89251 19 appa 4/4/07 1:08 pm page 759 asimultaneous...

759

Circuit analysis frequently involves the solution of linear simultaneousequations. Our purpose here is to review the use of determinants to solvesuch a set of equations.The theory of determinants (with applications) canbe found in most intermediate-level algebra texts. (A particularly goodreference for engineering students is Chapter 1 of E.A. Guillemin’s TheMathematics of Circuit Analysis [New York: Wiley, 1949]. In our reviewhere, we will limit our discussion to the mechanics of solving simultaneousequations with determinants.

A.1 Preliminary StepsThe first step in solving a set of simultaneous equations by determinants isto write the equations in a rectangular (square) format. In other words, wearrange the equations in a vertical stack such that each variable occupiesthe same horizontal position in every equation. For example, in Eqs. A.1,the variables , , and occupy the first, second, and third position,respectively, on the left-hand side of each equation:

(A.1)

Alternatively, one can describe this set of equations by saying that occupies the first column in the array, the second column, and thethird column.

If one or more variables are missing from a given equation, they canbe inserted by simply making their coefficient zero. Thus Eqs. A.2 can be“squared up” as shown by Eqs. A.3:

(A.2)

(A.3)

7v1 + 0v2 + 2v3 = 5.

0v1 + 4v2 + 3v3 = 16,

2v1 - v2 + 0v3 = 4,

7v1 + 2v3 = 5;

4v2 + 3v3 = 16,

2v1 - v2 = 4,

i3i2

i1

-8i1 - 4i2 + 22i3 = 50.

-3i1 + 6i2 - 2i3 = 3,

21i1 - 9i2 - 12i3 = -33,

i3i2i1

The Solution of LinearSimultaneous EquationsAAppendix

89251_19_AppA 4/4/07 1:08 PM Page 759

Electric Circuits, Eighth Edition, by James A. Nilsson and Susan A. Riedel.ISBN-13: 978-0-13-198922-1. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.

760 The Solution of Linear Simultaneous Equations

A.2 Cramer’s MethodThe value of each unknown variable in the set of equations is expressed asthe ratio of two determinants. If we let N, with an appropriate subscript,represent the numerator determinant and represent the denominatordeterminant, then the kth unknown is

(A.4)

The denominator determinant is the same for every unknown variableand is called the characteristic determinant of the set of equations. Thenumerator determinant varies with each unknown. Equation A.4 isreferred to as Cramer’s method for solving simultaneous equations.

A.3 The Characteristic DeterminantOnce we have organized the set of simultaneous equations into anordered array, as illustrated by Eqs. A.1 and A.3, it is a simple matter toform the characteristic determinant. This determinant is the square arraymade up from the coefficients of the unknown variables. For example, thecharacteristic determinants of Eqs. A.1 and A.3 are

(A.5)

and

(A.6)

respectively.

A.4 The Numerator DeterminantThe numerator determinant is formed from the characteristic determi-nant by replacing the kth column in the characteristic determinant withthe column of values appearing on the right-hand side of the equations.

Nk

¢ = 3 2 -1 00 4 37 0 2

3 ,

¢ = 3 21 -9 -12-3 6 -2-8 -4 22

3

Nk

¢

xk =

Nk

¢

.

xk

¢

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A.5 The Evaluation of a Determinant 761

For example, the numerator determinants for evaluating , , and inEqs. A.1 are

(A.7)

(A.8)

and

(A.9)

The numerator determinants for the evaluation of , , and inEqs. A.3 are

(A.10)

(A.11)

and

(A.12)

A.5 The Evaluation of a DeterminantThe value of a determinant is found by expanding it in terms of its minors.The minor of any element in a determinant is the determinant thatremains after the row and column occupied by the element have beendeleted. For example, the minor of the element 6 in Eq. A.7 is2 -33 -12

50 222 ,

N3 = 3 2 - 1 40 4 167 0 5

3 .

N2 = 3 2 4 00 16 37 5 2

3 , N1 = 3 4 -1 0

16 4 35 0 2

3 ,v3v2v1

N3 = 3 21 -9 -33-3 6 3-8 -4 50

3 .

N2 = 3 21 -33 -12-3 3 -2-8 50 22

3 , N1 = 3 -33 -9 -12

3 6 -250 - 4 22

3 ,i3i2i1

89251_19_AppA 4/4/07 1:08 PM Page 761




while the minor of the element 22 in Eq. A.7 is

The cofactor of an element is its minor multiplied by the sign-controlling factor

where i and j denote the row and column, respectively, occupied by theelement. Thus the cofactor of the element 6 in Eq. A.7 is

and the cofactor of the element 22 is

The cofactor of an element is also referred to as its signed minor.The sign-controlling factor will equal or depending on

whether is an even or odd integer. Thus the algebraic sign of a cofac-tor alternates between and as we move along a row or column. Fora 3 determinant, the plus and minus signs form the checkerboard pat-tern illustrated here:

A determinant can be expanded along any row or column. Thus the firststep in making an expansion is to select a row i or a column j. Once a rowor column has been selected, each element in that row or column is multi-plied by its signed minor, or cofactor. The value of the determinant is thesum of these products. As an example, let us evaluate the determinant inEq. A.5 by expanding it along its first column. Following the rules justexplained, we write the expansion as

(A.13)

The determinants in Eq. A.13 can also be expanded by minors.The minor of an element in a determinant is a single element. It fol-lows that the expansion reduces to multiplying the upper-left element bythe lower-right element and then subtracting from this product the product

2 * 22 * 2

¢ = 21(1) 2 6 -2-4 22

2 - 3(-1) 2 -9 -12-4 22

2 - 8(1) 2 -9 -126 -2

2

3 + - +

- + -

+ - +

3* 3

-1+1i + j

-1+1-1(i + j)

-1(3 + 3) 2 -33 -93 6

2 .-1(2 + 2) 2 -33 -12

50 222 ,

-1(i + j),

2 -33 -93 6

2 .

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A.5 The Evaluation of a Determinant 763

of the lower-left element times the upper-right element. Using this obser-vation, we evaluate Eq. A.13 to

(A.14)

Had we elected to expand the determinant along the second row of ele-ments, we would have written

(A.15)

The numerical values of the determinants , , and given byEqs. A.7, A.8, and A.9 are

(A.16)

(A.17)

and

(A.18)

It follows from Eqs.A.15 through A.18 that the solutions for , , and inEq. A.1 are

(A.19)

and

i3 =

N3

¢

= 3 A.

i2 =

N2

¢

= 2 A,

i1 =

N1

¢

= 1 A,

i3i2i1

N3 = 3438.

N2 = 2292,

N1 = 1146,

N3N2N1

= -738 + 2196 - 312 = 1146.

= 3(-198 - 48) + 6(462 - 96) + 2(-84 - 72)

¢ = -3(-1) 2 -9 -12-4 22

2 +6(+1) 2 21 -12-8 22

2 -2(-1) 2 21 -9-8 -4

2 = 2604 - 738 - 720 = 1146.

¢ = 21(132 - 8) + 3(-198 - 48) - 8(18 + 72)

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We leave you to verify that the solutions for , , and in Eqs. A.3 are

(A.20)

and

A.6 MatricesA system of simultaneous linear equations can also be solved usingmatrices. In what follows, we briefly review matrix notation, algebra, andterminology.1

A matrix is by definition a rectangular array of elements; thus

(A.21)

is a matrix with m rows and n columns.We describe A as being a matrix oforder m by n, or , where m equals the number of rows and n thenumber of columns. We always specify the rows first and the columns sec-ond. The elements of the matrix— —can be real numbers,complex numbers, or functions. We denote a matrix with a boldface capi-tal letter.

The array in Eq. A.21 is frequently abbreviated by writing

(A.22)

where is the element in the i th row and the jth column.If , A is called a row matrix, that is,

(A.23)A = [a11 a12 a13Á a1n].

m = 1aij

A = [aij]mn ,

a11, a12, a13, . . .

m * n

A = D a11 a12 a13Á a1n

a21 a22 a23Á a2n

Á Á Á Á Á

am1 am2 am3Á amn

T

v3 =

-184-5

= 36.8 V.

v2 =

118-5

= -23.6 V,

v1 =

49-5

= -9.8 V,

v3v2v1

1 An excellent introductory-level text in matrix applications to circuit analysis is Lawrence P.Huelsman, Circuits, Matrices, and Linear Vector Spaces (New York: McGraw-Hill, 1963).

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A.7 Matrix Algebra 765

If , A is called a column matrix, that is,

(A.24)

If , A is called a square matrix. For example, if , thesquare 3 by 3 matrix is

(A.25)

Also note that we use brackets to denote a matrix, whereas we usevertical lines to denote a determinant. It is important to know the dif-ference. A matrix is a rectangular array of elements. A determinant is afunction of a square array of elements.Thus if a matrix A is square, we candefine the determinant of A. For example, if

then

A.7 Matrix AlgebraThe equality, addition, and subtraction of matrices apply only to matricesof the same order. Two matrices are equal if, and only if, their correspon-ding elements are equal. In other words, if, and only if, forall i and j. For example, the two matrices in Eqs. A.26 and A.27 are equalbecause , , , and :

(A.26)

(A.27) B = c36 -204 16

d .

A = c36 -204 16

d ,

a22 = b22a21 = b21a12 = b12a11 = b11

aij = bijA = B

det A = 2 2 16 15

2 = 30 - 6 = 24.

A = c2 16 15

d ,

ƒ ƒ

[ ]

A = Ca11 a12 a13

a21 a22 a23

a31 a32 a33

S .

m = n = 3m = n

A = E a11

a21

a31

o

am1

U .

n = 1

89251_19_AppA 4/4/07 1:08 PM 65




If A and B are of the same order, then

(A.28)

implies

(A.29)

For example, if

(A.30)

and

(A.31)

then

(A.32)

The equation

(A.33)

implies

(A.34)

For the matrices in Eqs. A.30 and A.31, we would have

(A.35)

Matrices of the same order are said to be conformable for addition andsubtraction.

Multiplying a matrix by a scalar k is equivalent to multiplying eachelement by the scalar. Thus if, and only if, . It should benoted that k may be real or complex. As an example, we will multiply thematrix D in Eq. A.35 by 5. The result is

(A.36)5D = c -60 -80 200140 20 -95

d .

aij = kbijA = kB

D = c -12 -16 4028 4 -19

d .

dij = aij - bij.

D = A - B

C = c 20 4 -20-12 20 11

d .

B = c 16 10 -30-20 8 15

d ,

A = c4 -6 108 12 -4

d ,

cij = aij + bij .

C = A + B

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Matrix multiplication can be performed only if the number ofcolumns in the first matrix is equal to the number of rows in the secondmatrix. In other words, the product AB requires the number of columns inA to equal the number of rows in B. The order of the resulting matrix willbe the number of rows in A by the number of columns in B. Thus if

, where A is of order and B is of order , then willbe a matrix of order . When the number of columns in A equals thenumber of rows in B, we say A is conformable to B for multiplication.

An element in is given by the formula

(A.37)

The formula given by Eq. A.37 is easy to use if one remembers thatmatrix multiplication is a row-by-column operation. Hence to get the i th,j th term in , each element in the i th row of A is multiplied by the corre-sponding element in the j th column of B, and the resulting products aresummed.The following example illustrates the procedure.We are asked tofind the matrix when

(A.38)

and

(A.39)

First we note that will be a matrix and that each element in will require summing three products.

To find we multiply the corresponding elements in row 1 of matrix Awith the elements in column 1 of matrix B and then sum the products. Wecan visualize this multiplication and summing process by extracting thecorresponding row and column from each matrix and then lining them upelement by element. So to find we have

therefore

To find we visualize

Row 1 of AColumn 2 of B

6 2 � 3

3 � 2 -2

;

C12

C11 = 6 * 4 + 3 * 0 + 2 * 1 = 26.


6 4 � 3

0 � 2 1

;

C11

C11

C2 * 2C

B = C4 20 31 -2

S .

A = c6 3 21 4 6

d

C

C

cij = ap

k = 1aikbkj .

C

m * nCp * nm * pC = AB

89251_19_AppA 4/4/07 1:08 PM Page 767




thus

For we have

and

Finally, for we have

from which

It follows that

(A.40)

In general, matrix multiplication is not commutative, that is,. As an example, consider the product for the matrices in

Eqs.A.38 and A.39.The matrix generated by this multiplication is of order, and each term in the resulting matrix requires adding two products.

Therefore if , we have

(A.41)

Obviously, . We leave you to verify the elements in Eq. A.41.Matrix multiplication is associative and distributive. Thus

(A.42)

(A.43)

and

(A.44) (A + B)C = AC + BC.

A(B + C) = AB + AC,

(AB)C = A(BC),

C Z D

D = C26 20 203 12 184 -5 -10

S .

D = BA3 * 3

BAAB Z BA

C = AB = B26 1710 2

R .

C22 = 1 * 2 + 4 * 3 + 6 * (-2) = 2.


1 2 � 4

3 � 6 -2

;

C22

C21 = 1 * 4 + 4 * 0 + 6 * 1 = 10.


1 4 � 4

0 � 6 1

;

C21

C12 = 6 * 2 + 3 * 3 + 2 * (-2) = 17.

89251_19_AppA 4/4/07 1:08 PM Page 768




In Eqs.A.42, A.43, and A.44, we assume that the matrices are conformablefor addition and multiplication.

We have already noted that matrix multiplication is not commutative.There are two other properties of multiplication in scalar algebra that donot carry over to matrix algebra.

First, the matrix product does not imply either or. (Note: A matrix is equal to zero when all its elements are zero.) For

example, if

then

Hence the product is zero, but neither A nor B is zero.Second, the matrix equation does not imply . For

example, if

then

The transpose of a matrix is formed by interchanging the rows andcolumns. For example, if

The transpose of the sum of two matrices is equal to the sum of thetransposes, that is,

(A.45)

The transpose of the product of two matrices is equal to the productof the transposes taken in reverse order. In other words,

(A.46)[AB]T= BTAT.

(A + B)T= AT

+ BT.

A = C1 2 34 5 67 8 9

S , then AT= C1 4 7

2 5 83 6 9

S .

AB = AC = c3 46 8d , but B Z C.

A = c1 02 0d , B = c3 4

7 8d , and C = c3 4

5 6d ,

B = CAB = AC

AB = c0 00 0d = 0.

A = c1 02 0d and B = c0 0

4 8d ,

B = 0A = 0AB = 0

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Equation A.46 can be extended to a product of any number of matri-ces. For example,

(A.47)

If , the matrix is said to be symmetric. Only square matricescan be symmetric.

A.8 Identity, Adjoint, and InverseMatrices

An identity matrix is a square matrix where for , and for . In other words, all the elements in an identity matrix are zeroexcept those along the main diagonal, where they are equal to 1. Thus

are all identity matrices. Note that identity matrices are always square. Wewill use the symbol for an identity matrix.

The adjoint of a matrix A of order is defined as

(A.48)

where is the cofactor of . (See Section A.5 for the definition of acofactor.) It follows from Eq. A.48 that one can think of finding theadjoint of a square matrix as a two-step process. First construct a matrixmade up of the cofactors of A, and then transpose the matrix of cofactors.As an example we will find the adjoint of the matrix

The cofactors of the elements in A are

¢11 = 1(10 - 1) = 9,¢12 = -1(15 + 1) = -16,¢13 = 1(3 + 2) = 5,¢21 = -1(10 - 3) = -7,¢22 = 1(5 + 3) = 8,¢23 = -1(1 + 2) = -3,¢31 = 1(2 - 6) = -4,¢32 = -1(1 - 9) = 8,¢33 = 1(2 - 6) = -4.

A = C 1 2 33 2 1

-1 1 5S .

3 * 3

aij¢ij

adj A = [¢ji]n * n ,

n * nU

c1 00 1

d , C1 0 00 1 00 0 1

S , and D1 0 0 00 1 0 00 0 1 00 0 0 1

Ti = j

aij = 1i Z jaij = 0

A = AT

[ABCD]T= DTCTBTAT.

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A.8 Identity, Adjoint, and Inverse Matrices 771

The matrix of cofactors is

It follows that the adjoint of A is

One can check the arithmetic of finding the adjoint of a matrix byusing the theorem

(A.49)

Equation A.49 tells us that the adjoint of A times A equals the determi-nant of A times the identity matrix, or for our example,

If we let and use the technique illustrated in Section A.7,we find the elements of to be

Therefore

A square matrix A has an inverse, denoted as , if

(A.50)A-1A = AA-1= U.

A-1

= det A # U.

C = C -8 0 00 -8 00 0 -8

S = -8C1 0 00 1 00 0 1

S

c11 = 9 - 21 + 4 = -8,c12 = 18 - 14 - 4 = 0,c13 = 27 - 7 - 20 = 0,c21 = -16 + 24 - 8 = 0,c22 = -32 + 16 + 8 = -8,c23 = -48 + 8 + 40 = 0,c31 = 5 - 9 + 4 = 0,c32 = 10 - 6 - 4 = 0,c33 = 15 - 3 - 20 = -8.

CC = adj A # A

det A = 1(9) + 3(-7) - 1(-4) = -8.

adj A # A = det A # U.

adj A = BT= C 9 -7 -4

-16 8 85 -3 -4

S .

B = C 9 -16 5-7 8 -3-4 8 -4

S .

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Equation A.50 tells us that a matrix either premultiplied or postmultipliedby its inverse generates the identity matrix . For the inverse matrix toexist, it is necessary that the determinant of A not equal zero. Only squarematrices have inverses, and the inverse is also square.

A formula for finding the inverse of a matrix is

(A.51)

The formula in Eq. A.51 becomes very cumbersome if A is of an orderlarger than 3 by 3.2 Today the digital computer eliminates the drudgery ofhaving to find the inverse of a matrix in numerical applications of matrixalgebra.

It follows from Eq. A.51 that the inverse of the matrix A in the previ-ous example is

You should verify that .

A.9 Partitioned MatricesIt is often convenient in matrix manipulations to partition a given matrixinto submatrices. The original algebraic operations are then carried out interms of the submatrices. In partitioning a matrix, the placement of thepartitions is completely arbitrary, with the one restriction that a partitionmust dissect the entire matrix. In selecting the partitions, it is also neces-sary to make sure the submatrices are conformable to the mathematicaloperations in which they are involved.

For example, consider using submatrices to find the product, where

A = E 1 2 3 4 55 4 3 2 1

-1 0 2 -3 10 1 -1 0 10 2 1 -2 0

UC = AB

A-1A = AA-1= U

= C -1.125 0.875 0.52 -1 -1

-0.625 0.375 0.5S .

A-1= -1>8C 9 -7 -4

-16 8 85 -3 -4

S

A-1=

adj Adet A

.

U

2 You can learn alternative methods for finding the inverse in any introductory text onmatrix theory. See, for example, Franz E. Hohn, Elementary Matrix Algebra (New York:Macmillan, 1973).

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A.9 Partitioned Matrices 773

and

Assume that we decide to partition B into two submatrices, and ; thus

Now since B has been partitioned into a two-row column matrix, A must bepartitioned into at least a two-column matrix; otherwise the multiplicationcannot be performed.The location of the vertical partitions of the A matrixwill depend on the definitions of and . For example, if

then must contain three columns, and must contain two columns.Thus the partitioning shown in Eq. A.52 would be acceptable for execut-ing the product AB:

(A.52)

If, on the other hand, we partition the B matrix so that

then must contain two columns, and must contain three columns.In this case the partitioning shown in Eq.A.53 would be acceptable in exe-cuting the product :

(A.53)C = E 1 2 | 3 4 55 4 | 3 2 1

-1 0 | 2 -3 10 1 | -1 0 10 2 | 1 -2 0

U F20

Á

-130

V .

C = AB

A12A11

B11 = c20d and B21 = C -1

30S ,

C = E 1 2 3 | 4 55 4 3 | 2 1

-1 0 2 | -3 10 1 -1 | 0 10 2 1 | -2 0

U F20

-1Á

30

V .

A12A11

B11 = C 20

-1S and B21 = c3

0d ,

B21B11

B = cB11

B21d .

B21

B11

B = E 20

-130

U .

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For purposes of discussion, we will focus on the partitioning given inEq. A.52 and leave you to verify that the partitioning in Eq. A.53 leads tothe same result.

From Eq. A.52 we can write

(A.54)

It follows from Eqs. A.52 and A.54 that

and

The A matrix could also be partitioned horizontally once the verticalpartitioning is made consistent with the multiplication operation. In thissimple problem, the horizontal partitions can be made at the discretion ofthe analyst. Therefore could also be evaluated using the partitioningshown in Eq. A.55:

(A.55)C = F1 2 3 | 4 55 4 3 | 2 1

Á Á Á Á Á Á

-1 0 2 | -3 10 1 -1 | 0 10 2 1 | -2 0

V F

20

-1Á

30

V .

C

C = E 1113

-131

-7

U .

A12B21 = E 4 52 1

-3 10 1

-2 0

U c30d = E 12

6-9

0-6

U ,

A11B11 = E 1 2 35 4 3

-1 0 20 1 -10 2 1

U C 20

-1S = E -1

7-4

1-1

U ,

C = [A11 A12] cB11

B21d = A11B11 + A12B21.

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A.9 Partitioned Matrices 775

From Eq. A.55 it follows that

(A.56)

where

You should verify that

and

We note in passing that the partitioning in Eqs. A.52 and A.55 isconformable with respect to addition.

C = E 1113

-131

-7

U .

= C -41

-1S + C -9

0-6S = C -13

1-7S ,

C21 = C -1 0 20 1 -10 2 1

S C 20

-1S + C -3 1

0 1-2 0

S c30d

= c -17d + c12

6d = c11

13d ,

C11 = c1 2 35 4 3

d C 20

-1S + c4 5

2 1d c3

0d

C11 = A11B11 + A12B21,C21 = A21B11 + A22B21.

C = cA11 A12

A21 A22d cB11

B21d = cC11

C21d ,

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Use the matrix method to solve for the node volt-ages and in Eqs. 4.5 and 4.6.

SolutionThe first step is to rewrite Eqs. 4.5 and 4.6 in matrixnotation. Collecting the coefficients of and and at the same time shifting the constant terms tothe right-hand side of the equations gives us

(A.57)

It follows that in matrix notation, Eq.A.57 becomes

(A.58)

or

(A.59)

where

To find the elements of the matrix, we pre-multiply both sides of Eq. A.59 by the inverse ofA; thus

(A.60)

Equation A.60 reduces to

(A.61)

or

(A.62)V = A-1I.

UV = A-1I,

A-1AV = A-1I.

V

I = c102d .

V = cv1

v2d ,

A = c 1.7 -0.5-0.5 0.6

d ,

AV = I,

c 1.7 -0.5-0.5 0.6

d cv1

v2d = c10

2d ,

-0.5v1 + 0.6v2 = 2.

1.7v1 - 0.5v2 = 10,

v2v1

v2v1

It follows from Eq. A.62 that the solutions forand are obtained by solving for the matrix

product .To find the inverse of A, we first find the

cofactors of A. Thus

(A.63)

The matrix of cofactors is

(A.64)

and the adjoint of A is

(A.65)

The determinant of A is

adj A = BT= c0.6 0.5

0.5 1.7d .

B = c0.6 0.50.5 1.7

d ,

¢11 = (-1)2(0.6) = 0.6,¢12 = (-1)3(-0.5) = 0.5,¢21 = (-1)3(-0.5) = 0.5,¢22 = (-1)4(1.7) = 1.7.

A-1Iv2v1

Example A.1

A.10 ApplicationsThe following examples demonstrate some applications of matrix algebrain circuit analysis.

det A = 2 1.7 -0.5-0.5 0.6

2 = (1.7)(0.6) - (0.25) = 0.77.

(A.66)

From Eqs. A.65 and A.66, we can write the inverseof the coefficient matrix, that is,

(A.67)

Now the product is found:

(A.68)

It follows directly that

(A.69)

or and v2 = 10.91 V.v1 = 9.09 V

cv1

v2d = c 9.09

10.91d ,

=

10077c 78.4d = c 9.09

10.91d .

A-1I =

10077

c0.6 0.50.5 1.7

d c102d

A-1I

A-1=

10.77

c0.6 0.50.5 1.7

d .

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A.10 Applications 777

Use the matrix method to find the three mesh cur-rents in the circuit in Fig. 4.24.

SolutionThe mesh-current equations that describe the cir-cuit in Fig. 4.24 are given in Eq. 4.34. The constraintequation imposed by the current-controlled voltagesource is given in Eq. 4.35. When Eq. 4.35 is substi-tuted into Eq. 4.34, the following set of equationsevolves:

(A.70)

In matrix notation, Eqs. A.70 reduce to

(A.71)

where

and

It follows from Eq.A.71 that the solution for is

(A.72)

We find the inverse of A by using the relationship

(A.73)

To find the adjoint of A, we first calculate the cofac-tors of A. Thus

¢13 = (-1)4(20 + 50) = 70,

¢12 = (-1)3(-45 - 20) = 65,

¢11 = (-1)2(90 - 16) = 74,

A-1=

adj Adet A

.

I = A-1V.

I

V = C5000S .

I = C i1

i2

i3

S ,

A = C 25 -5 -20-5 10 -4-5 -4 9

S ,

AI = V,

-5i1 - 4i2 + 9i3 = 0.

-5ii + 10i2 - 4i3 = 0,

25ii - 5i2 - 20i3 = 50,The cofactor matrix is

(A.74)

from which we can write the adjoint of A:

(A.75)


adj A = BT= C74 125 220

65 125 20070 125 225

S .

B = C 74 65 70125 125 125220 200 225

S ,

¢33 = (-1)6(250 - 25) = 225.

¢32 = (-1)5(-100 - 100) = 200,

¢31 = (-1)4(20 + 200) = 220,

¢23 = (-1)5(-100 - 25) = 125,

¢22 = (-1)4(225 - 100) = 125,

¢21 = (-1)3(-45 - 80) = 125,

Example A.2

It follows from Eq. A.73 that

(A.76)

The solution for is

(A.77)

The mesh currents follow directly from Eq.A.77.Thus

(A.78)

or , , and . Example A.3illustrates the application of the matrix methodwhen the elements of the matrix are complexnumbers.

28 A=i326 A=i229.6 A=i1

C ii

i2

i3

S = C29.626.028.0S

I =

1125C74 125 220

65 125 20070 125 225

S C5000S = C29.60

26.0028.00

S .

I

A-1=

1125C74 125 220

65 125 20070 125 225

S .

= 25(90 - 16) + 5(-45 - 80) - 5(20 + 200) = 125.

det A = 3 25 -5 -20-5 10 -4-5 -4 9

3

89251_19_AppA 4/4/07 1:08 PM Page 777




Use the matrix method to find the phasor mesh cur-rents and in the circuit in Fig. 9.37.

SolutionSumming the voltages around mesh 1 generates theequation

(A.79)

Summing the voltages around mesh 2 produces theequation

(A.80)

The current controlling the dependent voltagesource is

(A.81)

After substituting Eq. A.81 into Eq. A.80, theequations are put into a matrix format by first collect-ing, in each equation, the coefficients of and ; thus

(A.82)

Now, using matrix notation, Eq. A.82 is written

(A.83)

where


(A.84)

The inverse of the coefficient matrix A is foundusing Eq. A.73. In this case, the cofactors of A are

¢11 = (-1)2(-26 - j13) = -26 - j13,¢12 = (-1)3(27 + j16) = -27 - j16,¢21 = (-1)3(-12 + j16) = 12 - j16,¢22 = (-1)4(13 - j14) = 13 - j14.

I = A-1V.

I = cI1

I2d , and V = c150l0 �

0d .

A = c13 - j14 -(12 - j16)27 + j16 -(26 + j13)

d ,

AI = V,

(13 - j14)I1 - (12 - j16)I2 = 150l0 � ,(27 + j16)I1 - (26 + j13)I2 = 0.

I2I1

Ix = (I1 - I2).

(12 - j16)(I2 - I1) + (1 + j3)I2 + 39Ix = 0.

(1 + j2)I1 + (12 - j16)(I1 - I2) = 150l0 � .

I2I1

The cofactor matrix B is

(A.85)

The adjoint of A is

(A.86)


adj A = BT= c(-26 - j13) (12 - j16)

(-27 - j16) (13 - j14)d .

B = c(-26 - j13) (-27 - j16)(12 - j16) (13 - j14)

d .

Example A.2Example A.3

In the first three examples, the matrix elements have been numbers—realnumbers in Examples A.1 and A.2, and complex numbers in Example A.3. Itis also possible for the elements to be functions. Example A.4 illustrates theuse of matrix algebra in a circuit problem where the elements in the coeffi-cient matrix are functions.

(A.87) = 60 - j45.

= -(13 - j14)(26 + j13) + (12 - j16)(27 + j16)

det A = 2 (13 - j14) -(12 - j16)(27 + j16) -(26 + j13)

2The inverse of the coefficient matrix is

(A.88)

Equation A.88 can be simplified to

(A.89)

Substituting Eq. A.89 into A.84 gives us

(A.90)


(A.91)I1 = (-26 - j52) = 58.14l -116.57 � A,I2 = (-24 - j58) = 62.77l -122.48 � A.

= c(-26 - j52)(-24 - j58)

d .

cI1

I2d =

1375c(-65 - j130) (96 - j28)(-60 - j145) (94 - j17)

d c150l0 �

0d

=

1375c -65 - j130 96 - j28-60 - j145 94 - j17

d .

A-1=

60 + j455625

c(-26 - j13) (12 - j16)(-27 - j16) (13 - j14)

d

A-1=

c(-26 - j13) (12 - j16)(-27 - j16) (13 - j14)

d(60 - j45)

.

89251_19_AppA 4/4/07 1:08 PM Page 778



A.10 Applications 779

Use the matrix method to derive expressions forthe node voltages and in the circuit in Fig. A.1.

SolutionSumming the currents away from nodes 1 and 2generates the following set of equations:

(A.92)

Letting and collecting the coefficients ofand gives us

(A.93)

Writing Eq. A.93 in matrix notation yields

(A.94)

where


(A.95)

As before, we find the inverse of the coefficientmatrix by first finding the adjoint of A and thedeterminant of A. The cofactors of A are

The cofactor matrix is

(A.96)

and therefore the adjoint of the coefficient matrix is

(A.97)adj A = BT= cG + 2sC sC

sC G + 2sCd .

B = cG + 2sC sC

sC G + 2sCd ,

¢11 = (-1)2[G + 2sC] = G + 2sC,¢12 = (-1)3(-sC) = sC,¢21 = (-1)3(-sC) = sC,¢22 = (-1)4[G + 2sC] = G + 2sC.

V = A-1I.

V = cV1

V2d , and I = c GVg

sCVgd .

A = cG + 2sC -sC

-sC G + 2sCd ,

AV = I ,

-sCV1 + (G + 2sC)V2 = sCVg.

(G + 2sC)V1 - sCV2 = GVg,

V2V1

G = 1>R

V1 - Vg

R+ V1sC + (V1 - V2)sC = 0,

V2

R+ (V2 - V1)sC + (V2 - Vg)sC = 0.

V2V1


det A = 2G + 2sC sC

sC G + 2sC2 = G2

+ 4sCG + 3s2C2.

Example A.4

vg v1

�

�

v2

�

�

1

1sC

1sC

1sCR

R�

�

Figure A.1 � The circuit for Example A.4.

(A.98)

The inverse of the coefficient matrix is

(A.99)


(A.100)

Carrying out the matrix multiplication called for inEq. A.100 gives

cV1

V2d =

cG + 2sC sC

sC G + 2sCd c GVg

sCVgd

(G2+ 4sCG + 3s2C2)

.

A-1=

cG + 2sC sC

sC G + 2sCd

(G2+ 4sCG + 3s2C2)

.

cV1

V2d =

1

(G2+ 4sCG + 3s2C2)

c(G2+ 2sCG + s2C2)Vg

(2sCG + 2s2C2)Vgd .

(A.101)

Now the expressions for and can be writtendirectly from Eq. A.101; thus

(A.102)

and

(A.103)V2 =

2(sCG + s2C2)Vg

(G2+ 4sCG + 3s2C2)

.

V1 =

(G2+ 2sCG + s2C2)Vg

(G2+ 4sCG + 3s2C2)

,

V2V1

89251_19_AppA 4/4/07 1:09 PM Page 779




In our final example, we illustrate how matrix algebra can be used toanalyze the cascade connection of two two-port circuits.

Show by means of matrix algebra how the inputvariables and can be described as functions ofthe output variables and in the cascade con-nection shown in Fig. 18.10.

SolutionWe begin by expressing, in matrix notation, therelationship between the input and output variablesof each two-port circuit. Thus

(A.104)

and

(A.105)

Now the cascade connection imposes the constraints

(A.106)V2œ

= V1œ and I2

œ

= -I1œ .

cV1œ

I1œd = ca11

fl

-a12fl

a21fl

-a22fld cV2

I2d ,

cV1

I1d = ca11

œ

-a12œ

a21œ

-a22œd cV2

œ

I2œd ,

I2V2

I1V1

These constraint relationships are substituted intoEq. A.104. Thus

(A.107)

The relationship between the input variables ( , )and the output variables ( , ) is obtained bysubstituting Eq. A.105 into Eq. A.107. The result is

(A.108)

After multiplying the coefficient matrices, we have

cV1

I1d = ca11

œ a12œ

a21œ a22

œd ca11

fl

-a12fl

a21fl

-a22fld cV2

I2d .

I2V2

I1V1

= ca11œ a12

œ

a21œ a22

œd cV1

œ

I1œd .

cV1

I1d = ca11

œ

-a12œ

a21œ

-a22œd c V1

œ

-I1œd

Example A.5

cV1

I1d = c(a11

œ a11fl

+ a12œ a21

fl ) -(a11œ a12

fl

+ a12œ a22

fl )(a21

œ a11fl

+ a22œ a21

fl ) -(a21œ a12

fl

+ a22œ a22

fl )d cV2

I2d .

(A.109)

Note that Eq.A.109 corresponds to writing Eqs. 18.72and 18.73 in matrix form.

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