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Stitz-Zeager_College_Algebra_e-book

To nd the associated x values we substitute each

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Unformatted text preview: = 0 = −5 = 0 The coefficient matrix for this system is 4 × 6 (4 equations with 6 unknowns) and is therefore not invertible. We do know, however, this system is consistent, since setting all the resistance values equal to 1 corresponds to our situation in problem 1a. This means we have an 8.4 Systems of Linear Equations: Matrix Inverses 503 underdetermined consistent system which is necessarily dependent. To solve this system, we encode it into an augmented matrix 5.25 0 4.375 0 0 1.25 −4.375 3.125 0 0 0 −3.125 −5.625 −1.25 0 0 0 0 10 0 0 0 0 −1.875 −5 0 5 1.875 and use the calculator to write in reduced row echelon form 1 0 0 0 0 0.7 1 −3.5 0 0 0 0 0 0 1 0 0 0 1.7 0 −1.5 −4 0 0.6 1.6 1 0 1 Decoding this system from the matrix, we get R1 + 0.7R3 R2 − 3.5R3 − 1.5R6 R4 + 0.6R6 R5 = 1.7 = −4 = 1.6 = 1 We have can solve for R1 , R2 , R4 and R5 leaving R3 and R6 as free variables. Labeling R3 = s and R6 = t, we have R1 = −0.7s + 1.7, R2 = 3.5s + 1.5t − 4, R4 = −0.6t + 1.6 and R5 = 1. Since resistance values are...
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