Explanation Given: The system of equations are, 2 x − 3 y + 4 z = 17 x + 2 y − 4 z = − 7 3 x − y + 2 z = 14 Approach: For finding the inverse of the matrix A of order n × n , the following steps are to be used 1. Adjoin the n × n identity matrix I to obtain the augmented matrix as follows, [ A | I ] . 2. Use a sequence of row operations to reduces [ A | I ] to the form [ I | B ] if possible. Then the matrix B is the inverse of A . If A X = B is a linear system of n equations in n unknowns and if A − 1 exists, then the unique solution of the system is given as, X = A − 1 B ... ( 1 ) Calculation: The given system of equations can be written in the form A X = C as, A = [ 2 − 3 4 1 2 − 4 3 − 1 2 ] , X = [ x y z ] , and C = [ 17 − 7 14 ] . The augmented matrix with a 3 × 3 identity matrix is [ A | I ] = [ 2 − 3 4 ∣ 1 0 0 1 2 − 4 ∣ 0 1 0 3 − 1 2 ∣ 0 0 1 ] Reduce the matrix to the identity matrix, using row operations as follows, Swap the two rows R 1 and R 2 . [ 1 2 − 4 ∣ 0 1 0 2 − 3 4 ∣ 1 0 0 3 − 1 2 ∣ 0 0 1 ] Eliminate the leading coefficients in row R 2 and R 3 by performing R 2 → R 2 − 2 R 1 and R 3 → R 3 − 3 R 1 . [ 1 2 − 4 ∣ 0 1 0 0 − 7 12 ∣ 1 − 2 0 0 − 7 14 ∣ 0 − 3 1 ] Again, eliminate the leading coefficients in row R 3 by performing R 3 → R 3 − R 2 . [ 1 2 − 4 ∣ 0 1 0 0 − 7 12 ∣ 1 − 2 0 0 0 2 ∣ − 1 − 1 1 ] Multiply matrix row by constant as R 3 → 1 2 R 3 . [ 1 2 − 4 ∣ 0 1 0 0 − 7 12 ∣ 1 − 2 0 0 0 1 ∣ − 1 2 − 1 2 1 2 ] Perform row operations R 2 → R 2 − 12 R 3 and R 1 = R 1 + 4 R 3 . [ 1 2 0 ∣ − 2 − 1 2 0 − 7 0 ∣ 7 4 − 6 0 0 1 ∣ − 1 2 − 1 2 1 2 ] Multiply matrix row by constant as R 2 → − 1 7 R 2

Finite Mathematics for the Managerial, Life, and Social Sciences

12th Edition

ISBN: 9781337405782

Author: Soo T. Tan

Publisher: Cengage Learning

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Chapter 2.CRE, Problem 40CRE

To determine

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The value of A−1 and the solution of the system of equations.

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Chapter 2.CRE, Problem 39CRE

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