Cramers Rule 2

8/2/2019 Cramers Rule 2

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Solution of Simultaneous

Linear Equations

Simultaneous linear equations appear often in the solution of problems in

electrical/electronics technology. The solutions of these equations has been

simplified by a branch of mathematics called linear algebra. Although the

actual theorems and proofs are well outside the scope of this textbook, we

will use some of the principles of linear algebra to solve simple linear

equations. The following is a set ofn simultaneous linear equations in n

unknowns:

a11x1 a12x2 a1nxn b1

a21x1 a22x2 a2nxn b2

.

.

.

an1x1 an2x2

annxn bn

The above equations may also be expressed in matrix form as

AX B

where

A , X , B Substitution

Althoughsimultaneouslinearequations maybe expressedin severalunknowns,

we begin with the most simple, namely two simultaneous linear equations in

two unknowns. Consider the equations below:

a11x1 a12x2 b1 (B1)

a21x1 a22x2 b2 (B2)

If we multiply Equation B1 by a22 and Equation B2 by a12, we have

a11a22x1 a12a22x2 a22b1

a12a21x1 a12a22x2 a12b2

b1

b2

.

.

.

bn

x1

x2

.

.

.

xn

a11a12 a1na21a22

a2n. . .

. . .

. .

an1an2

ann

1076

APPENDIX

B


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Subtracting, we obtain

a11a22x1 a12a21x1 a22b1 a12b2

which gives

x1 a

a

1

2

1

2

a

b

22

1

a

a

1

1

2

2

a

b

2

2

1

Similarly, we solve for the unknownx2 as

x2 a

a

11

1

a

1b

22

2

a

a

1

2

2

1b

a

1

21

Appendix B Solution of Simultaneous Linear Equations 1077

EXAMPLE B1 Use substitution to find the solutions for the following fol-

lowing linear equations:

2x1 8x2 2

x1 2x2 5

Solution Rewriting the first equation, we have

2x1 2 8x2

x1 1 4x2

Now, substituting the above expression into the second equation, we have

(1 4x2) 2x2 5

2x2 6

x2 3

Finally, we have

x1 1 4 (3) 11

Determinants

While substitution may be used for solving simultaneous linear equations in

two variables, it is lengthy and particularly complicated when solving for

more than two unknowns. An easier method used for solving simultaneous

linear equations involves using determinants. We begin by expressing the

simultaneous linear equations (B1) and (B2) as a product of matrices:

Column 1 Column 2 Column 3

(B3)b1

b2

x1

x2

a11 a12

a21 a22

A determinant is a set of coefficients which has the same number of

rows and columns and which may be expressed as a single value. The num-

ber of rows (or columns) defines the order of a determinant. The second-


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The solution of third-order simultaneous linear equations is similar to

the method used for solving second-order equations. Consider the following

third-order simultaneous linear equation:

a11x1 a12x2 a13x3 b1

a21x1 a22x2 a23x3 b2

a31x1 a32x2 a33x3 b3

The corresponding matrix equation is shown as follows:

The value of the third-order determinant may be found in one of several

ways. The first method works for only third-order determinants, while the

second method is a more general approach which evaluates any order of

determinant.

Method I. This method works only for third-order determinants:

1. Begin by writing the original columns of the third-order determinant.

2. Copy the first two columns, placing them to the right of the original

determinant.

3. Add the product of the elements of the principal diagonal to the products ofthe adjacent two parallel diagonals to the right of the principal diagonal.

4. Subtract the product of the elementsof the secondary diagonal and also sub-

tract theproducts of theelements alongthetwootherparalleldiagonals.

The resultant determinant is written asD a11a22a33 a12a23a31 a13a21a32 a31a22a13 a32a23a11 a33a21a12

add subtract

a11 a12 a13

a11 a12

a21 a22 a23 a21 a22

a31 a32 a33 a31 a32

b1b2

b3

x1x2

x3

a11a12a13a21a22a23

a31a32a33

and

2 21 5x2

(2)(5)

4

(1)(2)

12

4 3

4

EXAMPLE B3 Evaluate the following determinant:

D 2

3

2

1

2

3

3

1

2


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1080 Appendix B Solution of Simultaneous Linear Equations

Solution We begin by rewriting the first two columns as follows:

D

Now, adding the products of the principal diagonal and adjacent diagonalsand subtracting the products of the secondary diagonal and adjacent diago-

nals, we have

D (3)(2)(2) (1)(3)(2) (2)(1)(3)

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

49

1

2

3

3

1

2

2

3

2

1

2

3

3

1

2

b1 a12 a13

b2 a22 a23

b3 a32 a33x1

a11 a12 a13

a21 a22 a23

a31 a32 a33

b1 b2 b3 a12 a13a22 a23a12 a13

a32 a33

a22 a23

a32 a33x1

D

b1(a22a33 a23a32) b2(a12a33 a13a32) b3(a12a23 a13a22)

D

METHOD II. This evaluation of determinants is achieved by expansion by

minors. The minorof an element is the determinant which remains after

deleting the row and the column in which the element lies. The value of anynth-order determinant is found as follows:

1. For any row or column, find the product of each element and the determi-

nant of its minor.

2. A product is given a positive sign if the sum of the row and the column of

the element is even. The product is given a negative sign if the sum is

odd.

3. The value of the determinant is the sum of the resulting terms.

As before, Cramers rule is used to solve for the unknowns, x1, x2, and

x3, by using determinants and replacing the appropriate terms of the numera-

tor with the terms of the solution matrix. The resulting determinants and

solutions are given as follows:

By expansion of minors, the determinant of the denominator is found as

D a11 a21 a31 a11(a22a33 a23a32) a21(a12a33 a13a32) a31(a12a23 a13a22)

The solution forx1 is now found to be

a12 a13

a22 a23

a12 a13

a32 a33

a22 a23

a32 a33


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Similarly, forx2, we get

x2

a11 a12 a13

a21 a22 a23

a31 a32 a33

a11 b1 a13

a21 b2 a23

a31 b3 a33

b1(a21a33 a23a31) b2 (a11a33 a13a31) b3 (a11a23 a13a21)

D

and forx3 we have

a11 a12 b1

a21 a22 b2

a31 a32 b3

b1(a21a32 a22a31) b2(a11a32 a12a31) b3 (a11a22 a12a21)

D

x3

a11 a12 a13

a21 a22 a23

a31 a32 a33

EXAMPLE B4 Solve for x1 in the following system of linear equations

using minors.

3x1 x2 2x3 1

x1 2x2 3x3 11

2x

1

3x

2

2x

3

3

Solution The determinant of the denominator is evaluated as follows:

D 2

3

2

1

2

3

3

1

2

(3) (1) (2) (3)(4 9) (2 6) (2) (3 4)

49

1 2

2 3

1 2

3 2

2 3

3 2


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1082 Appendix B Solution of Simultaneous Linear Equations

(4 9) (11)(2 6) (3)(3 4)

49

2

and so the unknownx1 is calculated to be

2

3

2

1

2

3

1

11

3x1 49

(1) (11) (3) 1 22 31 2

3 2

2 3

3 2

49

PRACTICE

PROBLEM

Use expansion by minors to solve forx2 andx3 in Example B4.

Answers: x2 3, x3 1

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Cramers Rule 2