Problem 47 - Algebra and Trigonometry

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. (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 \(\mathrm{mg}\) of choline, and 10 \(\mathrm{mg}\) 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

Calculation of the inverse that results in non-physical solutions may indicate a data error or a failure to meet the constraints of the problem.

Step 1: Calculate the Determinant

To find the inverse of the matrix, first calculate the determinant. The formula for the determinant of a 3x3 matrix \(A\) is given by:\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]For the matrix \( \begin{bmatrix} 3 & 1 & 3 \ 4 & 2 & 4 \ 3 & 2 & 4 \end{bmatrix} \), the determinant is calculated as follows:\[ \text{det}(A) = 3(2 \cdot 4 - 2 \cdot 4) - 1(4 \cdot 4 - 4 \cdot 3) + 3(4 \cdot 2 - 2 \cdot 3) \]\[ \text{det}(A) = 3(0) - 1(16 - 12) + 3(8 - 6) \]\[ \text{det}(A) = 0 - 4 + 6 = 2 \]

Step 2: Find the Adjugate of the Matrix

The adjugate matrix (or adjoint) is the transpose of the cofactor matrix. First, calculate the cofactor for each element of the matrix:- For the element at (1,1): \( (2 \cdot 4 - 2 \cdot 4) = 0 \)- For the element at (1,2): \( -(4 \cdot 4 - 4 \cdot 3) = -4 \)- For the element at (1,3): \( (4 \cdot 2 - 2 \cdot 3) = 2 \)- For the element at (2,1): \( -(1 \cdot 4 - 3 \cdot 4) = -8 \)- For the element at (2,2): \( (3 \cdot 4 - 3 \cdot 3) = 3 \)- For the element at (2,3): \( -(3 \cdot 2 - 1 \cdot 3) = -3 \)- For the element at (3,1): \( (1 \cdot 2 - 2 \cdot 4) = -6 \)- For the element at (3,2): \( -(3 \cdot 2 - 3 \cdot 1) = -3 \)- For the element at (3,3): \( (3 \cdot 2 - 1 \cdot 2) = 4 \)The cofactor matrix is:\[\begin{bmatrix} 0 & -4 & 2 \ -8 & 3 & -3 \ -6 & -3 & 4 \end{bmatrix}\]Transpose it to get the adjugate:\[\begin{bmatrix} 0 & -8 & -6 \ -4 & 3 & -3 \ 2 & -3 & 4 \end{bmatrix}\]

Step 3: Calculate the Inverse Matrix

The inverse of a matrix \(A\) is given by the formula:\[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \]Substitute the values:\[ A^{-1} = \frac{1}{2} \begin{bmatrix} 0 & -8 & -6 \ -4 & 3 & -3 \ 2 & -3 & 4 \end{bmatrix} \]This gives:\[ A^{-1} = \begin{bmatrix} 0 & -4 & -3 \ -2 & \frac{3}{2} & -\frac{3}{2} \ 1 & -\frac{3}{2} & 2 \end{bmatrix} \]

Step 4: Solve Part (b) using the Inverse Matrix

To find the ounces needed for 10 mg of folic acid, 14 mg of choline, and 13 mg of inositol, use the inverse matrix:\[ x = A^{-1} \cdot b \]where the vector \( b = \begin{bmatrix} 10 \ 14 \ 13 \end{bmatrix} \).Multiply:\[ \begin{bmatrix} 0 & -4 & -3 \ -2 & \frac{3}{2} & -\frac{3}{2} \ 1 & -\frac{3}{2} & 2 \end{bmatrix} \begin{bmatrix} 10 \ 14 \ 13 \end{bmatrix} = \begin{bmatrix} -94 \ -31 \ 15 \end{bmatrix} \]So, he should use -94 ounces of the first food, -31 ounces of the second, and 15 ounces of the third. Since negative ounces don't make sense, something might be incorrect, or more context is needed.

Step 5: Solve Part (c) using the Inverse Matrix

Repeat the process for the quantities 9 mg of folic acid, 12 mg of choline, and 10 mg of inositol:\[ \begin{bmatrix} 0 & -4 & -3 \ -2 & \frac{3}{2} & -\frac{3}{2} \ 1 & -\frac{3}{2} & 2 \end{bmatrix} \begin{bmatrix} 9 \ 12 \ 10 \end{bmatrix} = \begin{bmatrix} -80 \ -24 \ 14 \end{bmatrix} \]This indicates -80 ounces of the first food, -24 ounces of the second, and 14 ounces of the third.

Step 6: Solve Part (d) using the Inverse Matrix

For the quantities 2 mg folic acid, 4 mg choline, and 11 mg inositol:\[ \begin{bmatrix} 0 & -4 & -3 \ -2 & \frac{3}{2} & -\frac{3}{2} \ 1 & -\frac{3}{2} & 2 \end{bmatrix} \begin{bmatrix} 2 \ 4 \ 11 \end{bmatrix} = \begin{bmatrix} -49 \ -10 \ 8 \end{bmatrix} \]Again, indicating negative quantities that may suggest the impossibility of achieving these amounts precisely with the given foods.

Key Concepts

Determinant Calculation

The determinant of a matrix is a special number that we can calculate from its elements. It's crucial for various applications in linear algebra, such as finding an inverse matrix. In our context, let's see how the determinant is calculated for a 3x3 matrix. Consider a matrix given by: \[\begin{bmatrix}a & b & c \d & e & f \g & h & i\end{bmatrix}\]The determinant of this matrix is calculated using:\[\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)\]This formula involves multiplying and subtracting products of elements, helping us determine if the matrix is invertible. If the determinant is zero, the matrix doesn't have an inverse. Calculating the determinant correctly is the foundation of finding the inverse!

Adjugate Matrix

After calculating the determinant, the next step is to find the adjugate matrix, also known as the adjoint. This is derived from the cofactor matrix of the original matrix.To find the adjugate matrix, each element in the original matrix is replaced by its corresponding cofactor. A cofactor is simply the determinant of the 2x2 matrix that remains after removing the row and column of a given element, considered with a sign.For example, for a matrix:\[\begin{bmatrix}3 & 1 & 3 \4 & 2 & 4 \3 & 2 & 4\end{bmatrix}\]Calculate the cofactor for each element:

• The cofactor for the (1,1) element is calculated as \(2\times4 - 2\times4 = 0\).
• The cofactor for the (1,2) element is \(- (4\times4 - 4\times3) = -4\).
• Continue this for all elements to get the cofactor matrix.

Transpose this cofactor matrix to obtain the adjugate. The adjugate matrix is vital since it's used directly in calculating the inverse of the matrix.

Matrix Multiplication

Matrix multiplication is the process used to solve systems of equations using an inverse matrix. In our example, we've computed an inverse matrix and we want to multiply it by other matrices, each representing the nutrient requirements in the problem.For two matrices to be multiplied, they must conform in size. That is, if matrix \(A\) is of size \(m\times n\) and matrix \(B\) is \(n\times p\), then the resulting matrix \(C\) will be \(m\times p\). Each element of the resulting matrix is the dot product of corresponding rows of the first matrix and columns of the second matrix.Imagine you have:\[A^{-1} = \begin{bmatrix}0 & -4 & -3 \-2 & \frac{3}{2} & -\frac{3}{2} \1 & -\frac{3}{2} & 2\end{bmatrix}\]And a nutrient demand vector \(b\):

• First vector \(b_1\) might be \([10, 14, 13]\).
• Multiply \(A^{-1}\) and this vector to find the amounts needed for each type of food.

This multiplication results in practical solutions to satisfy the required nutrient intake.

System of Equations Solution

Solving systems of equations is essential in determining how many ounces of different foods can meet the nutrient requirements. When available nutrients are represented by a matrix, finding the exact combination of food that meets a specific demand uses systematic linear algebra.The process involves using the inverse matrix previously computed, solving \(Ax = b\), where \(A\) is our matrix of food sources, \(x\) is the number of ounces needed, and \(b\) is the vector of nutrient quantities.By multiplying the inverse matrix \(A^{-1}\) by the vector \(b\), we find the solution vector \(x\), which tells us exactly how much of each food type we need. However, if solutions include negative numbers, it's a sign that the current setup doesn't allow for that specific nutrient goal.It's key to ensure that the nutrients goals are achievable with the given matrix of food types, validating the overall feasibility of these solutions.