Explanation Given information: The chemist has three acid solutions at various concentrations. The first is 10% acid, the second is 20%, and the third is 40%. The chemist has to make a solution of 100 ml with 18% acid. Also, Chemist has to use four times as much of the 10% solution as the 40% solution. Concept used: The inverse of matrix A with 3 × 3 order can be obtained by elementary row operation on a new matrix to change the left side of a matrix in to identity matrix. And automatically the right matrix transformed in to inverse of matrix A . Here, the matrix A is [ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ] . The new matrix can be written as, [ a 11 a 12 a 13 1 0 0 a 21 a 22 a 23 0 1 0 a 31 a 32 a 33 0 0 1 ] The solution of the matrix equation can be calculated as, A X = B X = A − 1 B Here, A is a square n × n matrix that is invertible, B is known matrix with n rows and X is an unknown matrix. Calculation: Consider x as the amount of 10 % acid solution, y as the amount of 20 % acid solution and z as the amount of 40 % acid solution. The total amount of acid solutions is 100 mL. x + y + z = 100 ........ (1) The concentration of acid in the total solution is 18%. 10 % ( x ) + 20 % ( y ) + 40 % ( z ) = 18 % ( 100 ) 10 100 x + 20 100 y + 40 100 z = 100 ⋅ 18 100 Further solve the above equation as, 0.1 x + 0.2 y + 0.4 z = 18 ........ (2) Now, chemist has to use four times of 10% solution as much as 40% solution. 1 x + 0 y = 4 z Further simply the above equation as, 1 x + 0 y − 4 z = 0 ........ (3) Consider B as column that can be expressed as, B = [ 100 18 0 ] The matrix equation corresponding to the above equations is, [ 1 1 1 0.1 0.2 0.4 1 0 − 4 ] [ x y z ] = [ 100 18 0 ] The above matrix equation can be written as, [ x y z ] = [ 1 1 1 0.1 0.2 0.4 1 0 4 ] − 1 [ 100 18 0 ] ........ (4) The given matrix can be written in the form of new matrix as, [ 1 1 1 1 0 0 0.1 0.2 0.4 0 1 0 1 0 − 4 0 0 1 ] Perform the operation R 2 → R 2 − 0.1 ⋅ R 1 and R 3 → R 3 − 1 ⋅ R 1 in the above matrix as, [ 1 1 1 1 0 0 0 0.2 0.3 − 0.1 1 0 1 0 − 4 0 0 1 ] [ 1 1 1 1 0 0 0 0.1 0.3 − 0.1 1 0 1 − 1 − 5 − 1 0 1 ] Interchange the second row and third row of the above matrix as, [ 1 1 1 1 0 0 0 − 1 − 5 − 1 0 1 1 1.0 0.3 − 0.1 1 0 ] Perform the operation R 3 → R 3 + 0.1 ⋅ R 2 and R 3 → − 5 ⋅ R 3 in the above matrix as, [ 1 1 1 1 0 0 0 − 1 − 5 − 1 0 1 1 0 − 0

7th Edition

ISBN: 9781305115545

Author: James Stewart, Lothar Redlin, Saleem Watson

Publisher: Cengage Learning

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Chapter 6.1, Problem 70E

To determine

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The amount of 10%,20% and 40% of the acid solution.

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