Problem 47 - Algebra and Trigonometry

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: (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

For part (b): 1, 3, 2 ounces; (c): 2, 2, 1 ounces; (d): no valid combination.

Step 1: Write the matrix form of the problem

We start by setting up the matrix equation \(AX = B\), where \(A\) is the matrix of the coefficients representing the nutrients' distribution in the foods, \(X\) is the column vector representing the ounces of each food type, and \(B\) is the column vector for the desired amounts of nutrients.

Step 2: Matrix Representation

The matrix \(A\) is given as: \[A = \begin{bmatrix} 3 & 1 & 3 \ 4 & 2 & 4 \ 3 & 2 & 4 \end{bmatrix}\]For the given problem, \(B\) would be the desired nutrition amounts in parts b, c, and d.

Step 3: Find the Inverse of Matrix A

Calculate the inverse of matrix \(A\). To find the inverse, apply the matrix inversion formula for a 3x3 matrix: \[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \]Calculate the determinant and the adjugate of \(A\) before computing \(A^{-1}\). The determinant \(\text{det}(A)\) is: \[\text{det}(A) = 3(2 \cdot 4 - 4 \cdot 2) - 1(4 \cdot 4 - 4 \cdot 3) + 3(4 \cdot 2 - 2 \cdot 3)\]Continuing with this process will give the adjugate, and finally:\[ A^{-1} = \begin{bmatrix} -2 & 0 & 1 \ 2 & 0 & -1 \ 1 & 1 & -2 \end{bmatrix} \]

Step 4: Solve Part (b)

To solve for part (b), use \(X = A^{-1}B\), where \(B = \begin{bmatrix} 10 \ 14 \ 13 \end{bmatrix}\).Multiply \(A^{-1}\) by \(B\): \[ X = \begin{bmatrix} -2 & 0 & 1 \ 2 & 0 & -1 \ 1 & 1 & -2 \end{bmatrix} \begin{bmatrix} 10 \ 14 \ 13 \end{bmatrix} = \begin{bmatrix} 1 \ 3 \ 2 \end{bmatrix} \]Thus, the nutritionist should feed 1, 3, and 2 ounces of each type of food, respectively, to get the required nutrients.

Step 5: Solve Part (c)

For part (c), let \(B = \begin{bmatrix} 9 \ 12 \ 10 \end{bmatrix}\), apply the inverse matrix to this new \(B\):\[ X = A^{-1}B = \begin{bmatrix} -2 & 0 & 1 \ 2 & 0 & -1 \ 1 & 1 & -2 \end{bmatrix} \begin{bmatrix} 9 \ 12 \ 10 \end{bmatrix} = \begin{bmatrix} 2 \ 2 \ 1 \end{bmatrix} \]The nutritionist needs to provide 2, 2, and 1 ounce of each food type, respectively.

Step 6: Solve Part (d)

For part (d), check if \(B = \begin{bmatrix} 2 \ 4 \ 11 \end{bmatrix}\) has a solution via \(X = A^{-1}B\): \[ X = \begin{bmatrix} -2 & 0 & 1 \ 2 & 0 & -1 \ 1 & 1 & -2 \end{bmatrix} \begin{bmatrix} 2 \ 4 \ 11 \end{bmatrix} = \begin{bmatrix} 7 \ -5 \ 0 \end{bmatrix} \]One component is negative, meaning one food type would need a negative quantity, which is not possible. Therefore, no combination can satisfy this requirement.

Key Concepts

Determinant of a Matrix

To understand how to find the inverse of a matrix, we first need to calculate its determinant. Think of the determinant as a special numeric value that gives us insight into whether the matrix can be inverted or not.
For a given 3x3 matrix like \[ A = \begin{bmatrix} 3 & 1 & 3 \ 4 & 2 & 4 \ 3 & 2 & 4 \end{bmatrix} \], it's composed by several calculations based on the elements of the matrix.

• First, choose any row or column (usually the first row is the easiest to work with).
• Next, multiply each element of the row by its corresponding minor's determinant (a 2x2 matrix derived by removing the row and column of the given element).
• Adjust these calculations for each element by its cofactor, which means alternating the sign based on its position.
• Finally, sum up all these values to get the determinant.

If the determinant is zero, the matrix has no inverse. For the matrix A, we calculated it and found a determinant value different from zero, meaning it can be inverted.

Adjugate Matrix

Once we have the determinant, the next step in finding the inverse is the adjugate matrix. The adjugate of a matrix is a special transpose of the cofactor matrix, which we use in the formula for the inverse. Here's a simple breakdown of how to compute it:

• Analyze each element of the original matrix.
• Compute the minor for each element by removing its row and column, then find the determinant of this minor.
• Apply the cofactor technique, which involves adjusting this minor value by a sign pattern: positive, negative, and so on.
• Once you have all cofactors, arrange them similarly to the original matrix positions but transpose them (switch rows and columns).

For our example matrix A, after finding each cofactor and transposing them, we derived the adjugate matrix. This adjugate, alongside the determinant, allows us to calculate the inverse by multiplying each adjugate element by the reciprocal of the determinant.

Matrix Equation

Matrix equations provide a concise way to express and solve various linear algebra problems, especially when dealing with systems of equations.
The general form of a matrix equation is \(AX = B\), where \(A\) is a square matrix, \(X\) is a column vector representing the unknowns, and \(B\) is another column vector representing the constants or outcomes.
Here's how we solve a system using an inverse matrix:

• Ensure the matrix \(A\) is invertible. This means its determinant does not equal zero.
• Calculate the inverse of \(A\), denoted as \(A^{-1}\).
• Multiply both sides of the equation by \(A^{-1}\) to isolate \(X\). Essentially, this step transforms the equation to \(X = A^{-1}B\).

The solution is contained in this vector \(X\). Each element of \(X\) tells us the required values to fulfill the conditions given by vector \(B\).
In our example, this allows the nutritionist to determine the precise quantities of each food type to reach the exact nutrient goals for folic acid, choline, and inositol.