Explanation Given Information: The system of equations, x + 2 z = 6 , − x + 2 y + 3 z = − 5 and x − y = 6 Formula used: A linear system of equations a 1 x + b 1 y + c 1 z = d 1 , a 2 x + b 2 y + c 2 z = d 2 , a 3 x + b 3 y + c 3 z = d 3 can be represented by the matrix form of the system as: [ a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 ] [ x y z ] = [ d 1 d 2 d 3 ] If the equation A X = B has a unique solution, then the solution of the equations is given by: X = A − 1 B Calculation: Consider the system of equations: x + 2 z = 6 − x + 2 y + 3 z = − 5 x − y = 6 The above system of equations can be written in the matrix form of system as: [ 1 0 2 − 1 2 3 1 − 1 0 ] [ x y z ] = [ 6 − 5 6 ] A X B The solution of the system is defined as X = A − 1 B . So, first find the inverse matrix A − 1 . Consider the matrix: A = [ 1 0 2 − 1 2 3 1 − 1 0 ] From the above matrix, form the augmented matrix of the form [ A | I 3 ] . So, [ A | I 3 ] = [ 1 0 2 − 1 2 3 1 − 1 0 | 1 0 0 0 1 0 0 0 1 ] Now, perform row operations in order to obtain a matrix of the form [ I 3 | B ] . This means that to the left of the vertical dividing bar, 1 s are required from upper left to the lower right down the diagonal and 0 s for all other elements. Replace row 2 by the operation 1 R 1 + R 2 and row 3 by the operation − 1 R 1 + R 3 . So, the matrix becomes: [ 1 0 2 1 0 0 − 1 2 3 0 1 0 1 − 1 0 0 0 1 ] → 1 R 1 + R 2 → R 2 → − 1 R 1 + R 3 → R 3 [ 1 0 2 1 0 0 1 ( 1 ) + ( − 1 ) 1 ( 0 ) + 2 1 ( 2 ) + 3 1 ( 1 ) + 0 1 ( 0 ) + 1 1 ( 0 ) + 0 − 1 ( 1 ) + 1 − 1 ( 0 ) + ( − 1 ) − 1 ( 2 ) + 0 − 1 ( 1 ) + 0 − 1 ( 0 ) + 0 − 1 ( 0 ) + 1 ] = [ 1 0 2 1 0 0 0 2 5 1 1 0 0 − 1 − 2 − 1 0 1 ] Replace row 2 by the operation R 3 + R 2 . So, the matrix changes to: [ 1 0 2 1 0 0 0 2 5 1 1 0 0 − 1 − 2 − 1 0 1 ] → R 3 + R 2 → R 2 [ 1 0 2 1 0 0 0 + 0 2 + ( − 1 ) 5 + ( − 2 ) 1 + ( − 1 ) 1 + 0 0 + 1 0 − 1 − 2 − 1 0 1 ] = [ 1 0 2 1 0 0 0 1 3 0 1 1 0 − 1 − 2 − 1 0 1 ] Replace row 3 by the operation 1 R 2 + R 3

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ISBN: 9780134469140

Author: Robert F. Blitzer

Publisher: PEARSON

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Chapter 8.4, Problem 5CP

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To calculate: The solution of the system of equations: x+2z=6,−x+2y+3z=−5 and x−y=6 with the use of the inverse of the coefficient matrix A−1.

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Chapter 8.4, Problem 4CP

Chapter 8.4, Problem 6CP

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Chapter 8 Solutions

Precalculus (6th Edition)

Ch. 8.1 - Check Point 1 Write the augmented matrix for the...Ch. 8.1 - Check Point 2 Use the matrix ...Ch. 8.1 - Check Point 3 Use matrices to solve the system:...Ch. 8.1 - Check Point 4 Use matrices to solve the system:...Ch. 8.1 - Check Point 5 Solve the system in Check Point. 3...Ch. 8.1 - Prob. 1CVC Ch. 8.1 - Fill in each blank so that the resulting statement...Ch. 8.1 - Prob. 3CVC Ch. 8.1 - Prob. 4CVC Ch. 8.1 - Fill in each blank so that the resulting statement...

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The solution of the system of equations: x + 2 z = 6 , − x + 2 y + 3 z = − 5 and x − y = 6 with the use of the inverse of the coefficient matrix A − 1 .

Solution Summary: The author explains how to calculate the solution of the system of equations: x+2z = 6,-x +2y+3z=-5andx-y=6 with the inverse

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To calculate: The solution of the system of equations: x+2z=6,−x+2y+3z=−5 and x−y=6 with the use of the inverse of the coefficient matrix A−1.

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