23

31 32 33

a a aa a aa a a

22 23 21 23 21 2211 12 13

31 3232 33 31 33. . .a a a a a aa a a a aa a a a

11 22 33 23 32 12 21 33 31 23 13 21 32 22 31.( ) .( ) .( )a a a a a a a a a a a a a a a

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

Expanding the given determinant about first row, we get

= 3.{(1).(-1) - (-5).(5)} – 0{(2).(1) – (-5).(2)} + 0.(2. 5 + 1. 2)

= 3.{-1 + 25} – 0 + 0 = 3 . 24 = 72

1 5 2 5 2 12 1 5 5 1 2 1 2 52

33

5

00

1

00

3 0 02 1 52 5 1

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

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.

= 1.(its cofactor) + 2.(its cofactor) + (-1).(its cofactor)

+ 3 .(its cofactor)

1 2 1 32 1 2 33 1 2 11 1 0 2

A

2 1 2 3| |

1

3 1 2 11 1 0 2

2 1 3

A

1. DETERMINANTS(Cont…)Solving further, we get,

= 1 . C11 + 2 . C12 + (-1) . C13 + 3 . C14

= 1 . M11 + 2 .(-M12) + (-1) . M13 + 3 . (-M)14 (Cij = (-1)i+j Mij).

= 1 . M11 - 2 .M12 + (-1) . M13 - 3 . M14

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 1 2 33 1 2 1

1

1 1 3

2

2

1 0

1 2 (1 2 3 2 2 3 2 1 3 2 1 2

. 1 2 1 3 2 1 3 1 1 3 1 21 1 21 0 2 1

1)0 2 1 1 0

3

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

a1x + b1y = c1

a2x + b2y = c2

is given by

where provided that D 0.

REMARK: Here, is the determinant of the coefficient matrix

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.

, yxDD

x yD D

1 1

2 2

a bD

a b

1 1

2 2

a b

a b

1 1 1 1

2 2

1 1

2 22 2

, ,x y

c c

c

a b b aD D D

a b b ca

Dx and Dy are obtained by replacing the x-coefficients and y-coefficients in D

respectively with the constants c1 and c2.

2. SOLVING A SYSTEM OF LINEAR EQUATIONS USING DETERMINANTS (Cont…)

2. Dx, the determinant in the numerator of x, is obtained by replacing the x-coefficients in D, a1 and a2, with the constants on the right side of the equations, c1 and c2.

3. Dy, the determinant in the numerator of y, is obtained by replacing the y-coefficients in D, b1 and b2, with the constants on the right side of the equations, c1 and c2.

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

If D = 0, Dx = 0, and Dy = 0, the system is consistent and has infinite number of solutions. The equations in the system are dependent.If D = 0, and one of Dx and Dy is non-zero, then the system is inconsistent.

, yxDD

x yD D

1 11

2 2 2

1

2

, x

b bD D

c

b ba c

a

11

2

1 1

2 2 2

, y

b aD D

cb

c

a a

a


(Cont…)Example: Solve the following system of equations with the help of determinants

2x – y = 17 3x + 5y = 6

We know , substituting the values from the given system of equation, we get,

Now, , substituting the values from the given system of equation, we get,

Next,

1 1

2 2

a bD

a b

1 1

2 2x

c bD

c b

1 1

2 2y

a cD

a c

2 1(2).(5) ( 1).(3) 10 3 13

3 5D

17 1(17).(5) ( 1).(6) 85 6 91

6 5xD


Substituting the values from the given system of equations, we have,

Now we have found the values of all the three determinants, using Cramer’s rule, we have,

Substituting the values, we get,

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.

, yxDD

x yD D

917,

13

393

13

x

y

Dx

DD

yD

2 172.6 (17).3 12 51 39

3 6yD


(Cont…)

Additional Examples:

SYSTEM OF LINEAR EQUATIONS

VALUES OF D, Dx AND Dy REMARKS

2x – y = 173x + 5y = 6

D = 13Dx = 91

Dy = -39

As D 0, the given system is consistent and has a unique solution given by x = 7 and y = -3

x + 2y = 34x + 8y = 12

D = 0Dx = 0

Dy = 0

As D = Dx = Dy = 0, the given system has infinite number of solutions

2x + y = 34x + 2y = 4

D = 0Dx = -1

As D = 0 and Dx 0, so the given system is inconsistent.


(Cont…)

2.2 SOLUTION OF SYSTEM OF LINEAR EQUATIONS IN THREE VARIABLES BY CRAMER’S RULE:The solution of the system of linear equations

a1x + b1y + c1z = d1

a2x + b2y + c2z = d2

a3x + b3y + c3z = d3

is given by

where

provided that D 0REMARK: Here D is the determinant of the coefficient matrix.

1 1 111 1 1 1 1 1

2 2 2 2 2 2 2 2 2

3 3 3 3 3 3 3

1 1

2

3 3 3

2 2

3 3

, , ,x y z

a b c b c a c a b

D a b c D b c D

d d d

a c D a bd d d

d d da b c b c a c a b

, ,yx zDD D

x y zD D D


(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

If D = 0 and Dx = Dy = Dz = 0, then the system of equations is consistent with infinitely many solutions.

If D = 0 and at least one of the determinants Dx , Dy , Dz is non-zero, then the given system of equations is inconsistent.

Example: Let us solve the following system of equations using Cramer’s rule:5x – 7y +z = 116x – 8y – z = 153x + 2y – 6z = 7

We know

(Substituting the values from the given system of equations)

11 1

2 2 2

3 3 3

5 7 1

6 8 1

3 2 6

a b c

D a b c

a b c

, ,yx zDD D

x y zD D D


(Cont…)

Expanding about first row, we get,

8 1 6 1 6 86 8 1

2 6 3 6 3 23 2 6

{( 8).( 6) ( 1).(2)} {(6).( 6) ( 1).(3)} {(6).(2) ( 8).(3)}

(48 2) (

5 7 1

5

36 3) (12 24)

(50) ( 33) 36

250 231

( 7) 1

5 ( 7) 1

5 7 1

5

5

7

36

5

D

D

D

D

D

D


(Cont…)

Similarly,

15 8 1 (48 2) ( 90 7) (30 56)

7 2 6

550 581 86 55

6 15 1 ( 90 7) ( 36 3) (42 45)

3 7 6

415 363 3 55

6 8 15 ( 56 30) (42 45) (12 24)

3 2 7

430

11 7 1

11 7 1

5 11 1

5 11 1

5 7 11

5 7 1

2 96 55

1

1 3

x

x

y

y

z

z

D

D

D

D

D

D


(Cont…)

So, by Cramer’s rule, we have,

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.

551

55

551

5555

155

x

y

z

Dx

DD

yDD

zD


(Cont…)

Additional Examples:

SYSTEM OF LINEAR EQUATIONS

VALUES OF D, Dx , Dy,

AND Dz

REMARKS

5x – 7y + z = 116x - 8y - z = 153x + 2y – 6z = 7

D = 55Dx = 55

Dy = -55

Dz = -55

As D 0, the given system is consistent and has a unique solution given by x = 1, y = -1 and z = -1

x + y + z = 1x + 2y + 3z = 4X + 3y + 5z = 7

D = 0Dx = 0

Dy = 0

Dz = 0

As D = Dx = Dy = Dz = 0, the given system has infinite number of solutions.

2x – y + z = 4x + 3y + 2z = 123x + 2y + 3z = 10

D = 0Dx = 30 0

As D = 0 and Dx 0, so the given system is inconsistent.

3. SUMMARYLet 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

Minor : The minor of an element aij 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 Mij .

Cofactor: The cofactor Cij of an element id defined as Cij = (-1)i+j Mij.

Value of the third order determinant

=

Cramer’s Rule: The solution of the system of simultaneous linear equations a1x + b1y = c1

a2x + b2y = c2 is given by

11 1211 22 12 21

21 22

a a a a a aa a

11 12 13

21 22 23

31 32 33

a a aa a aa a a

11 22 33 23 32 12 21 33 31 23 13 21 32 22 31.( ) .( ) .( )a a a a a a a a a a a a a a a

, yxDD

x yD D

3. SUMMARY(Cont…)Cramer’s Rule: The solution of the system of linear equations

a1x + b1y + c1z = d1

a2x + b2y + c2z = d2

a3x + b3y + c3z = d3 is given by 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

If D = 0, Dx = 0, and Dy = 0, the system is consistent and has infinite number of solutions.If D = 0, and one of Dx and Dy 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

If D = 0 and Dx = Dy = Dz = 0, then the system of equations is consistent with infinitely many solutions.If D = 0 and at least one of the determinants Dx , Dy , Dz is non-zero, then the given system of equations is inconsistent.

, yxDD

x yD D

, ,yx zDD D

x y zD D D

, ,yx zDD D

x y zD D D

week 8 systems of equations determinants and cramer’s rule

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