Stitz-Zeager_College_Algebra_e-book

A x2 2x2 20x 68 9 7x 8 x3 4x2 4x 16 x 4

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Unformatted text preview: 2i1 − i2 − i4 −i1 + 3i2 − i3 − i4 −i2 + 2i3 − i4 −i1 − i2 − i3 + 4i4 = = = = VB1 0 −VB2 0 This corresponds to the matrix equation AX = B where 2 −1 0 −1 i1 −1 i2 3 −1 −1 X= A= 0 −1 i3 2 −1 −1 −1 −1 4 i4 VB1 0 B= −VB2 0 When we input the matrix A into the calculator, we find from which we have A−1 1.625 1.25 = 1.125 1 1.25 1.125 1.5 1.25 1.25 1.625 1 1 1 1 . 1 1 To solve the four systems given to us, we find X = A−1 B where the value of B is determined by the given values of VB1 and VB2 1 (a) 10 0 B= −5 , 1 (b) 0 10 0 B= 0 , 1 (c) 0 0 0 B= −10 , 1 (d) 0 10 0 B= 10 0 (a) For VB1 = 10V and VB2 = 5V , the calculator gives i1 = 10.625 mA, i2 = 6.25 mA, i3 = 3.125 mA, and i4 = 5 mA. 502 Systems of Equations and Matrices (b) By keeping VB1 = 10V and setting VB2 = 0V , we are essentially removing the effect of the second battery. We get i1 = 16.25 mA, i2 = 12.5 mA, i3 = 11.25 mA, and i4 = 10 mA. Solution to 1(a) Solution to 1(b) (c) Part (c) is a symmetric situation to part (b) in...
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