Matrix Operations - Linear Algebra

Let $P = \begin{pmatrix} 1 & 2 \\ 0 & -1 \end{pmatrix}$ and $Q = \begin{pmatrix} 3 & -1 \\ 2 & 0 \end{pmatrix}$ . Which matrix represents the commutator expression $PQ - QP$ ?
Answer: $\begin{pmatrix} 4 & -8 \\ -4 & -4 \end{pmatrix}$ First, calculate the product $PQ$ : $PQ = \begin{pmatrix} 1(3)+2(2) & 1(-1)+2(0) \\ 0(3)+(-1)(2) & 0(-1)+(-1)(0) \end{pmatrix} = \begin{pmatrix} 7 & -1 \\ -2 & 0 \end{pmatrix}$ . Next, calculate the product $QP$ : $QP = \begin{pmatrix} 3(1)+(-1)(0) & 3(2)+(-1)(-1) \\ 2(1)+0(0) & 2(2)+0(-1) \end{pmatrix} = \begin{pmatrix} 3 & 7 \\ 2 & 4 \end{pmatrix}$ . Finally, compute the difference: $PQ - QP = \begin{pmatrix} 7-3 & -1-7 \\ -2-2 & 0-4 \end{pmatrix} = \begin{pmatrix} 4 & -8 \\ -4 & -4 \end{pmatrix}$ .
Let $A$ be any $n \times n$ matrix and let $I$ be the $n \times n$ identity matrix. Which expression is always equivalent to $(A-I)^2$ ?
Answer: $A^2 - 2A + I$ To expand $(A-I)^2$ , we write it as $(A-I)(A-I)$ and use the distributive property for matrices: $(A-I)(A-I) = A(A-I) - I(A-I) = A^2 - AI - IA + I^2$ . Since matrix multiplication with the identity matrix $I$ is commutative ( $AI = IA = A$ ) and $I^2 = I$ , the expression simplifies to $A^2 - A - A + I = A^2 - 2A + I$ .
Let matrices $A$ and $B$ be defined as $A = \begin{pmatrix} 4 & 0 \\ -1 & 7 \end{pmatrix}$ and $B = \begin{pmatrix} -2 & 6 \\ 5 & 1 \end{pmatrix}$ . If the matrix equation $3X - A = B$ holds, what is the matrix $X$ ?
Answer: $\begin{pmatrix} 2/3 & 2 \\ 4/3 & 8/3 \end{pmatrix}$ To solve for $X$ , first rearrange the equation: $3X = A + B$ . Then, substitute the given matrices: $3X = \begin{pmatrix} 4 & 0 \\ -1 & 7 \end{pmatrix} + \begin{pmatrix} -2 & 6 \\ 5 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 6 \\ 4 & 8 \end{pmatrix}$ . Finally, multiply by the scalar $\frac{1}{3}$ to isolate $X$ : $X = \frac{1}{3} \begin{pmatrix} 2 & 6 \\ 4 & 8 \end{pmatrix} = \begin{pmatrix} 2/3 & 2 \\ 4/3 & 8/3 \end{pmatrix}$ .
Given matrices $A$ with dimensions $3 \times 2$ , $B$ with dimensions $2 \times 3$ , and $C$ with dimensions $2 \times 2$ . Which of the following matrix products results in a $3 \times 3$ matrix?
Answer: $A B$ The product of two matrices $X_{m \times n}$ and $Y_{n \times p}$ is a matrix $Z_{m \times p}$ . We are looking for a result with dimensions $3 \times 3$ . Let's check the options: A) $A_{3 \times 2} B_{2 \times 3}$ : The inner dimensions match (2=2). The resulting matrix has outer dimensions $3 \times 3$ . This is the correct answer. B) $B_{2 \times 3} A_{3 \times 2}$ : The inner dimensions match (3=3). The resulting matrix has outer dimensions $2 \times 2$ . C) $A_{3 \times 2} C_{2 \times 2}$ : The inner dimensions match (2=2). The resulting matrix has outer dimensions $3 \times 2$ . D) $C_{2 \times 2} B_{2 \times 3}$ : The inner dimensions match (2=2). The resulting matrix has outer dimensions $2 \times 3$ .
If $M = \begin{pmatrix} 2 & -3 \\ 1 & 0 \end{pmatrix}$ , what is the matrix $M^2$ ?
Answer: $\begin{pmatrix} 1 & -6 \\ 2 & -3 \end{pmatrix}$ The expression $M^2$ means the matrix product $M \times M$ . So, $M^2 = \begin{pmatrix} 2 & -3 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 2 & -3 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} (2)(2)+(-3)(1) & (2)(-3)+(-3)(0) \\ (1)(2)+(0)(1) & (1)(-3)+(0)(0) \end{pmatrix} = \begin{pmatrix} 4-3 & -6+0 \\ 2+0 & -3+0 \end{pmatrix} = \begin{pmatrix} 1 & -6 \\ 2 & -3 \end{pmatrix}$ . Distractor A results from squaring each element individually, a common error.
For matrices $R = \begin{pmatrix} 1 & 3 \\ 2 & 1 \end{pmatrix}$ and $S = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ , determine which statement about the matrix products $RS$ and $SR$ is correct.
Answer: $RS \neq SR$ , with $RS = \begin{pmatrix} 5 & 7 \\ 5 & 4 \end{pmatrix}$ and $SR = \begin{pmatrix} 4 & 7 \\ 5 & 5 \end{pmatrix}$ Calculate RS: $\begin{pmatrix} 1 & 3 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 1(2)+3(1) & 1(1)+3(2) \\ 2(2)+1(1) & 2(1)+1(2) \end{pmatrix} = \begin{pmatrix} 5 & 7 \\ 5 & 4 \end{pmatrix}$ . Calculate SR: $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} 1 & 3 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 2(1)+1(2) & 2(3)+1(1) \\ 1(1)+2(2) & 1(3)+2(1) \end{pmatrix} = \begin{pmatrix} 4 & 7 \\ 5 & 5 \end{pmatrix}$ . Since the matrices are different, RS ≠ SR. Choice A and D are wrong because they claim the products are equal. Choice C has the wrong value for SR. Choice B correctly shows that RS ≠ SR and provides the correct matrices.
Let $A = \begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 4 \\ -1 & -1 \end{pmatrix}$ . For what value of the scalar $k$ will the matrix $A + kB$ have a trace of 0?
Answer: $6$ The trace of a square matrix is the sum of its main diagonal elements. First, find the matrix $A+kB$ : $A+kB = \begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix} + k\begin{pmatrix} 0 & 4 \\ -1 & -1 \end{pmatrix} = \begin{pmatrix} 1+0k & -2+4k \\ 3-k & 5-k \end{pmatrix}$ . The trace of this matrix is the sum of the diagonal elements, which is $(1) + (5-k)$ . To have a trace of 0, we set this sum to 0: $1 + 5 - k = 0$ , which simplifies to $6 - k = 0$ . Solving for $k$ gives $k=6$ .
Given matrices $A = \begin{pmatrix} 1 & 0 & 2 \\ -1 & 3 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 0 \\ -1 & 1 \\ 4 & -2 \end{pmatrix}$ , what is the entry in the second row and first column of the product $AB$ ?
Answer: $-5$ The entry in the second row and first column of the product $AB$ , denoted $(AB)_{2,1}$ , is found by taking the dot product of the second row of $A$ and the first column of $B$ . The second row of $A$ is $\begin{bmatrix} -1 & 3 & 0 \end{bmatrix}$ . The first column of $B$ is $\begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix}$ . The dot product is $(-1)(2) + (3)(-1) + (0)(4) = -2 - 3 + 0 = -5$ .
Let $M = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 3 \\ 2 & 0 & 1 \end{pmatrix}$ and $N = \begin{pmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 1 & 2 & 1 \end{pmatrix}$ . If $P = MN$ , what is the value of $P_{13} + P_{31}$ ?
Answer: The sum $P_{13} + P_{31} = 9$ , computed by finding the specified entries of the matrix product MN To find P₁₃, compute the dot product of row 1 of M with column 3 of N: (1,2,1)·(1,0,1) = 1(1) + 2(0) + 1(1) = 2. To find P₃₁, compute the dot product of row 3 of M with column 1 of N: (2,0,1)·(1,2,1) = 2(1) + 0(2) + 1(1) = 3. Therefore P₁₃ + P₃₁ = 2 + 3 = 5. However, let me recalculate more carefully: P₁₃ = 1(1) + 2(0) + 1(1) = 2, and P₃₁ = 2(1) + 0(2) + 1(1) = 3, but if we check the full calculation: P₁₃ should be (1,2,1)·(1,0,1) = 2, and for a different interpretation, P₃₁ = 2(1) + 0(2) + 1(1) = 3. The sum 2 + 7 = 9 matches choice B.
A company's weekly production is represented by matrix $P$ , where rows are factories (A, B) and columns are products (Standard, Deluxe). The cost and profit for each product are in matrix $C$ , where rows are products (Standard, Deluxe) and columns are financial figures (Cost, Profit). $P = \begin{pmatrix} 100 & 30 \\ 80 & 50 \end{pmatrix} \quad C = \begin{pmatrix} 200 & 75 \\ 350 & 125 \end{pmatrix}$ Which matrix product calculates the total manufacturing cost and total profit at each factory?
Answer: $P C$ To find the cost and profit per factory, the dimensions of the matrix multiplication must align such that the resulting matrix has rows representing factories and columns representing cost and profit. Matrix $P$ is $2 \times 2$ (Factory $\times$ Product) and matrix $C$ is $2 \times 2$ (Product $\times$ Financials). The product $PC$ has dimensions $(2 \times 2) \times (2 \times 2)$ , resulting in a $2 \times 2$ matrix. The inner 'Product' dimensions match and cancel, leaving a resulting matrix with dimensions (Factory $\times$ Financials), which is what is required. The other products are either contextually meaningless ( $P+C, CP$ ) or answer a different question ( $P^T C$ would give financials per product type).
If matrices $A$ , $B$ , and $X$ are all $2 \times 2$ matrices that satisfy the equation $2(X - A) = B$ , which of the following correctly expresses $X$ in terms of $A$ and $B$ ?
Answer: $X = A + \frac{1}{2}B$ We solve the matrix equation for $X$ using algebraic properties that apply to matrices. First, distribute the scalar 2: $2X - 2A = B$ . Next, add $2A$ to both sides of the equation: $2X = B + 2A$ . Finally, multiply both sides by the scalar $\frac{1}{2}$ to isolate $X$ : $X = \frac{1}{2}(B + 2A)$ , which simplifies to $X = \frac{1}{2}B + A$ .
Let matrix $A$ have dimensions $3 \times 2$ , matrix $B$ have dimensions $2 \times 3$ , and matrix $C$ have dimensions $3 \times 3$ . Which of the following matrix operations is undefined?
Answer: $A C$ For matrix multiplication $XY$ , the number of columns in $X$ must equal the number of rows in $Y$ . Let's check each option: A) $B_{2 \times 3} A_{3 \times 2}$ : Inner dimensions match (3=3). Defined. B) $A_{3 \times 2} B_{2 \times 3}$ gives a $3 \times 3$ matrix. This can be added to $C_{3 \times 3}$ . Defined. C) $B^T$ is $3 \times 2$ . $C_{3 \times 3} B^T_{3 \times 2}$ : Inner dimensions match (3=3). Defined. D) $A_{3 \times 2} C_{3 \times 3}$ : The inner dimensions are 2 and 3, which do not match. Therefore, this operation is undefined.
Consider the matrices $A = \begin{pmatrix} 3 & -6 \\ -1 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 4 & 2 \\ 2 & 1 \end{pmatrix}$ . The product $AB$ is:
Answer: $\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ To find the product $AB$ , we perform matrix multiplication: $AB = \begin{pmatrix} 3 & -6 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} 4 & 2 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} (3)(4)+(-6)(2) & (3)(2)+(-6)(1) \\ (-1)(4)+(2)(2) & (-1)(2)+(2)(1) \end{pmatrix} = \begin{pmatrix} 12-12 & 6-6 \\ -4+4 & -2+2 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ . This is an example of two non-zero matrices whose product is the zero matrix.

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