Explanation Procedure used: 1. Form an augmented matrix [ A | I ] where a matrix A is placed along with a same dimension identity matrix I separated by a line. 2. Apply row operations on [ A | I ] in order to obtain another augmented matrix [ I | B ] where I , is again an identity matrix and B is an inverse of A. 3. If the reduction process yields a row of zeros in the left half of the augmented matrix, then the inverse of A does not exist. Calculation: The given system of equation is { 2 x + 3 y + z = 20 3 x + 4 y − z = 40 2 x + 5 y + z = 60 . Write the system of equations in the matrix equation form A X = B where, A = [ 2 3 1 3 4 − 1 2 5 1 ] , X = [ x y z ] and B = [ 20 40 60 ] That is, [ 2 3 1 3 4 − 1 2 5 1 ] [ x y z ] = [ 20 40 60 ] Transform [ 2 3 1 3 4 − 1 2 5 1 ] [ x y z ] = [ 20 40 60 ] as follows. A X = [ 20 40 60 ] (1) Pre-multiplied equation (1) on both side by A − 1 , A − 1 A X = A − 1 [ 20 40 60 ] I X = A − 1 [ 20 40 60 ] [ ∵ A − 1 A = I ] X = A − 1 [ 20 40 60 ] [ ∵ I X = X ] Now, use the method of finding inverse of the matrix and obtain A − 1 as follows. The augment matrix [ A | I ] is [ A | I ] = [ 2 3 1 1 0 0 3 4 − 1 0 1 0 2 5 1 0 0 1 ] . Perform the row operations 3 R 3 − 2 R 2 → R 3 and obtain the row reduced equivalent matrix as, [ 2 3 1 1 0 0 3 4 − 1 0 1 0 0 7 5 0 − 2 3 ] Perform the row operations 2 R 2 − 3 R 1 → R 2 and obtain the equivalent row reduced matrix as, [ 2 3 1 1 0 0 0 − 1 − 5 − 3 2 0 0 7 5 0 − 2 3 ] Again perform row operation R 3 + 7 R 2 → R 2 and obtain the equivalent row reduced matrix as, [ 2 3 1 1 0 0 0 − 1 − 5 − 3 2 0 0 0 − 30 − 21 12 3 ] Perform the row operation ( − 1 30 ) R 3 → R 3 and obtain the equivalent row reduced matrix as, [ 2 3 1 1 0 0 0 − 1 − 5 − 3 2 0 0 0 1 7 10 − 4 10 − 1 10 ] Again perform row operation R 2 + 5 R 3 → R 2 and obtain the equivalent row reduced matrix as, [ 2 3 1 1 0 0 0 − 1 0 5 10 0

COLLEGE ALGEBRA IN CONTEXT W/ INT. REVIE

5th Edition

ISBN: 9780136482758

Author: HARSHBARGER

Publisher: PEARSON C

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Chapter 7, Problem 26RE

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To solve: The system of equations by using the inverse of the coefficient matrix.

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COLLEGE ALGEBRA IN CONTEXT W/ INT. REVIE

Ch. 7.1 - In Exercises 14, solve the systems of equations....Ch. 7.1 - Prob. 2E Ch. 7.1 - Prob. 3E Ch. 7.1 - Prob. 4E Ch. 7.1 - Prob. 5E Ch. 7.1 - Prob. 6E Ch. 7.1 - Prob. 7E Ch. 7.1 - Prob. 8E Ch. 7.1 - Prob. 9E Ch. 7.1 - Prob. 10E

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The system of equations by using the inverse of the coefficient matrix.

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