Problem 47 - Precalculus: Mathematical for Calculus

Question

Nutrition 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|ccc|} \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 \(10 \mathrm{mg}\) of folic acid, \(14 \mathrm{mg}\) of choline, and \(13 \mathrm{mg}\) of inositol? (c) How much of each food is needed to supply \(9 \mathrm{mg}\) of folic acid, 12 mg of choline, and 10 mg of inositol? (d) Will any combination of these foods supply \(2 \mathrm{mg}\) of folic acid, \(4 \mathrm{mg}\) of choline, and \(11 \mathrm{mg}\) of inositol?

Step-by-Step Solution

Answer

(b) 1 oz of A, 1 oz of B, 2 oz of C; (c) 0 oz of A, 0 oz of B, 2 oz of C; (d) Not possible.

Step 1: Compute the Determinant of the Matrix

To find the inverse of a 3x3 matrix, the first step is to calculate its determinant. The matrix is \( \begin{bmatrix} 3 & 1 & 3 \ 4 & 2 & 4 \ 3 & 2 & 4 \end{bmatrix} \).Use the formula for a 3x3 matrix determinant: \[ \text{det}(A) = a(ei-fh) - b(di-fg) + c(dh-eg) \] where \(a = 3, b = 1, c = 3, d = 4, e = 2, f = 4, g = 3, h = 2, i = 4\).Calculate: \[ \text{det}(A) = 3(2 \cdot 4 - 2 \cdot 4) - 1(4 \cdot 4 - 3 \cdot 4) + 3(4 \cdot 2 - 3 \cdot 2) = 3(8-8) - 1(16-12) + 3(8-6) \] \[ \text{det}(A) = 0 - 4 + 6 = 2 \]

Step 2: Find the Matrix of Minors

The matrix of minors is found by calculating the determinant of the 2x2 submatrices for each element in the 3x3 matrix. - For the element at position (1,1), ignore the first row and column: determinant is \(2 \cdot 4 - 2 \cdot 4 = 0\).- For (1,2): ignore the first row and second column: determinant is \(4 \cdot 4 - 3 \cdot 4 = 4\).- For (1,3), ignore the first row and third column: determinant is \(4 \cdot 2 - 3 \cdot 2 = 2\).- Continue this for all elements.

Step 3: Calculate the Matrix of Cofactors

The matrix of cofactors is formed by applying a checkerboard pattern of plus and minus signs to the matrix of minors:\[ \begin{bmatrix} +0 & -4 & +2 \ -4 & +4 & -2 \ +2 & -8 & +2 \end{bmatrix} \]

Step 4: Transpose the Matrix of Cofactors

Transposing the matrix of cofactors means swapping its rows and columns:\[ \begin{bmatrix} 0 & -4 & 2 \ -4 & 4 & -8 \ 2 & -2 & 2 \end{bmatrix} \]

Step 5: Find the Inverse Matrix

Divide every element of the transposed cofactor matrix by the determinant to find the inverse:The inverse matrix \( A^{-1} \) is:\[ \begin{bmatrix} 0 & -2 & 1 \ -2 & 2 & -4 \ 1 & -1 & 1 \end{bmatrix} \]

Step 6: Solve System for Part (b) using Inverse Matrix

To find the number of ounces needed for part (b):Multiply the inverse matrix by the vector of nutrients: \[ \begin{bmatrix} 0 & -2 & 1 \ -2 & 2 & -4 \ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 10 \ 14 \ 13 \end{bmatrix} = \begin{bmatrix} 1 \ 1 \ 2 \end{bmatrix} \]Therefore, 1 ounce of Type A, 1 ounce of Type B, and 2 ounces of Type C are needed.

Step 7: Solve System for Part (c)

For part (c), repeat the process with the new nutrient vector:Multiply the inverse matrix by the vector: \[ \begin{bmatrix} 0 & -2 & 1 \ -2 & 2 & -4 \ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 9 \ 12 \ 10 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 2 \end{bmatrix} \]Thus, 0 ounces of Type A, 0 ounces of Type B, and 2 ounces of Type C are needed.

Step 8: Solve System for Part (d)

For part (d), attempt to find a solution using the conditions \([2, 4, 11]\).Multiply the inverse matrix by: \[ \begin{bmatrix} 0 & -2 & 1 \ -2 & 2 & -4 \ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 2 \ 4 \ 11 \end{bmatrix} \]This results in a non-integer or negative solution indicating that it is impossible to achieve this combination with these foods.

Key Concepts

Determinant Calculation

To begin solving problems involving matrix inversion, calculating the determinant is crucial. The determinant is a special number that can provide important information about the matrix. For a 3x3 matrix, like our example matrix:\[ \begin{bmatrix} 3 & 1 & 3 \ 4 & 2 & 4 \ 3 & 2 & 4 \end{bmatrix} \]we use the formula:\[ \text{det}(A) = a(ei-fh) - b(di-fg) + c(dh-eg) \]In this formula, each letter represents an entry in the matrix. After substituting in the values, you calculate step-by-step. Here, the determinant ends up being 2, which is vital for finding the matrix inverse. A zero determinant indicates the matrix is not invertible, which means further calculations for the inverse would not be possible.
Calculating the determinant is a fundamental step when dealing with matrix equations as it directly affects the feasibility of inversion.

Matrix of Cofactors

After determining the determinant, the next step towards finding the inverse involves the matrix of cofactors. This matrix is derived from the matrix of minors. Each entry in the matrix of minors is the determinant of a 2x2 submatrix obtained by removing one row and one column of the original matrix.Once the matrix of minors is calculated, apply a pattern of alternating signs (often referred to as a "checkerboard" pattern) to generate the cofactor matrix. This involves multiplying each minor by "+" or "-" depending on its position:\[ \begin{bmatrix} +0 & -4 & +2 \ -4 & +4 & -2 \ +2 & -8 & +2 \end{bmatrix} \]The cofactor matrix plays a pivotal role in computing the inverse and is derived solely from the original matrix's elements. This process ensures each element considers the matrix's holistic properties for accurate inversion calculations.

Matrix Multiplication

Finally, matrix multiplication is used to solve systems of equations once the inverse matrix is determined. At this point, you'll already have your inverse matrix, such as:\[ \begin{bmatrix} 0 & -2 & 1 \ -2 & 2 & -4 \ 1 & -1 & 1 \end{bmatrix} \]To solve a system, multiply this inverse by the vector of desired nutrient outcomes. For example, to find how many ounces of each type of food are needed, you will use:\[ \begin{bmatrix} 0 & -2 & 1 \ -2 & 2 & -4 \ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} a \ b \ c \end{bmatrix} = \begin{bmatrix} x \ y \ z \end{bmatrix} \]Here, \( a, b, c \) are the nutrient requirements, and \( x, y, z \) will give you the amount of each type of food. This multiplication provides the exact quantities alongside the inverse concept, ensuring dietary needs are met accurately. Matrix multiplication transforms theoretical calculations into practical solutions.