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which indicates that the work is correct. 237. a At 2 " # ; answer: 4 5 9 2 !
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which is the same answer as above. 239. Since A is a (3 2) and B a (2 3 matrix, the product in the order AB does exist and will be a (3 3) matrix. From the de nition of matrix multiplication laid down in section 11.2, we have that 3 3 2 2 2 4 6 8 0 4 6 36 ! 0 3 7 3 7 6 6 AB 4 0 65 4 0 12 0 6 0 54 5 2 1 9 7 3 21 6 0 3 21 27 thus 2 2 30 3
6 AB 4 12 15
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and hence 2 6 AB t 4 then, next, 2 4 30 # 12 6 54
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thus proving that (AB t Bt At for the given matrix. 240. g11 has the dimension of admittance (mhos). g22 has the dimension of impedance (ohms). g12 and g21 are dimensionless ratios. a12 has dimensions of impedance (ohms). a21 has dimensions of admittance (mhos). a11 and a22 are dimensionless ratios. 241. y22 I2 V2 V1 0 g21 V2 V1 I2 0
242. Let us begin by writing eqs. (512) and (513) with the minus sign inside the a matrix, so that eq. (514) takes the equivalent form ! ! ! a11 a12 V2 V1 I1 a21 a22 I2 Now multiply both sides of the above equation by the inverse of the a matrix, which gives ! ! ! ! ! a11 a12 1 a11 a12 V2 a11 a12 1 V1 A a21 a22 I1 a21 a22 a21 a22 I2 Note that the right-hand side of the equation is of the form ! ! ! V2 V2 V2 1 a a I I2 I2 I2 by eqs. (489), (486), where I is the unit matrix of section 11.4. Therefore the preceding equation (eq. A) becomes ! ! ! a11 a12 1 V1 V2 a21 a22 I1 I2 which is a correct way of writing the answer.
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A more detailed answer, can, however, be written by taking the inverse of the 2 2 matrix as indicated. Following the procedure of problem 232, you should nd that the inverse of the above 2 2 matrix is ! ! a22 a12 a11 a12 1 1 a12 a21 a11 a22 a21 a11 a21 a22 and therefore the preferable nal answer, free of inverse notation, is ! ! ! V2 a22 a12 V1 1 a12 a21 a11 a22 a21 a11 I2 I1 243. Multiplication of both sides of eq. (482) by the constant c gives 3 2 ca11 ca12 ca1n 7 6 ca 6 21 ca22 ca2n 7 cA 6 . . 7 . 6 . . 7 . 4 . . . 5 can1 can2 cann We now wish to take the inverse of both sides of the above equation. The rst step in doing this is to nd the value of the determinant formed from the elements of the matrix. Note that, when regarded as a determinant, c factors from every row of the determinant, and thus we have determinant of cA cn D where D is the determinant value of the original square matrix A. Next, note that cn 1 will factor from every cofactor in A0 , eq. (483), and therefore from the transpose of A0 . It thus follows that 2 3 transpose n 1 c 6 7 1 1 of cA 1 n 4 5 A c c D matrix A0 244. As mentioned following eq. (504), z y 1 , so our problem is basically to nd the inverse of the 2 2 admittance matrix. Following the procedure of problem 232, you should nd that ! ! ! y11 y12 1 y22 y12 y22 =D y12 =D 1 y11 y22 y12 y21 y21 y21 =D y11 =D y21 y22 y11 where D y11 y22 y12 y21 . The z-matrix is equal to the last matrix to the right above, and thus, from the de nition of equal matrices in section 11.1, the answers to the problem are z11 y22 =D; z12 y12 =D; z21 y21 =D; z22 y11 =D
245. First, remembering that j 2 1, you should nd that D 20 4 j 10 6 . Then 1=D 2:941 4 j 103 approximately, and using this value with the other given values, in the equations found in problem 244, gives the following values in ohms: z11 17:65 j70:59 z21 735:3 j941:1 z12 20:59 j17:65 z22 191:2 j235:3
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