If we let C = adj A • A and use the technique illustrated in Section A.7, we find the elements of C to be Therefore C = = cu C\2 ^13 ^21 c 22 c 23 C31 C32 ^33 ~-8 0 0 Jet A = 9 - = 18 = 27 21 +4=- 8, - 14 - 4 = 0, - 7 - 20 = 0, = -16 + 24 - 8 = 0, = -32 + 16 + 8 = -8, = -48 + 8 + 40 = 0, = 5 - = 10 = 15 0 -8 0 •u. 9 + 4 = 0, -6-4 = 0, - 3 - 20 = -8. 0" 0 -8_ = -8 "l 0 0 1 _0 0 0 0 1 A square matrix A has an inverse, denoted as A ', if A -1 A = AA _1 = U. (A.50) Equation A.50 tells us that a matrix either premultiplied or postmultiplied by its inverse generates the identity matrix U. For the inverse matrix to exist, it is necessary that the determinant of A not equal zero. Only square matrices have inverses, and the inverse is also square. A formula for finding the inverse of a matrix is A- ! = detA (A.51) The formula in Eq. A.51 becomes very cumbersome if A is of an order larger than 3 by 3. 2 Today the digital computer eliminates the drudgery of having to find the inverse of a matrix in numerical applications of matrix algebra. It follows from Eq. A.51 that the inverse of the matrix A in the previ- ous example is -1 _ A -1 = -1/8 9 16 5 -7 8 -3 -4 8 ~4_ -1.125 0.875 2 -1 -0.625 0.375 -1A _ 1 i You should verify that A l A = AA = U. 0.5 -1 0.5 2 You can learn alternative methods for finding the inverse in any introductory text on matrix theory. See, for example, Franz E. Hohn, Elementary Matrix Algebra (New York: Macmillan, 1973). A.9 Partitioned Matrices It is often convenient in matrix manipulations to partition a given matrix into submatrices. The original algebraic operations are then carried out in terms of the submatrices. In partitioning a matrix, the placement of the partitions is completely arbitrary, with the one restriction that a partition must dissect the entire matrix. In selecting the partitions, it is also neces- sary to make sure the submatrices are conformable to the mathematical operations in which they are involved. For example, consider using submatrices to find the product C = AB, where A = 1 5 1 0 0 2 4 0 1 2 3 3 2 -1 1 4 2 -3 0 -2 5 1 1 1 0 and B 2 0 -1 3 0 Assume that we decide to partition B into two submatrices, B n and B 2 j; thus B B21 Now since B has been partitioned into a two-row column matrix, A must be partitioned into at least a two-column matrix; otherwise the multiplication cannot be performed. The location of the vertical partitions of the A matrix will depend on the definitions of B n and B 2 i. For example, if B, and B 21 then An must contain three columns, and A 12 must contain two columns. Thus the partitioning shown in Eq. A.52 would be acceptable for execut- ing the product AB: C = 1 2 5 4 -1 0 0 1 0 2 3 3 2 -1 1 4 5~ 2 1 3 1 0 1 2 0_ 2~ 0 -1 3 0_ (A.52) If, on the other hand, we partition the B matrix so that B n - 2 .0. and B?i = -1 3 0 then An must contain two columns, and A 12 must contain three columns. In this case the partitioning shown in Eq. A.53 would be acceptable in exe- cuting the product C = AB: C = 1 5 1 0 0 2 4 0 1 2 3 3 2 -1 1 4 2 -3 0 -2 5~ 1 1 1 0_ 2 0 -1 3 0 (A.53) For purposes of discussion, we will focus on the partitioning given in Eq. A.52 and leave you to verify that the partitioning in Eq. A.53 leads to the same result. From Eq. A.52 we can write C = [A n A 12 ] B„ = A n B n + A 12 B 21- (A. 54) It follows from Eqs. A.52 and A.54 that A,,B ll*»1l 1 2 5 4 -1 0 0 1 0 2 3~ 3 2 1 1 2~ 0 _-!_ = ~-l" 7 -4 1 _-!_ A12B21 - 4 5~ 2 1 -3 1 0 1 _~2 0_ "3" .0. = ~ 12" 6 -9 0 _-6_ and 11 13 -13 1 -7 The A matrix could also be partitioned horizontally once the vertical partitioning is made consistent with the multiplication operation. In this simple problem, the horizontal partitions can be made at the discretion of the analyst. Therefore C could also be evaluated using the partitioning shown in Eq.A.55: 1 5 1 0 0 ? 4 0 1 2 3 3 2 -1 1 5 1 1 1 0_ 2~ 0 -1 3 0_ (A.55) From Eq. A.55 it follows that An A 21 A 12 A,? B,i B„ LC 21 (A.56) where You should verify that Cj! - A u B n + A I2 B 2 |, C 2 i = A 21 B 11 + A22B21. C„ = 1 2 3 5 4 3 + 4 5 2 1 -1 7 + 12 6 = 11 13 c„ = -1 0 0 1 0 2 2~ 1 1_ 2~ 0 _-!_ + "-3 0 _-2 r 1 0_ "3* .0. -4 1 -1 + -9 0 -6 = -13 1 -7 and C = 11 13 -13 1 -7 We note in passing that the partitioning in Eqs. A.52 and A.55 is conformable with respect to addition. 720 The Solution of Linear Simultaneous Equations A.10 Applications The following examples demonstrate some applications of matrix algebra in circuit analysis. Example A.l Use the matrix method to solve for the node volt- ages V\ and v 2 in Eqs. 4.5 and 4.6. Solution The first step is to rewrite Eqs. 4.5 and 4.6 in matrix notation. Collecting the coefficients of «, and v 2 and at the same time shifting the constant terms to the right-hand side of the equations gives us 1.7«! - 0.5¾ = 10, (A.57) -0.5«] + 0.6« 2 = 2. It follows that in matrix notation, Eq. A.57 becomes 1.7 -0.5 •0.5 0.6 10 2 or where AV = I, (A.58) (A.59) A A. — V = I = 1.7 ().5 V '10" . 2. • • -0.5" 0.6. To find the elements of the V matrix, we pre- multiply both sides of Eq. A.59 by the inverse of A; thus or A _1 AV = A _1 I. Equation A.60 reduces to UV = A "T, V = A "I. (A.60) (A.61) (A.62) It follows from Eq. A.62 that the solutions for V\ and v 2 are obtained by solving for the matrix product A -1 1. To find the inverse of A, we first find the cofactors of A. Thus A H = (-1) 2 (0.6) = 0.6, A12 = (-l) 3 (-0.5) = 0.5, A 2 i = ("l) 3 (-0.5) = 0.5, A22 = (-l) 4 (l-7) = 1.7. The matrix of cofactors is (A.63) B 0.6 0.5 and the adjoint of A is adj A = B 7 = The determinant of A is 0.5 1.7 0.6 L0.5 0.5 1.7 (A.64) (A.65) detA 1.7 -0.5 -0.5 0.6 (1.7)(0.6) - (0.25) = 0.77. (A.66) From Eqs. A.65 and A.66, we can write the inverse of the coefficient matrix, that is, (A.67) A" 1 1 0.77 0.6 .0.5 0.5 1.7 Now the product A -1 1 is found: A-I = ^ 77 0.6 0.5" .0.5 1.7. loor 7" 77 U.4 _ " "10 . 2 " 9.09" .10.91. It follows directly that V v 2 - ' 9.09' .1 3.91 ' (A.68) (A.69) or vj = 9.09 V and v 2 = 10.91 V. A.10 Applications 721 Example A.2 Use the matrix method to find the three mesh cur- rents in the circuit in Fig. 4.24. Solution The mesh-current equations that describe the cir- cuit in Fig. 4.24 are given in Eq. 4.34. The constraint equation imposed by the current-controlled voltage source is given in Eq. 4.35. When Eq. 4.35 is substi- tuted into Eq. 4.34, the following set of equations evolves: 25/,- - 5/ 2 - 20/ 3 = 50, -5/, - 4z 2 + 9/ 3 = 0. In matrix notation, Eqs. A.70 reduce to AI = V, where A = (A.70) (A.71) and 25 -5 -5 10 -5 -4 '*] h h_ 1 V = r 50" 0 0 -20 -4 9 . It follows from Eq. A.71 that the solution for I is I = A 1 V. (A.72) We find the inverse of A by using the relationship A - ' = _ adjA detA (A.73) To find the adjoint of A, we first calculate the cofac- tors of A. Thus An = (-1) 2 (90 - 16) = 74, A12 = (-l) 3 (-45 - 20) = 65, A13 = (-1)^(20 + 50) = 70, A21 = (-l) 3 (-45 - 80) = 125, A22 = A 23 = A 3 1 = I A 32 = < A33 = ( ;-l) 4 (225 - 100) = 125, ;-l) 5 (-100 - 25) = 125, ;-l) 4 (20 + 200) = 220, -1) 5 (-100 - 100) = 200, -1) 6 (25() - 25) = 225. The cofactor matrix is B 74 65 70 125 125 125 220 200 225 (A.74) from which we can write the adjoint of A: adj A = B r 74 65 _70 The determinant of A is detA = 25 -5 -20 -5 10 -4 -5 -4 9 125 125 125 220 200 225 (A.75) = 25(90 -16) + 5(-45 - 80) - 5(20 + 200) = 125. It follows from Eq. A.73 that -1 _ 1 125 74 125 220 65 125 200 70 125 225 (A.76) The solution for I is 1 125 74 125 65 125 70 125 220 200 225 50 0 0 = 29.60 26.00 28.00 . (A.77) The mesh currents follow directly from Eq. A.77. Thus (A.78) or /j = 29.6 A, i 2 = 26 A, and / 3 = 28 A. Example A.3 illustrates the application of the matrix method when the elements of the matrix are complex numbers. h h _*3_ = "29.6" 26.0 _28.0_ 722 The Solution of Linear Simultaneous Equations Example A.3 Use the matrix method to find the phasor mesh cur- rents I, and I 2 in the circuit in Fig. 9.37. Solution Summing the voltages around mesh 1 generates the equation (1 + /2)1, + (12 - /16)(1, - I 2 ) = 150/0". (A.79) Summing the voltages around mesh 2 produces the equation (12 - /16)(I 2 - Ii) + (1+ /3)I 2 + 39I V = 0.(A.80) The current controlling the dependent voltage source is I, = (Ii - I 2 )- (A.81) After substituting Eq. A.81 into Eq. A.80, the equations are put into a matrix format by first collect- ing, in each equation, the coefficients of I, and I 2 : thus (13 - /14)1, - (12 - /16)I 2 = 150/0°, (27 + /16)1, - (26 + /13)¾ = 0. Now, using matrix notation, Eq. A.82 is written (A.82) where A = I AI = V, 13 - /14 -(12 - /16) [27 + /16 -(26 + /13) (A.83) and V 150/0 0 It follows from Eq. A.83 that I = A V. (A.84) The inverse of the coefficient matrix A is found using Eq. A.73. In this case, the cofactors of A are ^ n = (-l) 2 (-26-/13) = -26-/13, A 12 = (-l) 3 (27 + /16) = -27-/16, A 2 i = (-1) 3 (-12 + /16) = 12 - /16, A 22 = (-1) 4 (13 - /14) = 13 - /14. The cofactor matrix B is B = The adjoint of A is adj A = B 7 = The determinant of A is detA = (-26 - /13) (-27 - /16) (12-/16) (13-/14) (-26 - /13) (12 - /16) L(-27-/16) (13-/14)J (A.85) (A.86) (13 - /14) (27 + /16) (12 - /16) (26 + /13) = -(13 - /14)(26 + /13) + (12 - /16)(27 + /16) = 60 - /45. (A.87) The inverse of the coefficient matrix is A~ ! = (-26-/13) (12-/16) L(-27 -/16) (13 -/14)J (60 - /45) Equation A.88 can be simplified to (A.88) 60 + /45 5625 (-26 - /13) (12 - /16) (-27 - /16) (13 - /14) 1 375 -65 - /130 96 - /28 -60 - /145 94 - /17 (A.89) Substituting Eq. A.89 into A.84 gives us 1 375 (-65 - /130) (96 - /28) (-60 - /145) (94 - /17) 150/0 C 0 (-26 - /52) (-24 - /58) It follows from Eq. A.90 that I, = (-26 - /52) = 58.14/-116.57° A, h = (-24 - /58) = 62.77/-122.48° A. (A.90) (A.91) In the first three examples, the matrix elements have been numbers—real numbers in Examples A.l and A.2, and complex numbers in Example A.3. It is also possible for the elements to be functions. Example A.4 illustrates the use of matrix algebra in a circuit problem where the elements in the coeffi- cient matrix are functions. A.10 Applications 723 Example A.4 Use the matrix method to derive expressions for the node voltages V x and V 2 in the circuit in Fig. A. 1. Solution Summing the currents away from nodes 1 and 2 generates the following set of equations: R + V x sC + (K, - V 2 )sC = 0, f + (½ ~ VfisC + (V 2 - V g )sC = 0. (A.92) Letting G = \/R and collecting the coefficients of V] and V 2 gives us (G + 2sC)Vi - sCV 2 = GVp -sCVi + (G + 2sC)V 2 = sCV R . Writing Eq. A.93 in matrix notation yields AV = I, where (A.93) (A.94) G + 2sC -sC v 2 . , and —5 C G + 2sC_ I - _sCV i V = It follows from Eq. A.94 that V - A'l. (A.95) As before, we find the inverse of the coefficient matrix by first finding the adjoint of A and the determinant of A. The cofactors of A are A n = (-1) 2 [G + 2sC] = G + 2sC, A 12 = (-l)\sC) = sC, A 2 i = {-l)\sC) - sC, A22 = (~1) 4 [G + 2sC] = G + 2sC. The cofactor matrix is G + 2sC B - sC sC G + 2sC (A.96) and therefore the adjoint of the coefficient matrix is I G + 2sC sC sC G + 2sC adj A = B 7 (A.97) Figure A.l • The circuit for Example A.4. The determinant of A is G + 2sC sC detA = „ _ „ _ sC G + 2sC = G 2 + 4sCG + 3.v 2 C 2 . (A.98) The inverse of the coefficient matrix is A~ l = G + 2sC sC sC G + 2sC (G 2 + AsCG + 3.v 2 C 2 ) (A.99) It follows from Eq. A.95 that G + 2sC sC sC G + 2sC sCV a {G z + AsCG + 3s l C l ) (A.100) Carrying out the matrix multiplication called for in Eq.A.100 gives V 2 J (G 2 + 4.vCG + 3.v 2 C 2 ) (G 2 + 2sCG + s 2 C 2 )V g (2sCG + 2s*C 2 )V n (A.101) Now the expressions for V\ and V 2 can be written directly from Eq. A. 101; thus (G 2 + 2sCG + s 2 C 2 )V K V\ = ,w> . . ^ . „ ^ » ( A - 102 ) (G 2 + 4sCG + 3s 2 C 2 ) ' and 2^ v> = 2(sCG + s l C z )V<, (G 2 + 4sCG + 3s 2 C 2 ) (A.103) 724 The Solution of Linear Simultaneous Equations In our final example, we illustrate how matrix algebra can be used to analyze the cascade connection of two two-port circuits. Example A.5 Show by means of matrix algebra how the input variables V x and Iy can be described as functions of the output variables V 2 and I 2 in the cascade con- nection shown in Fig. 18.10. Solution We begin by expressing, in matrix notation, the relationship between the input and output variables of each two-port circuit. Thus (A.104) and (A.105) Vi /J v\ /', «il .«21 r«u ki -«12 ~«22- -«12] ~«22 J v 2 L/2 \v 2 lh Now the cascade connection imposes the constraints V' 2 = V\ and 1' 2 = -J\. (A.106) These constraint relationships are substituted into Eq. A.104. Thus «h «21 «ii «21 -«12 -«22 «12 «22- -I'u L/'I (A.107) The relationship between the input variables (V h /j) and the output variables (V 2 , J 2 ) is obtained by substituting Eq. A.105 into Eq. A.107. The result is «11 «23 «12 «22 «11 L« 2 'i -«12 "«22J v 2 L/2 (A.108) After multiplying the coefficient matrices, we have V 2 Vi LA («il«ll + «12«2l) («21«11 + «22«2l) - («11«']2 + «12«22) -(«21«12 + «22«22)J L/2 (A.109) Note that Eq.A.109 corresponds to writing Eqs. 18.72 and 18.73 in matrix form. Appendix Q Complex Numbers Complex numbers were invented to permit the extraction of the square roots of negative numbers. Complex numbers simplify the solution of problems that would otherwise be very difficult. The equation x 2 + 8x + 41 = 0, for example, has no solution in a number system that excludes complex numbers. These numbers, and the ability to manipulate them algebraically, are extremely useful in circuit analysis. B.l Notation There are two ways to designate a complex number: with the cartesian, or rectangular, form or with the polar, or trigonometric, form. In the rectangular form, a complex number is written in terms of its real and imaginary components; hence n = a + jb, (B.l) where a is the real component, b is the imaginary component, and ; is by definition V-l. 1 In the polar form, a complex number is written in terms of its magni- tude (or modulus) and angle (or argument); hence n = ce je (B.2) where c is the magnitude, 6 is the angle, e is the base of the natural loga- rithm, and, as before, j = V-T. In the literature, the symbol /6° is fre- quently used in place of e jB \ that is, the polar form is written n = c/6°. (B.3) Although Eq. B.3 is more convenient in printing text material, Eq. B.2 is of primary importance in mathematical operations because the rules for manipulating an exponential quantity are well known. For example, because ( 7 7> = y *»,then(e^)" = e jn6 \ because v" v = l/y\ then e' 10 = l/e^;and so forth. Because there are two ways of expressing the same complex number, we need to relate one form to the other. The transition from the polar to the rectangular form makes use of Euler's identity: e. ±ie = cos 6 ± /sin 6. (B.4) 1 You may be more familiar with the notation i = y/^. In electrical engineering, / is used as the svmbol for current, and hence in electrical engineering literature,/ is used to denote . 1 You may be more familiar with the notation i = y/^. In electrical engineering, / is used as the svmbol for current, and hence in electrical engineering literature,/ is used to denote