Explanation Given: Total amount in three accounts is $400,000. The amount invested at 8% is twice the amount invested at 6%. The total interest earned on all three accounts is $36,000. 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: Let x , y and z be the amounts of money in each account that pays interest of 6%, 8% and 10%, respectively. Since the total amount in three accounts is $400,000, the first equation is x + y + z = $ 400 , 000 . The total interest earned on all three accounts is $36,000, the second equation is 0.06 x + 0.08 y + 0.1 z = $ 36 , 000 . Since the amount invested at 8% is twice the amount invested at 6%, the third equation is formulated as y = 2 x or 2 x − y = 0 . Thus, the system of equations becomes, { x + y + z = $ 400 , 000 0.06 x + 0.08 y + 0.1 z = $ 36 , 000 2 x − y = 0 Convert the system of equation into a matrix notation. [ 1 1 1 0.06 0.08 0.1 2 − 1 0 ] [ x y z ] = [ 400 , 000 36 , 000 0 ] Take A = [ 1 1 1 0.06 0.08 0.1 2 − 1 0 ] and X = [ x y z ] and transform [ 1 1 1 0.06 0.08 0.1 2 − 1 0 ] [ x y z ] = [ 400 , 000 36 , 000 0 ] as follows. A − 1 A X = A − 1 [ 400 , 000 36 , 000 0 ] ( Multiply with A − 1 on RHS of both sides ) I X = A − 1 [ 400 , 000 36 , 000 0 ] ( ∵ A − 1 A = I ) X = A − 1 [ 400 , 000 36 , 000 0 ] Obtain the augmented matrix as follows. [ 1 1 1 1 0 0 0.06 0.08 0.1 0 1 0 2 − 1 0 0 0 1 ] Apply the row operation 100 R 2 → R 2 and reduce the matrix. [ 1 1 1 1 0 0 6 8 10 0 100 0 2 − 1 0 0 0 1 ] Apply the row operation 6 R 1 − R 2 → R 2 and 2 R 1 − R 3 → R 3 and reduce the matrix. [ 1 1 1 1 0 0 0 − 2 − 4 6 − 100 0 0 3 2 2 0 − 1 ] Apply the row operation 2 R 1 + R 2 → R 1 and 3 R 2 + 2 R 3 → R 3 and reduce the matrix

College Algebra in Context with Integrated Review and Worksheets Plus MyLab Math with Pearson eText-- Access Card Package

1st Edition

ISBN: 9780134380155

Author: Ronald J. Harshbarger, Lisa S. Yocco

Publisher: PEARSON

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Chapter 7.4, Problem 31E

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Chapter 7.4, Problem 30E

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Chapter 7 Solutions

College Algebra in Context with Integrated Review and Worksheets Plus MyLab Math with Pearson eText-- Access Card Package

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 amount of money she has invested in each account.

Solution Summary: The author calculates the total amount of money in three accounts. The first equation is x+y+z = 400,000, and the second equation y = 2xor2x-y=0

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To find: The amount of money she has invested in each account.

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