Stitz-Zeager_College_Algebra_e-book

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online