Matrix-Vector Products - Linear Algebra

Let $A = \begin{pmatrix} k & 2 \\ 1 & -3 \end{pmatrix}$ and $\mathbf{x} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$ . The product $A\mathbf{x}$ is the vector $\begin{pmatrix} -6 \\ 7 \end{pmatrix}$ . What is the value of $k$ ?
Answer: $k = -2$ The matrix-vector product $A\mathbf{x}$ is calculated as $\begin{pmatrix} k & 2 \\ 1 & -3 \end{pmatrix} \begin{pmatrix} 1 \\ -2 \end{pmatrix} = \begin{pmatrix} k(1) + 2(-2) \\ 1(1) + (-3)(-2) \end{pmatrix} = \begin{pmatrix} k-4 \\ 1+6 \end{pmatrix} = \begin{pmatrix} k-4 \\ 7 \end{pmatrix}$ We are given that this product equals $\begin{pmatrix} -6 \\ 7 \end{pmatrix}$ . The second components match ( $7=7$ ). Equating the first components gives the equation $k-4 = -6$ . Adding 4 to both sides yields $k = -2$ . (A) is incorrect; it results from a sign error in the calculation, such as $k+4=-6$ . (B) is incorrect; it could result from an error in setting up the product, such as using elements from the first column of $A$ as weights: $k(1) + 1(-2) = -6$ , which gives $k=-4$ . (D) is incorrect; it results from a sign error on the right side of the equation, such as solving $k-4=6$ .
Let $\mathbf{e}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ , $\mathbf{e}_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$ , and $\mathbf{e}_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ . If matrix $P$ satisfies $P\mathbf{e}_1 = \mathbf{e}_2$ , $P\mathbf{e}_2 = \mathbf{e}_3$ , and $P\mathbf{e}_3 = \mathbf{e}_1$ , what is $P^3\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ ?
Answer: $\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ When you see a matrix defined by how it transforms the standard basis vectors, you're dealing with a linear transformation. The key insight is recognizing what the matrix $P$ actually does and using properties of matrix powers. First, let's understand what $P$ does. Since $P\mathbf{e}_1 = \mathbf{e}_2$ , $P\mathbf{e}_2 = \mathbf{e}_3$ , and $P\mathbf{e}_3 = \mathbf{e}_1$ , the matrix $P$ creates a cyclic permutation of the standard basis vectors. This means $P = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$ . Now, what happens when we apply $P$ three times? Following the cycle: $\mathbf{e}_1 \to \mathbf{e}_2 \to \mathbf{e}_3 \to \mathbf{e}_1$ . After three applications, each basis vector returns to itself! This means $P^3 = I$ (the identity matrix), so $P^3\mathbf{v} = \mathbf{v}$ for any vector $\mathbf{v}$ . Therefore, $P^3\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ . Answer C is correct because $P^3$ is the identity transformation. Answer A gives $\begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix}$ , which would be $P^2$ applied to the vector (two steps in the cycle). Answer B gives $\begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix}$ , which would be $P$ applied once to the vector. Answer D incorrectly assumes the transformation is nilpotent (eventually gives zero). Study tip: When you see cyclic permutations, always check if the power equals the cycle length—this often gives you the identity matrix, making calculations much simpler.
Let $A = \begin{pmatrix} 2 & 5 \\ -1 & -3 \end{pmatrix}$ . The product $A\mathbf{x}$ results in the vector $\mathbf{b} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ . What is vector $\mathbf{x}$ ?
Answer: $\begin{pmatrix} 3 \\ -1 \end{pmatrix}$ The equation $A\mathbf{x} = \mathbf{b}$ corresponds to the system of linear equations: 1) $2x_1 + 5x_2 = 1$ 2) $-x_1 - 3x_2 = 0$ From equation (2), we can express $x_1$ in terms of $x_2$ : $x_1 = -3x_2$ . Now, substitute this expression for $x_1$ into equation (1): $2(-3x_2) + 5x_2 = 1$ $-6x_2 + 5x_2 = 1$ $-x_2 = 1$ , which means $x_2 = -1$ . Finally, substitute $x_2 = -1$ back into the expression for $x_1$ : $x_1 = -3(-1) = 3$ . Thus, the vector is $\mathbf{x} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}$ . (A) is incorrect; this vector yields $\begin{pmatrix} -1 \\ 0 \end{pmatrix}$ , likely due to a sign error in the final result. (B) is incorrect; it results from a sign error in solving for $x_2$ . (D) is incorrect; it results from swapping the calculated values of $x_1$ and $x_2$ .
Let $A = \begin{pmatrix} 1 & -2 \\ -3 & 6 \end{pmatrix}$ . For which of the following vectors $\mathbf{b}$ does the equation $A\mathbf{x} = \mathbf{b}$ have no solution?
Answer: $\begin{pmatrix} 3 \\ -1 \end{pmatrix}$ The equation $A\mathbf{x} = \mathbf{b}$ has a solution if and only if $\mathbf{b}$ can be written as a linear combination of the columns of $A$ . The columns of $A$ are $\mathbf{a}_1 = \begin{pmatrix} 1 \\ -3 \end{pmatrix}$ and $\mathbf{a}_2 = \begin{pmatrix} -2 \\ 6 \end{pmatrix}$ . Notice that $\mathbf{a}_2 = -2\mathbf{a}_1$ . This means the columns are linearly dependent (collinear). Any linear combination of these columns will also be a scalar multiple of $\mathbf{a}_1$ . So, a solution exists only if $\mathbf{b}$ is a scalar multiple of $\begin{pmatrix} 1 \\ -3 \end{pmatrix}$ . (A) $\begin{pmatrix} 2 \\ -6 \end{pmatrix} = 2\begin{pmatrix} 1 \\ -3 \end{pmatrix}$ . A solution exists. (B) $\begin{pmatrix} -1 \\ 3 \end{pmatrix} = -1\begin{pmatrix} 1 \\ -3 \end{pmatrix}$ . A solution exists. (C) $\begin{pmatrix} 3 \\ -1 \end{pmatrix}$ is not a scalar multiple of $\begin{pmatrix} 1 \\ -3 \end{pmatrix}$ , because $3/1 \neq -1/(-3)$ . Therefore, no solution exists for this $\mathbf{b}$ . (D) $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ always has the trivial solution $\mathbf{x} = \mathbf{0}$ , since $A\mathbf{0} = \mathbf{0}$ .
The vector $\mathbf{v} = \begin{pmatrix} 7 \\ -1 \end{pmatrix}$ is expressed as the linear combination $3\begin{pmatrix} 1 \\ 1 \end{pmatrix} - 2\begin{pmatrix} -2 \\ 2 \end{pmatrix}$ . Which of the following matrix-vector products is equivalent to this expression?
Answer: $\begin{pmatrix} 1 & -2 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ A linear combination of vectors can be written as a matrix-vector product $A\mathbf{x}$ . The columns of the matrix $A$ are the vectors in the linear combination, and the vector $\mathbf{x}$ contains the scalar weights. In the expression $3\begin{pmatrix} 1 \\ 1 \end{pmatrix} - 2\begin{pmatrix} -2 \\ 2 \end{pmatrix}$ , the vectors are $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} -2 \\ 2 \end{pmatrix}$ , and the weights are $3$ and $-2$ . Therefore, the matrix $A$ is $\begin{pmatrix} 1 & -2 \\ 1 & 2 \end{pmatrix}$ and the vector $\mathbf{x}$ is $\begin{pmatrix} 3 \\ -2 \end{pmatrix}$ . The product is $\begin{pmatrix} 1 & -2 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ . (A) is incorrect because it constructs the matrix using one column vector and one row vector from the combination. (C) is incorrect because it incorrectly assigns the scalars and vectors to the matrix and vector components. (D) is incorrect because the dimensions are incompatible for matrix multiplication in this order.
Let $A$ be an $m \times n$ matrix and let $\mathbf{u}, \mathbf{v}$ be vectors in $\mathbb{R}^n$ . Given $A\mathbf{u} = \begin{pmatrix} 1 \\ 5 \end{pmatrix}$ and $A\mathbf{v} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}$ , what is the result of the product $A(2\mathbf{u} - \mathbf{v})$ ?
Answer: $\begin{pmatrix} 4 \\ 7 \end{pmatrix}$ Matrix multiplication is a linear transformation, which means it has the properties $A(c\mathbf{x}) = c(A\mathbf{x})$ and $A(\mathbf{x} + \mathbf{y}) = A\mathbf{x} + A\mathbf{y}$ . Combining these properties, we get $A(2\mathbf{u} - \mathbf{v}) = A(2\mathbf{u}) - A(\mathbf{v}) = 2(A\mathbf{u}) - A(\mathbf{v})$ . Substituting the given values: $2(A\mathbf{u}) - A(\mathbf{v}) = 2\begin{pmatrix} 1 \\ 5 \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \\ 10 \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 - (-2) \\ 10 - 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 7 \end{pmatrix}$ . (A) is incorrect; it results from misplacing the scalar and calculating $A\mathbf{u} - 2A\mathbf{v}$ . (B) is incorrect; it results from a sign error, calculating $2A\mathbf{u} + A\mathbf{v}$ . (C) is incorrect; it results from an incorrect order of operations, calculating $2(A\mathbf{u} - A\mathbf{v})$ .
Let $A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}$ and $\mathbf{x} = \begin{pmatrix} 5 \\ -2 \end{pmatrix}$ . The product $A\mathbf{x}$ can be expressed as a linear combination of the columns of $A$ . Which of the following correctly represents this linear combination?
Answer: $5\begin{pmatrix} 2 \\ 3 \end{pmatrix} - 2\begin{pmatrix} -1 \\ 4 \end{pmatrix}$ The definition of a matrix-vector product $A\mathbf{x}$ is the linear combination of the columns of $A$ with weights given by the corresponding entries of $\mathbf{x}$ . For this problem, the columns of $A$ are $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and $\begin{pmatrix} -1 \\ 4 \end{pmatrix}$ , and the entries of $\mathbf{x}$ are $5$ and $-2$ . Therefore, the correct linear combination is $5\begin{pmatrix} 2 \\ 3 \end{pmatrix} + (-2)\begin{pmatrix} -1 \\ 4 \end{pmatrix}$ , which simplifies to $5\begin{pmatrix} 2 \\ 3 \end{pmatrix} - 2\begin{pmatrix} -1 \\ 4 \end{pmatrix}$ . (B) is incorrect because it forms a linear combination of the vector $\mathbf{x}$ using scalars from the first row of $A$ . (C) is incorrect because it uses the rows of $A$ as vectors, not the columns. (D) is incorrect as it represents a misunderstanding of how the scalars and vectors are combined.
Let $A$ be a $3 \times 4$ matrix and $B$ be a $4 \times 2$ matrix. If $\mathbf{v}$ is a vector such that $AB\mathbf{v}$ is defined and equals $\begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}$ , what are the possible dimensions of $\mathbf{v}$ ?
Answer: $\mathbf{v}$ must be $2 \times 1$ Since $A$ is $3 \times 4$ and $B$ is $4 \times 2$ , the product $AB$ is $3 \times 2$ . For $(AB)\mathbf{v}$ to be defined with result $3 \times 1$ , we need $\mathbf{v}$ to be $2 \times 1$ . Choice A confuses the inner dimension of $A$ . Choice C uses the dimension of the result vector. Choice D incorrectly suggests multiple possibilities.
Let $A$ be a $3 \times 2$ matrix with columns $\mathbf{a}_1$ and $\mathbf{a}_2$ , so that $A = \begin{bmatrix} \mathbf{a}_1 & \mathbf{a}_2 \end{bmatrix}$ . If $\mathbf{x} = \begin{pmatrix} -4 \\ 3 \end{pmatrix}$ , which expression represents the product $A\mathbf{x}$ ?
Answer: $-4\mathbf{a}_1 + 3\mathbf{a}_2$ By definition, the product of a matrix $A$ and a vector $\mathbf{x}$ is the linear combination of the columns of $A$ using the corresponding entries of $\mathbf{x}$ as scalar weights. Here, the columns are $\mathbf{a}_1$ and $\mathbf{a}_2$ , and the weights from $\mathbf{x}$ are $-4$ and $3$ . Therefore, $A\mathbf{x}$ is equal to $(-4)\mathbf{a}_1 + (3)\mathbf{a}_2$ , or $-4\mathbf{a}_1 + 3\mathbf{a}_2$ . (A) is incorrect because it swaps the scalar weights for the two columns. (C) is incorrect because the dimensions are compatible. A $3 \times 2$ matrix can be multiplied by a $2 \times 1$ vector, resulting in a $3 \times 1$ vector. (D) is incorrect because it does not represent the sum of the scaled column vectors, which is fundamental to the definition of a matrix-vector product.
Let $A = \begin{bmatrix} \mathbf{a}_1 & \mathbf{a}_2 & \mathbf{a}_3 \end{bmatrix}$ be a $2 \times 3$ matrix. If the equation $A\mathbf{x} = \mathbf{0}$ has a solution $\mathbf{x} = \begin{pmatrix} 4 \\ -1 \\ -2 \end{pmatrix}$ , what must be true about the columns of $A$ ?
Answer: The vector $\mathbf{a}_1$ can be written as a linear combination of $\mathbf{a}_2$ and $\mathbf{a}_3$ . When you encounter a homogeneous equation $A\mathbf{x} = \mathbf{0}$ with a nontrivial solution, you're looking at a fundamental concept about linear dependence. The key insight is that any nontrivial solution to this equation reveals a linear dependence relationship among the columns. Given that $\mathbf{x} = \begin{pmatrix} 4 \\ -1 \\ -2 \end{pmatrix}$ satisfies $A\mathbf{x} = \mathbf{0}$ , we can write this as: $4\mathbf{a}_1 + (-1)\mathbf{a}_2 + (-2)\mathbf{a}_3 = \mathbf{0}$ Rearranging: $4\mathbf{a}_1 = \mathbf{a}_2 + 2\mathbf{a}_3$ , so $\mathbf{a}_1 = \frac{1}{4}\mathbf{a}_2 + \frac{1}{2}\mathbf{a}_3$ This shows that $\mathbf{a}_1$ can indeed be written as a linear combination of $\mathbf{a}_2$ and $\mathbf{a}_3$ , confirming answer A. Answer B is wrong because the existence of a nontrivial solution to $A\mathbf{x} = \mathbf{0}$ proves the columns are linearly dependent, not independent. Answer C is incorrect because the columns don't need to be zero vectors—they just need to be linearly dependent. Our solution shows a specific relationship between non-zero columns. Answer D is false because linearly dependent vectors cannot form a basis. Additionally, since $A$ is $2 \times 3$ , having three vectors in $\mathbb{R}^2$ automatically means they're linearly dependent. Study tip: Remember that $A\mathbf{x} = \mathbf{0}$ having nontrivial solutions is equivalent to the columns being linearly dependent. The solution vector's coefficients directly show you the dependence relationship.
Given that $\begin{pmatrix} 2 & -1 & 3 \\ 1 & 0 & 2 \\ 4 & -2 & 6 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ , which of the following relationships must hold between $x$ , $y$ , and $z$ ?
Answer: $2x - y + 3z = 0$ and $x + 2z = 0$ The matrix equation gives us the system: $2x - y + 3z = 0$ , $x + 2z = 0$ , $4x - 2y + 6z = 0$ . Notice the third equation is twice the first, so we have two independent equations: $2x - y + 3z = 0$ and $x + 2z = 0$ . Choice A assumes the matrix is invertible. Choice B solves incorrectly for $z$ . Choice C incorrectly solves for $y$ but gets the second equation right.
Consider the matrix product $\begin{pmatrix} a & b & c \\ d & e & f \end{pmatrix} \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ . Which of the following expressions represents the second component of the resulting vector?
Answer: $2d - e + 3f$ The second component of a matrix-vector product is found by taking the dot product of the second row with the vector. The second row is $(d, e, f)$ and the vector is $(2, -1, 3)^T$ , so the second component is $2d - e + 3f$ . Choice A uses the first row instead. Choice C incorrectly multiplies corresponding elements from both rows. Choice D misapplies the distributive property.
Consider vectors $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$ , $\mathbf{v}_2 = \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}$ , and $\mathbf{v}_3 = \begin{pmatrix} 0 \\ -1 \\ 2 \end{pmatrix}$ . If $\mathbf{w} = 2\mathbf{v}_1 - 3\mathbf{v}_2 + \mathbf{v}_3$ , what is the sum of the components of $\mathbf{w}$ ?
Answer: -2 When you encounter vector linear combinations, you're working with one of the most fundamental operations in linear algebra. This type of problem tests your ability to perform scalar multiplication and vector addition systematically. To find $\mathbf{w} = 2\mathbf{v}_1 - 3\mathbf{v}_2 + \mathbf{v}_3$ , you need to multiply each vector by its coefficient and then add the results component-wise: $2\mathbf{v}_1 = 2\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ 0 \end{pmatrix}$ $-3\mathbf{v}_2 = -3\begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} -9 \\ -3 \\ 3 \end{pmatrix}$ $\mathbf{v}_3 = \begin{pmatrix} 0 \\ -1 \\ 2 \end{pmatrix}$ Adding these together: $\mathbf{w} = \begin{pmatrix} 2 + (-9) + 0 \\ 4 + (-3) + (-1) \\ 0 + 3 + 2 \end{pmatrix} = \begin{pmatrix} -7 \\ 0 \\ 5 \end{pmatrix}$ The sum of components is $-7 + 0 + 5 = -2$ , making (D) correct. Looking at the wrong answers: (A) 1 likely comes from sign errors in the scalar multiplication. (B) -6 might result from incorrectly handling the coefficient of $\mathbf{v}_3$ or making arithmetic mistakes. (C) 0 could stem from assuming the middle component (which is 0) represents the entire sum. Always work through linear combinations step-by-step: scalar multiply first, then add component-wise. Double-check your arithmetic, especially with negative coefficients, as sign errors are the most common mistake in these problems.
Let $\mathbf{u} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ , $\mathbf{v} = \begin{pmatrix} 3 \\ 0 \\ 2 \end{pmatrix}$ , and $\mathbf{w} = \begin{pmatrix} -2 \\ 1 \\ 4 \end{pmatrix}$ . If $A$ is a $2 \times 3$ matrix such that $A\mathbf{u} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}$ and $A\mathbf{v} = \begin{pmatrix} 6 \\ -2 \end{pmatrix}$ , what is $A(2\mathbf{u} - \mathbf{v})$ ?
Answer: $\begin{pmatrix} 4 \\ 4 \end{pmatrix}$ Using linearity of matrix multiplication: $A(2\mathbf{u} - \mathbf{v}) = 2A\mathbf{u} - A\mathbf{v} = 2\begin{pmatrix} 5 \\ 1 \end{pmatrix} - \begin{pmatrix} 6 \\ -2 \end{pmatrix} = \begin{pmatrix} 10 \\ 2 \end{pmatrix} - \begin{pmatrix} 6 \\ -2 \end{pmatrix} = \begin{pmatrix} 4 \\ 4 \end{pmatrix}$ . Choice B forgets to subtract $A\mathbf{v}$ . Choice C incorrectly adds instead of subtracting. Choice D uses incorrect signs throughout.
The system of linear equations $\begin{cases} 3x_1 - 2x_2 = 5 \\ x_1 + 4x_2 = 7 \end{cases}$ can be written in the matrix form $A\mathbf{x} = \mathbf{b}$ . How can the vector $\mathbf{b}$ be expressed as a linear combination of the columns of $A$ ?
Answer: $\mathbf{b} = x_1\begin{pmatrix} 3 \\ 1 \end{pmatrix} + x_2\begin{pmatrix} -2 \\ 4 \end{pmatrix}$ First, we identify the matrix $A$ , vector $\mathbf{x}$ , and vector $\mathbf{b}$ from the system. $A = \begin{pmatrix} 3 & -2 \\ 1 & 4 \end{pmatrix}$ , $\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$ , and $\mathbf{b} = \begin{pmatrix} 5 \\ 7 \end{pmatrix}$ . The matrix equation $A\mathbf{x} = \mathbf{b}$ is interpreted as a linear combination of the columns of $A$ where the weights are the components of $\mathbf{x}$ . The columns of $A$ are $\begin{pmatrix} 3 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} -2 \\ 4 \end{pmatrix}$ . Thus, the linear combination that equals $\mathbf{b}$ is $x_1\begin{pmatrix} 3 \\ 1 \end{pmatrix} + x_2\begin{pmatrix} -2 \\ 4 \end{pmatrix}$ . (A) is incorrect because it uses the components of $\mathbf{b}$ as the weights, not the variables $x_1$ and $x_2$ . (B) is incorrect because it uses the rows of $A$ as vectors, not the columns. (C) is incorrect because it uses scalars from the first row of $A$ as weights for the vector $\mathbf{x}$ .
A linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^2$ is represented by matrix $A$ . If $T\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}$ , $T\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}$ , and $T\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -5 \\ 1 \end{pmatrix}$ , what is $T\begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}$ ?
Answer: $\begin{pmatrix} -1 \\ -15 \end{pmatrix}$ Since $T$ is linear, $T\begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} = 2T\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} - 3T\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + T\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = 2\begin{pmatrix} 3 \\ -1 \end{pmatrix} - 3\begin{pmatrix} 2 \\ 4 \end{pmatrix} + \begin{pmatrix} -5 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ -2 \end{pmatrix} + \begin{pmatrix} -6 \\ -12 \end{pmatrix} + \begin{pmatrix} -5 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ -15 \end{pmatrix}$ . Choice A has sign errors in the calculation. Choice C forgets the third component contribution. Choice D omits the first component contribution.
Let $M = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$ and $N = \begin{pmatrix} 2 & 1 \\ 0 & 3 \\ -1 & 2 \end{pmatrix}$ . If $\mathbf{x} = \begin{pmatrix} a \\ b \end{pmatrix}$ , which expression equals the first component of $M(N\mathbf{x})$ ?
Answer: $a + 2(2a + b) + 3(-a + 2b)$ First compute $N\mathbf{x} = \begin{pmatrix} 2a + b \\ 3b \\ -a + 2b \end{pmatrix}$ . Then the first component of $M(N\mathbf{x})$ is the dot product of the first row of $M$ with $N\mathbf{x}$ : $1(2a + b) + 2(3b) + 3(-a + 2b) = 2a + b + 6b - 3a + 6b = -a + 13b$ . This matches choice C when expanded. Choice A omits the coefficient 2 in the middle term. Choice B incorrectly expands the matrix multiplication. Choice D uses wrong coefficients from the matrix $M$ .
Let $A = \begin{pmatrix} 1 & 3 \\ -2 & -6 \\ 0 & 1 \end{pmatrix}$ . Which statement best describes the set of all possible vectors $A\mathbf{x}$ where $\mathbf{x} \in \mathbb{R}^2$ ?
Answer: A plane in $\mathbb{R}^3$ passing through the origin. The set of all vectors $A\mathbf{x}$ is the set of all linear combinations of the columns of $A$ . This set is also known as the column space of $A$ . The columns of $A$ are $\mathbf{a}_1 = \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}$ and $\mathbf{a}_2 = \begin{pmatrix} 3 \\ -6 \\ 1 \end{pmatrix}$ . Since $\mathbf{a}_2$ is not a scalar multiple of $\mathbf{a}_1$ , the two column vectors are linearly independent. Two linearly independent vectors in $\mathbb{R}^3$ span a plane. Since a linear combination with weights of zero (i.e., $\mathbf{x} = \mathbf{0}$ ) results in the zero vector, this plane must pass through the origin. (A) is incorrect because the columns are not parallel, so they span more than just a line. (C) is incorrect because two vectors can span at most a plane in $\mathbb{R}^3$ , not the entire space. (D) is incorrect because setting $\mathbf{x} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$ gives $A\mathbf{x} = \mathbf{0}$ , so the set must contain the origin.
Let $A = \begin{pmatrix} 2 & -1 & 3 \\ 1 & 4 & -2 \\ 0 & 3 & 1 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$ . If $A\mathbf{v} = \begin{pmatrix} 7 \\ -3 \\ 5 \end{pmatrix}$ , what is the value of $2x - y + 3z$ ?
Answer: 7 The first row of the matrix equation $A\mathbf{v} = \begin{pmatrix} 7 \\ -3 \\ 5 \end{pmatrix}$ gives us $2x - y + 3z = 7$ . This is exactly what we're asked to find, so the answer is 7. Choice B confuses this with the third component of the result vector. Choice C incorrectly adds the first and third components. Choice D mistakenly uses the second component.

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