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 1 + 2 x 2 + x 4 = 4 2 x 1 + x 2 + 2 x 4 = 12 x 1 + x 2 + x 3 + x 4 = 4 x 2 + x 4 = 8 . Write the system of equations in the matrix equation A X = B where, A = [ 2 2 0 1 2 1 0 2 1 1 1 1 0 1 0 1 ] , X = [ x 1 x 2 x 3 x 4 ] and B = [ 4 12 4 8 ] That is, [ 2 2 0 1 2 1 0 2 1 1 1 1 0 1 0 1 ] [ x 1 x 2 x 3 x 4 ] = [ 4 12 4 8 ] Transform [ 2 2 0 1 2 1 0 2 1 1 1 1 0 1 0 1 ] [ x 1 x 2 x 3 x 4 ] = [ 4 12 4 8 ] as follows. A X = [ 4 12 4 8 ] (1) Pre-multiplied equation (1) on both sides by A − 1 , A − 1 A X = A − 1 [ 4 12 4 8 ] I X = A − 1 [ 4 12 4 8 ] [ ∵ A − 1 A = I ] X = A − 1 [ 4 12 4 8 ] [ ∵ 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 2 0 1 1 0 0 0 2 1 0 2 0 1 0 0 1 1 1 1 0 0 1 0 0 1 0 1 0 0 0 1 ] . Perform the row operations R 2 − R 1 → R 2 and 2 R 3 − R 1 → R 3 and obtain the row reduced equivalent matrix as, [ 2 2 0 1 1 0 0 0 0 − 1 0 1 − 1 1 0 0 0 0 2 1 − 1 0 2 0 0 1 0 1 0 0 0 1 ] Perform the row operation R 4 + R 2 → R 4 and obtain the equivalent row reduced matrix as, [ 2 2 0 1 1 0 0 0 0 − 1 0 1 − 1 1 0 0 0 0 2 1 − 1 0 2 0 0 0 0 2 − 1 1 0 1 ] Perform the row operation ( 1 2 ) R 4 → R 4 and obtain the equivalent row reduced matrix as, [ 2 2 0 1 1 0 0 0 0 − 1 0 1 − 1 1 0 0 0 0 2 1 − 1 0 2 0 0 0 0 1 − 1 2 1 2 0 1 2 ] Again perform row operations R 3 − R 4 → R 3 , R 2 − R 4 → R 2 and R 1 − R 4 → R 1 and obtain the equivalent row reduced matrix as, [ 2 2 0 0 3 2 − 1 2 0 − 1 2 0 − 1 0 0 − 1 2 1

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ISBN: 9780135823002

Author: HARSHBARGER

Publisher: PEARSON

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Chapter 7, Problem 28RE

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Chapter 7, Problem 27RE

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

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