Problem 47 - Precalculus Mathematics for Calculus

Question

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. $$\begin{array}{|l|c|c|c|} \hline & \text { Type A } & \text { Type B } & \text { Type C } \\ \hline \text { Folic acid (mg) } & 3 & 1 & 3 \\ \text { Choline (mg) } & 4 & 2 & 4 \\ \text { Inositol (mg) } & 3 & 2 & 4 \\ \hline \end{array}$$ (a) Find the inverse of the matrix $$ \left[\begin{array}{lll} 3 & 1 & 3 \\ 4 & 2 & 4 \\ 3 & 2 & 4 \end{array}\right] $$ and use it to solve the remaining parts of this problem. (b) How many ounces of each food should the nutritionist feed his laboratory rats if he wants their daily diet to contain 10 mg of folic acid, 14 mg of choline, and 13 mg of inositol? (c) How much of each food is needed to supply 9 mg of folic acid, 12 mg of choline, and 10 mg of inositol? (d) Will any combination of these foods supply 2 mg of folic acid, 4 mg of choline, and 11 mg of inositol?

Step-by-Step Solution

Answer

No, the provided nutrients cannot be obtained for part (d) due to inconsistent equations.

Step 1: Set Up the Problem

We need to find the inverse of the given matrix \( A \) which represents the nutrient contents of Type A, Type B, and Type C foods.\[A = \begin{bmatrix} 3 & 1 & 3 \4 & 2 & 4 \3 & 2 & 4 \end{bmatrix}\]

Step 2: Find the Determinant of Matrix A

Compute the determinant \( \det(A) \) of matrix \( A \). Use the formula: \[\det(A) = a(ei-fh) - b(di-fg) + c(dh-eg)\]Substituting the respective values, \[\det(A) = 3(2 \times 4 - 2 \times 4) - 1(4 \times 4 - 4 \times 3) + 3(4 \times 2 - 2 \times 3)\]Thus, \[\det(A) = 3(0) - 1(4) + 3(2)\]\[\det(A) = -4 + 6 = 2\]

Step 3: Compute the Inverse of Matrix A

The inverse of a matrix \( A \) is given by \[A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A)\]where \( \text{adj}(A) \) is the adjugate of \( A \). First, find the minor, cofactor, and then the adjugate matrix of \( A \).The minors are:\[M_{11} = \begin{vmatrix} 2 & 4 \ 2 & 4 \end{vmatrix} = 0, M_{12} = \begin{vmatrix} 4 & 4 \ 3 & 4 \end{vmatrix} = 4 - 12 = -8, M_{13} = \begin{vmatrix} 4 & 2 \ 3 & 2 \end{vmatrix} = 4 - 12 = -4\]\[M_{21} = \begin{vmatrix} 1 & 3 \ 2 & 4 \end{vmatrix} = 4 - 6 = -2, M_{22} = \begin{vmatrix} 3 & 3 \ 3 & 4 \end{vmatrix} = 12 - 9 = 3, M_{23} = \begin{vmatrix} 3 & 1 \ 3 & 2 \end{vmatrix} = 6 - 3 = 3\]\[M_{31} = \begin{vmatrix} 1 & 3 \ 2 & 4 \end{vmatrix} = 0, M_{32} = \begin{vmatrix} 3 & 3 \ 4 & 4 \end{vmatrix} = 12 - 12 = 0, M_{33} = \begin{vmatrix} 3 & 1 \ 4 & 2 \end{vmatrix} = 6 - 4 = 2\]The cofactors are:\[C_{ij} = (-1)^{i+j}M_{ij}\]Calculating, we obtain \[C_{11} = 0, C_{12} = 8, C_{13} = -4 \C_{21} = 2, C_{22} = 3, C_{23} = 3 \C_{31} = 0, C_{32} = 0, C_{33} = 2\]The adjugate of \( A \) is the transpose of the cofactor matrix, given by:\[\text{adj}(A) = \begin{bmatrix} 0 & 2 & 0 \8 & 3 & 0 \-4 & 3 & 2 \end{bmatrix}\]Finally, the inverse is:\[A^{-1} = \frac{1}{2}\begin{bmatrix} 0 & 2 & 0 \8 & 3 & 0 \-4 & 3 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \4 & \frac{3}{2} & 0 \-2 & \frac{3}{2} & 1 \end{bmatrix}\]

Step 4: Use the Inverse to Solve for Specific Nutrient Requirements

To determine how many ounces are needed, we will multiply the inverse matrix by the desired nutrients vector for each subtask. For part (b), the vector is:\[\begin{bmatrix} 10 \14 \13 \end{bmatrix}\]Multiply by \( A^{-1} \):\[\begin{bmatrix} 0 & 1 & 0 \4 & \frac{3}{2} & 0 \-2 & \frac{3}{2} & 1 \end{bmatrix} \begin{bmatrix} 10 \14 \13 \end{bmatrix} = \begin{bmatrix} 14 \74 \times \frac{1}{2} \22 - 13 \end{bmatrix} = \begin{bmatrix} 14 \37 \9 \end{bmatrix}\]Therefore, 0 ounces of Type A, 5 ounces of Type B, and 1 ounce of Type C are needed for part (b). For (c), with a similar calculation, solve for the vector \[\begin{bmatrix} 9 \12 \10 \end{bmatrix} \] and for (d) solve for vector \[\begin{bmatrix} 2 \4 \11 \end{bmatrix}\]. Check the possibility with determinant analysis.

Key Concepts

Determinant Calculation

The determinant of a matrix is a special number that can be calculated from its elements. It is crucial in determining whether a matrix is invertible. If the determinant is zero, the matrix does not have an inverse. To find the determinant of a 3x3 matrix like \[A = \begin{bmatrix} 3 & 1 & 3 \ 4 & 2 & 4 \ 3 & 2 & 4 \end{bmatrix}\]we use the formula: \[\det(A) = a(ei-fh) - b(di-fg) + c(dh-eg)\]This involves calculating several sub-determinants of smaller 2x2 matrices. For matrix \(A\), the determinant is calculated as follows:

• First, multiply along the diagonals and subtract them: \3(2 \times 4 - 2 \times 4)\ - \1(4 \times 4 - 4 \times 3)\ + \3(4 \times 2 - 2 \times 3)
• Perform the arithmetic: \3(0) - 1(4) + 3(2) = -4 + 6 = 2

Since the determinant is 2, which is not zero, the matrix is invertible.

Cofactor Matrix

Once we have the determinant and confirm that the matrix is invertible, the next step in finding the inverse is calculating the cofactor matrix. The cofactor of an element is obtained by:

• Taking the minor of the element, obtained by removing the row and column of that element.
• Applying a sign based on the element's position, positive if the sum of row and column indices is even, negative if odd: \(\text{Cofactor}_{ij} = (-1)^{i+j} \times \text{Minor}_{ij}\)

For example, the cofactor of the element at position (1,1), which is 3, requires the minor matrix: \[\begin{vmatrix} 2 & 4 \ 2 & 4 \end{vmatrix} = 0\]Thus, the cofactor of this element is:\[C_{11} = (-1)^{1+1} \times 0 = 0\]Compute similar cofactor values for the other elements to build the cofactor matrix.

Adjugate Matrix

The adjugate of a matrix, also known as the adjoint, is the transpose of the cofactor matrix. This means we flip rows and columns of the cofactor matrix. Given our example matrix \(A\), the cofactor matrix is:\[\begin{bmatrix} 0 & 8 & -4 \ 2 & 3 & 3 \ 0 & 0 & 2 \end{bmatrix}\]The adjugate matrix is then:\[\text{adj}(A) = \begin{bmatrix} 0 & 2 & 0 \ 8 & 3 & 0 \ -4 & 3 & 2 \end{bmatrix}\]Calculating the adjugate is a critical step towards finding the inverse of the matrix. Once you have the adjugate matrix, the next step is straightforward: divide each element by the determinant calculated earlier to obtain the inverse matrix. This process is essential in solving systems of linear equations, especially when analyzing the nutrient content presented in different food types.

Nutrient Content Analysis

Nutrient content analysis using matrix inversion involves determining the quantities of available food types needed to meet certain nutritional goals. In this problem, after finding the inverse of the matrix representing nutrient contents of food types, we can easily decide how much of each food type is needed to meet desired nutrient levels.
Let's consider the matrix \[A = \begin{bmatrix} 3 & 1 & 3 \ 4 & 2 & 4 \ 3 & 2 & 4 \end{bmatrix}\] and its inverse. To achieve specific nutrient intake e.g., 10 mg of folic acid, 14 mg of choline, and 13 mg of inositol, set up a vector for these goals: \[\begin{bmatrix} 10 \ 14 \ 13 \end{bmatrix}\]You multiply the inverse of the matrix \(A\) by this vector to determine the food quantities. Key aspects include:

• Using the inverse matrix to transform nutrient goals into food quantities.
• Checking for solutions to verify feasibility and correctness within realistic ranges.

This method provides a precise, mathematical way to calculate dietary needs based on available food nutrients, optimizing diet plans efficiently.