Explanation Given information: The alloy 1 contains 80% copper and 20% Nickel. The alloy 2 contains 60% copper, 20% nickel, and 20% zinc. The alloy 3 contains 30% copper, 60% nickel, and 10% iron. The alloy four contains 20% nickel, 40% zinc, and 40% iron. The final alloy contains 56 g copper, 28 g nickel, 10g zinc, and 6 g iron. Calculation: Consider x , y , z , and w are represents the fractions in alloy1, 2, 3, and 4 respectively. Write the general form of equations using given information. Use the percentage in numerical. Consider ratio of copper in each alloy. 80 % x + 60 % y + 30 % z + 0 w = 56 0.8 x + 0.6 y + 0.3 z = 56 Consider ratio of nickel in each alloy. 20 % x + 20 % y + 60 % z + 20 % w = 28 0.2 x + 0.2 y + 0.6 z + 0.2 w = 28 Consider ratio of zinc in each alloy. 0 x + 20 % y + 0 % z + 40 % w = 10 0.2 y + 0.4 w = 10 Consider ratio of iron in each alloy. 0 x + 0 y + 10 % w + 40 % w = 6 0.1 z + 0.4 w = 6 Write the given equations in general matrix form [ A ] . [ A ] = [ 0.8 0.6 0.3 0 0.2 0.2 0.6 0.2 0 0.2 0 0.4 0 0 0.1 0.4 ] Find the determinant | A | of matrix A . | A | = | 0.8 0.6 0.3 0 0.2 0.2 0.6 0.2 0 0.2 0 0.4 0 0 0.1 0.4 | Expand matrix A by first column. | A | = 0.8 | 0.2 0.6 0.2 0.2 0 0.4 0 0.1 0.4 | − 0.2 | 0.6 0.3 0 0.2 0 0.4 0 0.1 0.4 | + 0 − 0 = { 0.8 [ 0.2 ( 0 − 0.04 ) − 0.2 ( 0.24 − 0.02 ) + 0 ] − 0.2 [ 0.6 ( 0 − 0.04 ) − 0.2 ( 0.12 − 0 ) + 0 ] } = 0.8 ( − 0.008 − 0.044 ) − 0.2 ( − 0.024 − 0.024 ) = − 0.0416 + 0.0096 = − 0.032 Replace the coefficients of x (elements of column 1) with constant [ X ] to form matrix [ X A ] . [ X A ] = [ 56 0.6 0.3 0 28 0.2 0.6 0.2 10 0.2 0 0.4 6 0 0.1 0.4 ] Find the determinant | X A | of matrix X A . | X A | = | 56 0.6 0.3 0 28 0.2 0.6 0.2 10 0.2 0 0.4 6 0 0.1 0.4 | Expand matrix X A by first row. | X A | = 56 | 0.2 0.6 0.2 0.2 0 0.4 0 0.1 0.4 | − 0.6 | 28 0.6 0.2 10 0 0.4 6 0.1 0.4 | + 0.3 | 28 0.2 0.2 10 0.2 0.4 6 0 0.4 | − 0 = { 56 [ 0.2 ( 0 − 0.04 ) − 0.2 ( 0.24 − 0.02 ) ] − 0.6 [ − 10 ( 0.24 − 0.02 ) + 0 − 0.4 ( 2.8 − 3.6 ) ] + 0.3 [ 6 ( 0.08 − 0.04 ) − 0 + 0.4 ( 5.6 − 2 ) ] } = 56 ( − 0.008 − 0.044 ) − 0.6 ( − 2.2 + 0.32 ) + 0.3 ( 0.24 + 1.44 ) = 56 ( − 0.052 ) − 0.6 ( − 1.88 ) + 0.3 ( 1.68 ) = − 2.912 + 1.128 + 0.504 = − 1.28 Find the weight of copper ( x ) using the equation: x = | X A | | A | Substitute − 1.28 for | X A | and − 0.032 for | A | . x = − 1.28 − 0.032 = 40 g Therefore, the weight of copper is 40 g _ . Replace the coefficients of y (elements of column 2) with constant [ X ] to form matrix [ X B ] . [ X B ] = [ 0.8 56 0.3 0 0.2 28 0.6 0.2 0 10 0 0.4 0 6 0.1 0.4 ] Find the determinant | X B | of matrix X B . | X B | = | 0.8 56 0.3 0 0.2 28 0

11th Edition

ISBN: 9780134437705

Author: Washington

Publisher: PEARSON

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Chapter 16.6, Problem 36E

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Find the weight of copper, nickel, zinc, and iron in the final alloy.

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Chapter 16.6, Problem 35E

Chapter 16.6, Problem 37E

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Chapter 16 Solutions

Basic Technical Mathematics

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