How Do You Use Matrix Inversion to Determine Nutrient Ratios in Food?

AI Thread Summary
Matrix inversion is used to solve nutrient ratio problems by setting up the equation Ax = y, where A represents the nutrient content of the foods, x is the amount of each food, and y is the desired nutrient intake. The inverse matrix A^-1 allows for the calculation of x by multiplying both sides of the equation by A^-1. The discussion emphasizes the importance of correctly interpreting the matrix setup and applying the inverse to find solutions for specific nutrient requirements. Participants clarify the steps needed to approach the problem, leading to successful resolution of the nutrient intake questions. Understanding matrix operations is crucial for solving such nutritional equations effectively.
uselessjack
Messages
4
Reaction score
0

Homework Statement



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:

70qfV.png
a) Find the inverse of the matrix

4erZX.png


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?

Homework Equations



A * A^-1 = Identity

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 1I do not know how to approach the rest of the problem!
 
Physics news on Phys.org
For a, you want to solve the matrix equation Ax = y
\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]

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.
 
Last edited:
Mark44 said:
For a, you want to solve the matrix equation Ax = y
\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]

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.

Thank you very much, but I'm not sure I understand.

Would I solve for x by setting up the equation "5x + 3y + 5z = 23" and solving?
 
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-1Ax = A-1y

Why do you think they asked you to find the inverse?
 
Ah, I see now! Thank you very much for your time and help. I was finally able to find the answers!
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top