Matrix Equations for Systems - Finite Mathematics

1. A dietician is planning a meal consisting of three foods: Food 1, Food 2, and Food 3. Each unit of Food 1 contains 10g of protein, 20g of carbohydrates, and 5g of fat. Each unit of Food 2 contains 15g of protein, 10g of carbohydrates, and 10g of fat. Each unit of Food 3 contains 20g of protein, 25g of carbohydrates, and 8g of fat. The meal must provide exactly 215g of protein, 290g of carbohydrates, and 121g of fat. Let $x_1, x_2$, and $x_3$ be the number of units of each food, respectively. If this system is modeled by the matrix equation $AX = B$, what is the coefficient matrix $A$?
   Answer: $A = \begin{pmatrix} 10 & 15 & 20 \\ 20 & 10 & 25 \\ 5 & 10 & 8 \end{pmatrix}$ The system of linear equations is derived by considering each nutrient. Let $x_1, x_2, x_3$ be the amounts of Food 1, Food 2, and Food 3. The equations are: Protein: $10x_1 + 15x_2 + 20x_3 = 215$ Carbohydrates: $20x_1 + 10x_2 + 25x_3 = 290$ Fat: $5x_1 + 10x_2 + 8x_3 = 121$ In the matrix equation $AX=B$, the coefficient matrix $A$ consists of the coefficients of the variables. Each row corresponds to an equation (a nutrient), and each column corresponds to a variable (a food). Thus, the correct matrix is $A = \begin{pmatrix} 10 & 15 & 20 \\ 20 & 10 & 25 \\ 5 & 10 & 8 \end{pmatrix}$.
2. A furniture company manufactures chairs, tables, and cabinets. A chair requires 1 hour in cutting, 2 in assembly, and 1 in finishing. A table requires 2 hours in cutting, 3 in assembly, and 1 in finishing. A cabinet requires 3 hours in cutting, 4 in assembly, and 2 in finishing. The cutting department has 350 hours available, assembly has 500, and finishing has 225. Let $x, y$, and $z$ be the number of chairs, tables, and cabinets produced, respectively. Which matrix equation would be solved to determine the number of each product to manufacture to use all available departmental hours?
   Answer: $\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 1 & 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 350 \\ 500 \\ 225 \end{pmatrix}$ The system of equations is set up by considering the constraints of each department. Each row in the coefficient matrix will represent a department. Cutting: $1x + 2y + 3z = 350$ Assembly: $2x + 3y + 4z = 500$ Finishing: $1x + 1y + 2z = 225$ This system translates to the matrix equation $AX=B$, where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix. This corresponds exactly to the equation shown in option C.
3. A company produces two models of bicycles, Model A and Model B. Assembly of Model A requires 2 hours of labor and 3 kg of steel. Assembly of Model B requires 4 hours of labor and 1 kg of steel. The company has 800 hours of labor and 450 kg of steel available per week. Let $x$ be the number of Model A bicycles and $y$ be the number of Model B bicycles produced. The situation is modeled by $CZ = D$. To find the number of each model to produce to use all resources, one would calculate the solution matrix $Z = C^{-1}D$. What does the matrix $Z$ represent?
   Answer: A $2 \times 1$ matrix where the first entry is the number of Model A bicycles and the second is the number of Model B bicycles. The system of equations is: Labor: $2x + 4y = 800$ Steel: $3x + 1y = 450$ In the matrix equation $CZ=D$, $Z$ is the variable matrix. Given that $x$ is the number of Model A bicycles and $y$ is the number of Model B bicycles, the variable matrix is set up as $Z = \begin{pmatrix} x \\ y \end{pmatrix}$. This is a $2 \times 1$ matrix where the first entry represents the quantity of Model A and the second represents the quantity of Model B. The solution $Z = C^{-1}D$ gives the specific values for $x$ and $y$.
4. A chemist needs to create 100 liters of a solution that is 27% acid. Three stock solutions are available. Stock A is 50% acid, Stock B is 20% acid, and Stock C is 10% acid. Let $a, b$, and $c$ be the number of liters used from stocks A, B, and C, respectively. In modeling this problem, one of the equations will relate the total volume, and another will relate the total amount of acid. If the system of equations for this mixture problem is written as $MZ = K$, where the first row of matrix $M$ corresponds to the total volume equation and the second row corresponds to the total acid equation, what are the first two rows of $M$?
   Answer: $\begin{pmatrix} 1 & 1 & 1 \\ 0.5 & 0.2 & 0.1 \end{pmatrix}$ The problem describes two relationships that can be turned into equations. 1. Total Volume: The sum of the volumes from each stock solution must equal the total desired volume. $a + b + c = 100$. 2. Total Acid: The sum of the amount of acid from each stock solution must equal the amount of acid in the final mixture. The amount of acid in the final mixture is $27\%$ of 100 liters, which is 27 liters. So, $0.50a + 0.20b + 0.10c = 27$. When setting up the matrix equation $MZ=K$, the coefficient matrix $M$ contains the coefficients of the variables $a, b, c$. The coefficients from the first equation are $[1, 1, 1]$. The coefficients from the second equation are $[0.5, 0.2, 0.1]$. Therefore, the first two rows of $M$ are as shown in option A.
5. A system of linear equations is given as: $2x + 5y - z = 4$ $3y + 4z = -1$ $x - y + 2z = 6$ The system is modeled by a matrix equation $AX = B$. However, for a particular computational purpose, the variable matrix is defined as $X = \begin{pmatrix} y \\ z \\ x \end{pmatrix}$. Which of the following is the correct coefficient matrix $A$ for this specific model?
   Answer: $A = \begin{pmatrix} 5 & -1 & 2 \\ 3 & 4 & 0 \\ -1 & 2 & 1 \end{pmatrix}$ When you encounter a matrix equation $AX = B$ representing a system of linear equations, the key is understanding that the coefficient matrix $A$ must align with the order of variables in your variable matrix $X$. Since the variable matrix is defined as $X = \begin{pmatrix} y \\ z \\ x \end{pmatrix}$ (not the usual $x, y, z$ order), you need to rearrange the coefficients in matrix $A$ to match this sequence. Each row of $A$ represents one equation, and the columns must correspond to the coefficients of $y$, $z$, and $x$ respectively. Let's rewrite each equation with variables in $y, z, x$ order: - First equation: $2x + 5y - z = 4$ becomes $5y - z + 2x = 4$ - Second equation: $3y + 4z = -1$ becomes $3y + 4z + 0x = -1$ - Third equation: $x - y + 2z = 6$ becomes $-y + 2z + x = 6$ Therefore, the coefficient matrix is: $A = \begin{pmatrix} 5 & -1 & 2 \\ 3 & 4 & 0 \\ -1 & 2 & 1 \end{pmatrix}$ This matches option C. Option A uses the standard $x, y, z$ ordering, ignoring the given variable arrangement. Option B has the correct coefficients but in wrong positions—it appears to swap columns incorrectly. Option D also rearranges coefficients incorrectly, mixing up the variable correspondence. Study tip: Always check the order of variables in the variable matrix first. The coefficient matrix columns must match this exact order, so rearrange the original equations accordingly before constructing your matrix.
6. Consider the following system of linear equations: $5x + 2y - z = 9$ $3x + 4z = -1$ $-x + y + 6z = 0$ If this system is represented by the matrix equation $AX = B$ where $X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$, which of the following is the coefficient matrix $A$?
   Answer: $A = \begin{pmatrix} 5 & 2 & -1 \\ 3 & 0 & 4 \\ -1 & 1 & 6 \end{pmatrix}$ When you encounter a system of linear equations that needs to be written as a matrix equation $AX = B$, you're extracting the coefficient matrix $A$ by identifying the coefficients of each variable in their proper positions. To find the coefficient matrix $A$, examine each equation and record the coefficients of $x$, $y$, and $z$ in that order for each row. Let's work through this systematically: From $5x + 2y - z = 9$: coefficients are $5, 2, -1$ From $3x + 4z = -1$: coefficients are $3, 0, 4$ (note that $y$ has coefficient $0$) From $-x + y + 6z = 0$: coefficients are $-1, 1, 6$ This gives us the coefficient matrix: $A = \begin{pmatrix} 5 & 2 & -1 \\ 3 & 0 & 4 \\ -1 & 1 & 6 \end{pmatrix}$ Looking at the wrong answers: Choice A incorrectly places $1$ instead of $0$ as the coefficient of $y$ in the second equation. Choice B places the $3$ (coefficient of $x$) in the $y$ position of the second row, shifting everything incorrectly. Choice D omits the $y$ column entirely, which would only work if $y$ appeared in no equations—but it appears in two of the three equations. The key insight is that missing variables in an equation correspond to zero coefficients in the matrix, not missing positions. When writing coefficient matrices, always maintain the same column structure for each variable, using zeros where variables don't appear. This systematic approach prevents the positioning errors seen in the incorrect choices.
7. A system of linear equations is defined in terms of a constant $k$: $2x + ky = 5$ $x - 3y = 7z$ $4x + y - kz = 3$ If the system is represented by the matrix equation $AX = B$ with $X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$, which expression correctly represents the coefficient matrix $A$?
   Answer: $A = \begin{pmatrix} 2 & k & 0 \\ 1 & -3 & -7 \\ 4 & 1 & -k \end{pmatrix}$ First, we must write all equations in the standard form $ax+by+cz=d$. 1. $2x + ky = 5 \Rightarrow 2x + ky + 0z = 5$ 2. $x - 3y = 7z \Rightarrow x - 3y - 7z = 0$ 3. $4x + y - kz = 3$ (already in standard form) The coefficient matrix $A$ is formed by the coefficients of the variables $x, y, z$ for each row. The first row is $[2, k, 0]$. The second row is $[1, -3, -7]$. The third row is $[4, 1, -k]$. Combining these gives the correct matrix $A$.
8. An investor divides $50,000 among three different mutual funds: a stock fund (S), a bond fund (B), and a money market fund (M). The annual yields are 8% for the stock fund, 5% for the bond fund, and 2% for the money market fund. The total annual income from the investments is $2,710. The amount invested in the stock fund is $5,000 more than the amount invested in the bond fund. If $s, b$, and $m$ represent the amounts invested in each fund, and the resulting system of equations is written in the form $AX = K$, where $X = \begin{pmatrix} s \\ b \\ m \end{pmatrix}$, what is the constant matrix $K$?
   Answer: $K = \begin{pmatrix} 50000 \\ 2710 \\ 5000 \end{pmatrix}$ We must set up the system of three linear equations. 1. Total investment: $s + b + m = 50000$ 2. Total annual income: $0.08s + 0.05b + 0.02m = 2710$ 3. Relationship between investments: $s = b + 5000$, which can be rewritten in standard form as $s - b + 0m = 5000$. In the matrix equation $AX = K$, the matrix $K$ contains the constant terms from the right side of these equations. Therefore, $K = \begin{pmatrix} 50000 \\ 2710 \\ 5000 \end{pmatrix}$.
9. A system of linear equations is given by: $3x - 5z = 12 - 2y$ $x + 4y - z = 7$ $6z - x = 2y - 9$ Which matrix equation of the form $AX=B$ correctly represents this system, assuming $X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$?
   Answer: $\begin{pmatrix} 3 & 2 & -5 \\ 1 & 4 & -1 \\ -1 & -2 & 6 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 12 \\ 7 \\ -9 \end{pmatrix}$ First, rewrite each equation in the standard form $ax + by + cz = d$. 1) $3x - 5z = 12 - 2y \Rightarrow 3x + 2y - 5z = 12$ 2) $x + 4y - z = 7$ (already in standard form) 3) $6z - x = 2y - 9 \Rightarrow -x - 2y + 6z = -9$ The coefficient matrix $A$ is formed by the coefficients of $x, y, z$ in each equation. The constant matrix $B$ is formed by the constants on the right side. This gives: $A = \begin{pmatrix} 3 & 2 & -5 \\ 1 & 4 & -1 \\ -1 & -2 & 6 \end{pmatrix}$ and $B = \begin{pmatrix} 12 \\ 7 \\ -9 \end{pmatrix}$. The correct matrix equation is option B.
10. The matrix equation below models a system of linear equations in variables $x, y$, and $z$. $\begin{pmatrix} 2 & 0 & -3 \\ 1 & 5 & 1 \\ -1 & 4 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 7 \\ -2 \\ 1 \end{pmatrix}$ Which of the following equations is part of the corresponding system?
    Answer: $x + 5y + z = -2$ To convert the matrix equation $AX=B$ back into a system of linear equations, we perform the matrix multiplication on the left side. Each row of the resulting matrix corresponds to one equation. Row 1: $(2)(x) + (0)(y) + (-3)(z) = 2x - 3z = 7$ Row 2: $(1)(x) + (5)(y) + (1)(z) = x + 5y + z = -2$ Row 3: $(-1)(x) + (4)(y) + (0)(z) = -x + 4y = 1$ Comparing these derived equations with the answer choices, we find that $x + 5y + z = -2$ is a part of the system.
11. A bakery sells three types of pastries: croissants ($c$), danishes ($d$), and muffins ($m$). The production is constrained by the available amounts of flour, sugar, and butter. The matrix equation $AX = B$ models this situation, where $X = \begin{pmatrix} c \\ d \\ m \end{pmatrix}$ represents the number of each pastry to produce. The matrix $A$ is $A = \begin{pmatrix} 0.1 & 0.15 & 0.2 \\ 0.05 & 0.1 & 0.12 \\ 0.08 & 0.06 & 0.05 \end{pmatrix}$ and represents the kilograms of flour (row 1), sugar (row 2), and butter (row 3) needed per pastry. If $B = \begin{pmatrix} 200 \\ 120 \\ 80 \end{pmatrix}$, what is the physical meaning of the entry $120$?
    Answer: The total amount of sugar available for production is 120 kilograms. In the matrix equation $AX=B$, the product $AX$ calculates the total resources used. Each row in this calculation corresponds to a specific resource. The problem states that row 2 of matrix $A$ represents the sugar required per pastry. Therefore, the second row of the equation $AX=B$ represents the total sugar constraint: $0.05c + 0.1d + 0.12m = 120$. The value $120$ on the right side of this equation is the total amount of that resource (sugar) available, measured in kilograms.

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