Matrix Operations - Finite Mathematics

1. Let $A = \begin{pmatrix} 3 & 1 \\ 4 & 2 \end{pmatrix}$ and $I$ be the $2 \times 2$ identity matrix. Find the value of the scalar $k$ such that $A^2 - 5A = kI$.
   Answer: -2 This problem involves matrix multiplication, scalar multiplication, and subtraction to solve for an unknown scalar. Step 1: Compute $A^2$. $A^2 = \begin{pmatrix} 3 & 1 \\ 4 & 2 \end{pmatrix} \begin{pmatrix} 3 & 1 \\ 4 & 2 \end{pmatrix} = \begin{pmatrix} 3(3)+1(4) & 3(1)+1(2) \\ 4(3)+2(4) & 4(1)+2(2) \end{pmatrix} = \begin{pmatrix} 13 & 5 \\ 20 & 8 \end{pmatrix}$ Step 2: Compute $5A$. $5A = 5 \begin{pmatrix} 3 & 1 \\ 4 & 2 \end{pmatrix} = \begin{pmatrix} 15 & 5 \\ 20 & 10 \end{pmatrix}$ Step 3: Compute the left side of the equation, $A^2 - 5A$. $A^2 - 5A = \begin{pmatrix} 13 & 5 \\ 20 & 8 \end{pmatrix} - \begin{pmatrix} 15 & 5 \\ 20 & 10 \end{pmatrix} = \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix}$ Step 4: Set the result equal to $kI$ and solve for $k$. $kI = k \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$ $\begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} = \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$ By comparing the matrices, we see that $k = -2$. Distractor C comes from a sign error or from using the determinant of $A$, which is $3(2)-1(4)=2$. Distractors A and D come from misinterpreting the scalar in the equation.
2. Based on the information, what is the total profit for Store 2?
   Answer: $3,740 To find the total profit, we first need to calculate the total wholesale cost and total retail revenue for each store. This can be done by computing the matrix product $QP$. The resulting matrix will have rows representing stores and columns representing total cost and total revenue. $QP = \begin{pmatrix} 100 & 150 & 200 \\ 120 & 180 & 170 \end{pmatrix} \begin{pmatrix} 10 & 15 \\ 20 & 28 \\ 30 & 40 \end{pmatrix}$ $(QP)_{11} = 100(10) + 150(20) + 200(30) = 1000 + 3000 + 6000 = 10000$ $(QP)_{12} = 100(15) + 150(28) + 200(40) = 1500 + 4200 + 8000 = 13700$ $(QP)_{21} = 120(10) + 180(20) + 170(30) = 1200 + 3600 + 5100 = 9900$ $(QP)_{22} = 120(15) + 180(28) + 170(40) = 1800 + 5040 + 6800 = 13640$ So, the resulting matrix is $QP = \begin{pmatrix} 10000 & 13700 \\ 9900 & 13640 \end{pmatrix}$. The second row corresponds to Store 2. The total wholesale cost for Store 2 is $9,900, and the total retail revenue is $13,640. The profit is the difference: Profit = Revenue - Cost = $13,640 - $9,900 = $3,740. Distractor A is the profit for Store 1 ($13,700 - $10,000 = $3,700). Distractor C is the total wholesale cost for Store 2. Distractor D is the total retail revenue for Store 2.
3. How many units of Raw Material 2 are required to produce one unit of Final Product 1?
   Answer: 380 To find the number of units of each raw material needed for each final product, we need to multiply the matrices $R$ and $P$. The resulting matrix, $M = RP$, will show the units of raw material $i$ (row) needed for final product $k$ (column). $M = RP = \begin{pmatrix} 10 & 5 & 8 \\ 4 & 6 & 12 \end{pmatrix} \begin{pmatrix} 50 & 60 \\ 20 & 10 \\ 5 & 30 \end{pmatrix}$ We need the entry corresponding to Raw Material 2 (row 2) and Final Product 1 (column 1), which is $M_{21}$. $M_{21} = (\text{row 2 of } R) \cdot (\text{column 1 of } P)$ $M_{21} = (4)(50) + (6)(20) + (12)(5) = 200 + 120 + 60 = 380$ So, 380 units of Raw Material 2 are needed for one unit of Final Product 1. Distractor A, 240, comes from calculating the product in the wrong order, $PR$, which is not meaningful in this context. Distractor C, 640, is the amount of Raw Material 1 needed for Final Product 1 ($M_{11}$). Distractor D, 660, is the amount of Raw Material 2 needed for Final Product 2 ($M_{22}$).
4. Let matrices $A = \begin{pmatrix} 1 & 5 \\ -2 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix}$. If $X$ is a $2 \times 2$ matrix such that $2X + A = B^2$, which of the following is matrix $X$?
   Answer: $\begin{pmatrix} 1 & -3.5 \\ 2 & -2 \end{pmatrix}$ The problem requires solving the matrix equation $2X + A = B^2$ for $X$. First, rearrange the equation to isolate $X$: $X = \frac{1}{2}(B^2 - A)$. Step 1: Calculate $B^2$. $B^2 = B \times B = \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} (2)(2)+(-1)(1) & (2)(-1)+(-1)(0) \\ (1)(2)+(0)(1) & (1)(-1)+(0)(0) \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ 2 & -1 \end{pmatrix}$ Step 2: Calculate $B^2 - A$. $B^2 - A = \begin{pmatrix} 3 & -2 \\ 2 & -1 \end{pmatrix} - \begin{pmatrix} 1 & 5 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} 3-1 & -2-5 \\ 2-(-2) & -1-3 \end{pmatrix} = \begin{pmatrix} 2 & -7 \\ 4 & -4 \end{pmatrix}$ Step 3: Multiply by the scalar $\frac{1}{2}$. $X = \frac{1}{2} \begin{pmatrix} 2 & -7 \\ 4 & -4 \end{pmatrix} = \begin{pmatrix} 1 & -3.5 \\ 2 & -2 \end{pmatrix}$ Distractor B comes from incorrectly solving for $X$ as $X = \frac{1}{2}(A - B^2)$, a sign error. Distractor C results from incorrectly calculating $B^2$ by squaring each element of $B$. Distractor D arises from incorrectly calculating $X = \frac{1}{2}(B^2 + A)$.
5. A transformation matrix $T = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ satisfies $T \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and $T \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$. What is the value of $\det(2T)$?
   Answer: 56 From the given conditions, T(1,0) = (3,-2) gives us the first column of T is (3,-2). T(0,1) = (1,4) gives us the second column of T is (1,4). Therefore T = (3 1; -2 4). The determinant of T is (3)(4) - (1)(-2) = 12 + 2 = 14. For matrix 2T, det(2T) = 2² × det(T) = 4 × 14 = 56.
6. Let $A$ be a $3 \times 2$ matrix, $B$ be a $2 \times 3$ matrix, $C$ be a $3 \times 3$ matrix, and $D$ be a $2 \times 2$ matrix. Which of the following matrix expressions is defined?
   Answer: $D + BA$ To determine if a matrix expression is defined, we must check the dimensions for each operation. A) For $A(B+C)$, the addition $B+C$ requires matrices of the same dimensions. $B$ is $2 \times 3$ and $C$ is $3 \times 3$, so they cannot be added. The expression is undefined. B) For $C^2 - B$, the operation $C^2 = C \times C$ is defined because $C$ is square ($3 \times 3$). The result is a $3 \times 3$ matrix. However, to subtract $B$, which is $2 \times 3$, the dimensions must match. Since $3 \times 3 \neq 2 \times 3$, the subtraction is not possible. The expression is undefined. C) For $D + BA$, first consider the product $BA$. The inner dimensions must match. $B$ is $2 \times 3$ and $A$ is $3 \times 2$. The inner dimensions (3 and 3) match, so $BA$ is defined. The resulting matrix $BA$ has dimensions $2 \times 2$. The addition $D + BA$ is defined because both $D$ and $BA$ are $2 \times 2$ matrices. This expression is defined. D) For $AC+D$, consider the product $AC$. $A$ is $3 \times 2$ and $C$ is $3 \times 3$. The inner dimensions (2 and 3) do not match, so the product $AC$ is undefined. The entire expression is undefined.
7. Consider the matrices $A = \begin{pmatrix} 1 & 0 & 2 \\ -1 & 3 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 0 & 1 & 2 \\ 4 & 1 & 0 & -1 \\ 0 & 2 & 3 & 1 \end{pmatrix}$, and $C = \begin{pmatrix} 5 & 1 \\ 0 & 2 \\ -1 & 0 \\ 2 & 3 \end{pmatrix}$. What is the entry in the second row, first column of the product matrix $ABC$?
   Answer: 45 We need to find the entry $(ABC)_{21}$. This can be found without computing the entire matrix product. Let $D = AB$. The entry $(ABC)_{21}$ is the same as $(DC)_{21}$, which is the dot product of the second row of $D$ and the first column of $C$. Step 1: Find the second row of $D = AB$. The second row of $AB$ is found by multiplying the second row of $A$ by the matrix $B$. $(\text{row 2 of } A) \times B = \begin{pmatrix} -1 & 3 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 1 & 2 \\ 4 & 1 & 0 & -1 \\ 0 & 2 & 3 & 1 \end{pmatrix}$ This results in a $1 \times 4$ row vector: $\begin{pmatrix} -1(1)+3(4)+1(0) & -1(0)+3(1)+1(2) & -1(1)+3(0)+1(3) & -1(2)+3(-1)+1(1) \end{pmatrix}$ $= \begin{pmatrix} 11 & 5 & 2 & -4 \end{pmatrix}$ This is the second row of the matrix $D = AB$. Step 2: Find the dot product of this row with the first column of $C$. $(\text{row 2 of } AB) \cdot (\text{col 1 of } C) = \begin{pmatrix} 11 & 5 & 2 & -4 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ 0 \\ -1 \\ 2 \end{pmatrix}$ $= 11(5) + 5(0) + 2(-1) + (-4)(2) = 55 + 0 - 2 - 8 = 45$ Thus, the entry $(ABC)_{21}$ is 45. Distractor A, 9, is $(ABC)_{22}$. Distractor B, 21, is $(ABC)_{12}$. Distractor D, 61, is a calculation error, likely a sign error in the final step ($55-2+8$).
8. Let $A = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 0 & 1 \\ -1 & 2 \end{pmatrix}$, and $C = \begin{pmatrix} 2 & 3 \\ 1 & 8 \end{pmatrix}$. If $x$ and $y$ are scalars such that $xA + yB = C$, what is the value of $x+y$?
   Answer: 5 The matrix equation $xA + yB = C$ translates into a system of linear equations for the scalars $x$ and $y$. $x \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} + y \begin{pmatrix} 0 & 1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 1 & 8 \end{pmatrix}$ Performing the scalar multiplication and addition on the left side gives: $\begin{pmatrix} x & 0 \\ 2x & x \end{pmatrix} + \begin{pmatrix} 0 & y \\ -y & 2y \end{pmatrix} = \begin{pmatrix} x & y \\ 2x-y & x+2y \end{pmatrix}$ Setting this equal to matrix $C$ gives a system of four equations: 1) $x = 2$ 2) $y = 3$ 3) $2x - y = 1$ 4) $x + 2y = 8$ From the first two equations, we get the potential solution $x=2$ and $y=3$. We must check if these values satisfy the other two equations. For equation 3: $2(2) - 3 = 4 - 3 = 1$. This is true. For equation 4: $2 + 2(3) = 2 + 6 = 8$. This is also true. Since the solution $x=2, y=3$ is consistent for all entries, the scalars exist. The question asks for the value of $x+y$, which is $2+3=5$. Distractors A and B result from errors in solving the system. Distractor D would be correct if the system were inconsistent.
9. Let $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $I$ be the $2 \times 2$ identity matrix. If $B = I + A$, what is the matrix $B^5$?
   Answer: $\begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix}$ This problem can be solved by observing a pattern in the powers of $B$. First, note the properties of matrix $A$: $A^2 = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = Z$ (the zero matrix). All higher powers $A^k$ for $k \ge 2$ will also be the zero matrix. Now, let's compute the first few powers of $B = I+A$. Since $I$ and $A$ commute ($IA=AI=A$), we can use the binomial expansion. $B^2 = (I+A)^2 = I^2 + 2IA + A^2 = I + 2A + Z = I + 2A$. $B^3 = (I+A)^3 = I^3 + 3I^2A + 3IA^2 + A^3 = I + 3A + 3Z + Z = I + 3A$. The pattern is $B^n = I + nA$. Therefore, for $n=5$: $B^5 = I + 5A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + 5 \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 0 & 5 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix}$ Distractor B is $B$ itself. Distractor C represents $5B$. Distractor D is the identity matrix, which might be chosen if one mistakenly thought $(I+A)$ somehow simplifies to $I$.
10. Let $A = \begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 0 \\ 1 & 4 \end{pmatrix}$. What is the matrix product $AB - BA$?
    Answer: $\begin{pmatrix} 2 & 4 \\ -8 & -2 \end{pmatrix}$ This question tests matrix multiplication and subtraction, and highlights that matrix multiplication is not commutative ($AB \neq BA$). First, compute $AB$: $AB = \begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 1 & 4 \end{pmatrix} = \begin{pmatrix} 1(2)+2(1) & 1(0)+2(4) \\ 3(2)+(-1)(1) & 3(0)+(-1)(4) \end{pmatrix} = \begin{pmatrix} 4 & 8 \\ 5 & -4 \end{pmatrix}$ Next, compute $BA$: $BA = \begin{pmatrix} 2 & 0 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix} = \begin{pmatrix} 2(1)+0(3) & 2(2)+0(-1) \\ 1(1)+4(3) & 1(2)+4(-1) \end{pmatrix} = \begin{pmatrix} 2 & 4 \\ 13 & -2 \end{pmatrix}$ Finally, compute the difference $AB - BA$: $AB - BA = \begin{pmatrix} 4 & 8 \\ 5 & -4 \end{pmatrix} - \begin{pmatrix} 2 & 4 \\ 13 & -2 \end{pmatrix} = \begin{pmatrix} 4-2 & 8-4 \\ 5-13 & -4-(-2) \end{pmatrix} = \begin{pmatrix} 2 & 4 \\ -8 & -2 \end{pmatrix}$ Distractor A is chosen by students who incorrectly assume matrix multiplication is commutative, making $AB - BA$ the zero matrix. Distractor B is the result of adding $AB+BA$. Distractor C is the result of $BA - AB$, which is a sign error.
11. Given matrices $A = \begin{pmatrix} 1 & 0 \\ 3 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} -1 & 4 \\ 1 & 0 \end{pmatrix}$, what is the result of the operation $(A + B^T)^2$?
    Answer: $\begin{pmatrix} 7 & 2 \\ 14 & 11 \end{pmatrix}$ This is a multi-step problem involving matrix transpose, addition, and multiplication. Step 1: Find the transpose of $B$, denoted $B^T$. $B^T = \begin{pmatrix} -1 & 1 \\ 4 & 0 \end{pmatrix}$ Step 2: Add $A$ and $B^T$. $A + B^T = \begin{pmatrix} 1 & 0 \\ 3 & 2 \end{pmatrix} + \begin{pmatrix} -1 & 1 \\ 4 & 0 \end{pmatrix} = \begin{pmatrix} 1+(-1) & 0+1 \\ 3+4 & 2+0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 7 & 2 \end{pmatrix}$ Step 3: Square the resulting matrix. $(A + B^T)^2 = \begin{pmatrix} 0 & 1 \\ 7 & 2 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 7 & 2 \end{pmatrix} = \begin{pmatrix} 0(0)+1(7) & 0(1)+1(2) \\ 7(0)+2(7) & 7(1)+2(2) \end{pmatrix} = \begin{pmatrix} 7 & 2 \\ 14 & 11 \end{pmatrix}$ Distractor B is the result of computing $(A+B)^2$, failing to take the transpose of $B$. Distractor C is the result of incorrectly assuming that $(X+Y)^2 = X^2 + Y^2$ for matrices and calculating $A^2+(B^T)^2$. Distractor D is the result of squaring each element of $(A+B^T)$ instead of performing matrix multiplication.
12. Let $A$ and $B$ be $n \times n$ matrices. If the equation $(A+B)^2 = A^2 + 2AB + B^2$ holds true, which of the following statements must be true?
    Answer: $A$ and $B$ must commute. This question tests the understanding of matrix multiplication properties, specifically its non-commutativity. Let's expand the left side of the equation: $(A+B)^2 = (A+B)(A+B) = A(A+B) + B(A+B) = A^2 + AB + BA + B^2$ Now, set this expanded form equal to the right side of the given equation: $A^2 + AB + BA + B^2 = A^2 + 2AB + B^2$ Subtract $A^2$ and $B^2$ from both sides: $AB + BA = 2AB$ Subtract $AB$ from both sides: $BA = AB$ This final equation is the definition of commutativity for matrices $A$ and $B$. Therefore, $A$ and $B$ must commute. Distractors A and B are sufficient conditions for the equation to hold, but they are not necessary. For example, if $A = 2I$ and $B = 3I$, neither is the zero matrix, but they commute, and the equation holds. Distractor C is irrelevant; invertibility is not required for the property to hold.
13. Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$. If the matrices commute, meaning $AB = BA$, which of the following conditions on the entries of $A$ must be true?
    Answer: $a=d$ and $b=c$ To find the condition for commutativity, we compute both products $AB$ and $BA$ and set them equal to each other. First, compute $AB$: $AB = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} a(1)+b(1) & a(1)+b(1) \\ c(1)+d(1) & c(1)+d(1) \end{pmatrix} = \begin{pmatrix} a+b & a+b \\ c+d & c+d \end{pmatrix}$ Next, compute $BA$: $BA = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1(a)+1(c) & 1(b)+1(d) \\ 1(a)+1(c) & 1(b)+1(d) \end{pmatrix} = \begin{pmatrix} a+c & b+d \\ a+c & b+d \end{pmatrix}$ Now, set $AB = BA$: $\begin{pmatrix} a+b & a+b \\ c+d & c+d \end{pmatrix} = \begin{pmatrix} a+c & b+d \\ a+c & b+d \end{pmatrix}$ For these matrices to be equal, their corresponding entries must be equal. This gives a system of equations: 1) $a+b = a+c \implies b=c$ 2) $a+b = b+d \implies a=d$ 3) $c+d = a+c \implies d=a$ 4) $c+d = b+d \implies c=b$ All four equations lead to the same two conditions: $a=d$ and $b=c$. This means the diagonal elements must be equal, and the off-diagonal elements must be equal. Distractors A and B represent other forms of symmetry in the matrix that are not the required condition. Distractor D is a condition related to the trace, but it is not sufficient for commutativity with this specific matrix $B$.

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