Explanation Given: The constraint of the system is as follows, { x + y + z = ( total investment ) 0.045 x + 0.05 y + 0.09 z = ( annual return ) 2 y − z = 0 The total amount invested is $ 12 , 000 . The annual return is $ 835 . Approach: If A is an invertible matrix, then the system of linear equations represented by A X = B has a unique solution given by X = A − 1 B . Calculation: Let the amount invested in AAA bond be x , in A be y, and in B be z . The system of linear constraint is, x + y + z = 12 , 000 0.045 x + 0.05 y + 0.09 z = 835 2 y − z = 0 Write the system of equation in the form A X = B . [ 1 1 1 0.045 0.05 0.09 0 2 − 1 ] [ x y z ] = [ 12000 835 0 ] ... ( 1 ) Substitute A for [ 1 1 1 0.045 0.05 0.09 0 2 − 1 ] in equation ( 1 ) . A [ x y z ] = [ 12000 835 0 ] [ x y z ] = A − 1 [ 12000 835 0 ] ... ( 2 ) Calculate A − 1 Proceed by adjoining the identity matrix to form the matrix, [ A ⋮ I ] = [ 1 1 1 ⋮ 1 0 0 0.045 0.05 0.09 ⋮ 0 1 0 0 2 − 1 ⋮ 0 0 1 ] Use the elementary row operation to obtain the form of [ I ⋮ A − 1 ] . Apply operation R 2 → R 2 − 9 200 R 1 . [ A ⋮ I ] = [ 1 1 1 ⋮ 1 0 0 0 1 / 200 9 / 200 ⋮ − 9 / 200 1 0 0 2 − 1 ⋮ 0 0 1 ] Apply operation R 1 → R 1 − 200 R 2 and R 3 → R 3 − 400 R 2 . [ A ⋮ I ] = [ 1 0 − 8 ⋮ 10 − 200 0 0 1 / 200 9 / 200 ⋮ − 9 / 200 1 0 0 0 − 19 ⋮ 18 − 400 1 ] Apply operation R 2 → 200 R 2 [ A ⋮ I ] = [ 1 0 − 8 ⋮ 10 − 200 0 0 1 9 ⋮ − 9 200 0 0 0 − 19 ⋮ 18 − 400 1 ] Apply operation R 3 → − 1 19 R 3

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Chapter 8.3, Problem 58E

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The amount invested in each type of bond using inverse matrix if total amount invested is $12,000.

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Chapter 8.3, Problem 57E

Chapter 8.3, Problem 59E

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The amount invested in each type of bond using inverse matrix if total amount invested is $ 12 , 000 .

Solution Summary: The author analyzes the system of linear equations represented by AX=B with a unique solution.

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The amount invested in each type of bond using inverse matrix if total amount invested is $12,000.

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Chapter 8 Solutions

The amount invested in each type of bond using inverse matrix if total amount invested is $ 12 , 000 .

Solution Summary: The author analyzes the system of linear equations represented by AX=B with a unique solution.

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To find: The amount invested in each type of bond using inverse matrix if total amount invested is $12,000.

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Chapter 8 Solutions

To find:

The amount invested in each type of bond using inverse matrix if total amount invested is $12,000.