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Help please... Precalculus problem using matricesby uselessjack
Tags: matrix matrices 
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#1
Mar1410, 11:42 PM

P: 4

1. The problem statement, all variables and given/known data
A nutritionist is studying the effects of the nutrients folic acid, choline, and inositol. He has three types of food available, and each type contains the following amounts of these nutrients per ounce: a) Find the inverse of the matrix and use it to solve the remaining parts of this problem. A calculator may be used. 1) How many ounces of each food should the nutritionist feed his laboratory rats if he wants their daily diet to contain 23 mg of folic acid, 28 mg of choline, and 27 mg of inositol? 2) How much of each food is needed to supply 20 mg of folic acid, 24 mg of choline, and 21 mg of inositol? 3) Will any combination of these foods supply 6 mg of folic acid, 8 mg of choline, and 13 mg of inositol? 2. Relevant equations A * A^1 = Identity 3. The attempt at a solution I have only been able to yield the inverse of the matrix: 0 1 1 3 5/2 0 2 5/2 1 I do not know how to approach the rest of the problem! 


#2
Mar1510, 01:11 AM

Mentor
P: 21,215

For a, you want to solve the matrix equation Ax = y
[tex]\left[ \begin{array} {c c c } 5 & 3 & 5 \\ 6 & 4 & 6 \\ 5 & 4 & 6 \end{array} \right]\left[ \begin{array}{c}x_1\\x_2\\x_3\end{array} \right] = \left[ \begin{array}{c}23\\28\\27\end{array} \right][/tex] Using your inverse, A^{1}, can you figure out how to solve for the vector x? For b, similar setup, but the vector on the right uses the three values of this part of the problem. 


#3
Mar1510, 01:24 AM

P: 4

Would I solve for x by setting up the equation "5x + 3y + 5z = 23" and solving? 


#4
Mar1510, 01:33 AM

Mentor
P: 21,215

Help please... Precalculus problem using matrices
No. In the matrix equation I showed, the 3 x 3 matrix is A, the column vector in the middle represents the amounts of foods A, B, and C, and the column vector on the right represents the desired amounts of folic acid, choline, and inisotol.
If A is an invertible matrix, then the equation Ax = y can be solved by multiplying the left and right sides by A^{1}. Ax = y ==> A^{1}Ax = A^{1}y Why do you think they asked you to find the inverse? 


#5
Mar1510, 03:03 AM

P: 4

Ah, I see now! Thank you very much for your time and help. I was finally able to find the answers!



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