Explanation Given information: The given equations are, 2 p + 3 r + s = 4 p − 2 r − 3 s + 4 t = − 1 3 p + r + s − 5 t = 3 − p + 2 r + s + 3 t = 2 Calculation: Write the given equations in general matrix form [ A ] . [ A ] = [ 2 3 1 0 1 − 2 − 3 4 3 1 1 − 5 − 1 2 1 3 ] Find the determinant | A | of matrix A . | A | = | 2 3 1 0 1 − 2 − 3 4 3 1 1 − 5 − 1 2 1 3 | Expand matrix A by first row. | A | = 2 | − 2 − 3 4 1 1 − 5 2 1 3 | − 3 | 1 − 3 4 3 1 − 5 − 1 1 3 | + 1 | 1 − 2 4 3 1 − 5 − 1 2 3 | − 0 = { 2 [ ( − 2 ) ( 3 + 5 ) − 1 ( − 9 − 4 ) + 2 ( 15 − 4 ) ] − 3 [ 1 ( 3 + 5 ) − 3 ( − 9 − 4 ) + ( − 1 ) ( 15 − 4 ) ] + 1 [ 1 ( 3 + 10 ) − 3 ( − 6 − 8 ) + ( − 1 ) ( 10 − 4 ) ] } = 2 ( − 16 + 13 + 22 ) − 3 ( 8 + 39 − 11 ) + 1 ( 13 + 42 − 6 ) = 2 ( 19 ) − 3 ( 36 ) + 1 ( 49 ) = − 21 Replace the coefficients of p (elements of column 1) with constant [ X ] to form matrix [ X A ] . [ X A ] = [ 4 3 1 0 − 1 − 2 − 3 4 3 1 1 − 5 2 2 1 3 ] Find the determinant | X A | of matrix X A . | X A | = | 4 3 1 0 − 1 − 2 − 3 4 3 1 1 − 5 2 2 1 3 | Expand matrix X A by first row. | X A | = 4 | − 2 − 3 4 1 1 − 5 2 1 3 | − 3 | − 1 − 3 4 3 1 − 5 2 1 3 | + 1 | − 1 − 2 4 3 1 − 5 2 2 3 | − 0 = { 4 [ ( − 2 ) ( 3 + 5 ) − 1 ( − 9 − 4 ) + 2 ( 15 − 4 ) ] − 3 [ ( − 1 ) ( 3 + 5 ) − 3 ( − 9 − 4 ) + 2 ( 15 − 4 ) ] + 1 [ ( − 1 ) ( 3 + 10 ) − 3 ( − 6 − 8 ) + 2 ( 10 − 4 ) ] } = 4 ( − 16 + 13 + 22 ) − 3 ( − 8 + 39 + 22 ) + 1 ( − 13 + 42 + 12 ) = 4 ( 19 ) − 3 ( 53 ) + 1 ( 41 ) = − 42 Find the value of p using the equation: p = | X A | | A | Substitute − 42 for | X A | and − 21 for | A | . p = − 42 − 21 = 2 Therefore, value of p is 2 _ . Replace the coefficients of r (elements of column 2) with constant [ X ] to form matrix [ X B ] . [ X B ] = [ 2 4 1 0 1 − 1 − 3 4 3 3 1 − 5 − 1 2 1 3 ] Find the determinant | X B | of matrix X B . | X B | = | 2 4 1 0 1 − 1 − 3 4 3 3 1 − 5 − 1 2 1 3 | Expand matrix X B by first row. | X B | = 2 | − 1 − 3 4 3 1 − 5 2 1 3 | − 4 | 1 − 3 4 3 1 − 5 − 1 1 3 | + 1 | 1 − 1 4 3 3 − 5 − 1 2 3 | − 0 = { 2 [ ( − 1 ) ( 3 + 5 ) − 3 ( − 9 − 4 ) + 2 ( 15 − 4 ) ] − 4 [ 1 ( 3 + 5 ) − 3 ( − 9 − 4 ) + ( − 1 ) ( 15 − 4 ) ] + 1 [ 1 ( 9 + 10 ) − 3 ( − 3 − 8 ) + ( − 1 ) ( 5 − 12 ) ] } = 2 ( − 8 + 39 + 22 ) − 4 ( 8 + 39 − 11 ) + 1 ( 19 + 33 + 7 ) = 106 − 144 + 59 = 21 Find the value of r using the equation: r = | X B | | A | Substitute 21 for | X B | and − 21 for | A |

Mylab Math With Pearson Etext -- 18 Week Standalone Access Card -- For Basic Technical Mathematics With Calculus

11th Edition

ISBN: 9780135902912

Author: Allyn J. Washington

Publisher: PEARSON

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Chapter 16.6, Problem 24E

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Chapter 16.6, Problem 23E

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Chapter 16 Solutions

Mylab Math With Pearson Etext -- 18 Week Standalone Access Card -- For Basic Technical Mathematics With Calculus

Ch. 16.1 - For matrices A and B, find A + B. Ch. 16.1 - Prob. 2PE Ch. 16.1 - Prob. 3PE Ch. 16.1 - Prob. 1E Ch. 16.1 - Prob. 2E Ch. 16.1 - Prob. 3E Ch. 16.1 - Prob. 4E Ch. 16.1 - In Exercises 3–10, determine the value of the...Ch. 16.1 - Prob. 6E Ch. 16.1 - Prob. 7E

Ch. 16.1 - Prob. 8E Ch. 16.1 - In Exercises 3–10, determine the value of the...Ch. 16.1 - Prob. 10E Ch. 16.1 - Prob. 11E Ch. 16.1 - Prob. 12E Ch. 16.1 - Prob. 13E Ch. 16.1 - Prob. 14E Ch. 16.1 - Prob. 15E Ch. 16.1 - Prob. 16E Ch. 16.1 - Prob. 17E Ch. 16.1 - Prob. 18E Ch. 16.1 - Prob. 19E Ch. 16.1 - Prob. 20E Ch. 16.1 - Prob. 21E Ch. 16.1 - Prob. 22E Ch. 16.1 - Prob. 23E Ch. 16.1 - Prob. 24E Ch. 16.1 - Prob. 25E Ch. 16.1 - Prob. 26E Ch. 16.1 - Prob. 27E Ch. 16.1 - Prob. 28E Ch. 16.1 - Prob. 29E Ch. 16.1 - Prob. 30E Ch. 16.1 - Prob. 31E Ch. 16.1 - Prob. 32E Ch. 16.1 - Prob. 33E Ch. 16.1 - Prob. 34E Ch. 16.1 - Prob. 35E Ch. 16.1 - Prob. 36E Ch. 16.1 - Prob. 37E Ch. 16.1 - Prob. 38E Ch. 16.1 - Prob. 39E Ch. 16.1 - Prob. 40E Ch. 16.1 - Prob. 41E Ch. 16.1 - Prob. 42E Ch. 16.1 - Prob. 43E Ch. 16.1 - Prob. 44E Ch. 16.2 - Prob. 1PE Ch. 16.2 - Prob. 2PE Ch. 16.2 - Prob. 3PE Ch. 16.2 - Prob. 1E Ch. 16.2 - Prob. 2E Ch. 16.2 - Prob. 3E Ch. 16.2 - Prob. 4E Ch. 16.2 - Prob. 5E Ch. 16.2 - Prob. 6E Ch. 16.2 - 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