| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To determine if the vectors [1, 2], [2, 3], and [3, 4] are linearly independent, we need to see if they can be expressed as a linear combination of each other. In other words, we need to see if there exists a non-trivial solution (i.e., a solution where not all coefficients are zero) to the equation: c1 * [1, 2] + c2 * [2, 3] + c3 * [3, 4] = [0, 0] where c1, c2, and c3 are constants. We can write this equation in matrix form as: | 1 2 3 | | c1 | | 0 | | 2 3 4 | | c2 | = | 0 | We can solve this system of equations by finding the determinant of the matrix: | 1 2 3 | | 2 3 4 | The determinant of this matrix is: 1 * (3 * 4 - 4 * 3) - 2 * (2 * 4 - 4 * 2) + 3 * (2 * 3 - 3 * 2) = 0 Since the determinant is zero, the matrix is singular and the vectors are linearly dependent. Therefore, the vectors [1, 2], [2, 3], and [3, 4] are not linearly independent. |
False |
| deepseekr1 | The vectors [1, 2], [2, 3], and [3, 4] are in \(\mathbb{R}^2\). Since there are three vectors in a two-dimensional space, they must be linearly dependent. To confirm, we set up the equation \(a[1, 2] + b[2, 3] + c[3, 4] = [0, 0]\), leading to the system: \[ \begin{cases} a + 2b + 3c = 0 \\ 2a + 3b + 4c = 0 \end{cases} \] Row reducing the augmented matrix: \[ \left[\begin{array}{ccc|c} 1 & 2 & 3 & 0 \\ 2 & 3 & 4 & 0 \end{array}\right] \rightarrow \left[\begin{array}{ccc|c} 1 & 2 & 3 & 0 \\ 0 & -1 & -2 & 0 \end{array}\right] \] From the reduced matrix, we find \(b = -2c\) and \(a = c\). Choosing \(c = 1\) gives a non-trivial solution: \(a = 1\), \(b = -2\), \(c = 1\). Thus, \(1\cdot[1,2] - 2\cdot[2,3] + 1\cdot[3,4] = [0,0]\), proving linear dependence. **Answer:** No, the vectors are not linearly independent. |
False |
| gpto1 | To determine whether the vectors \(\mathbf{v}_1 = [1, 2]\), \(\mathbf{v}_2 = [2, 3]\), and \(\mathbf{v}_3 = [3, 4]\) are linearly independent, we need to check if there exist scalars \(a\), \(b\), and \(c\), not all zero, such that: \[ a\mathbf{v}_1 + b\mathbf{v}_2 + c\mathbf{v}_3 = \mathbf{0} \] Let's set up the equation component-wise: For the first component: \[ a(1) + b(2) + c(3) = 0 \] \[ \Rightarrow a + 2b + 3c = 0 \quad \text{(1)} \] For the second component: \[ a(2) + b(3) + c(4) = 0 \] \[ \Rightarrow 2a + 3b + 4c = 0 \quad \text{(2)} \] Now, we have a system of two equations with three unknowns: \[ \begin{cases} a + 2b + 3c = 0 \quad \text{(1)} \\ 2a + 3b + 4c = 0 \quad \text{(2)} \end{cases} \] **Step 1: Eliminate one variable** First, we'll eliminate one variable to find a relationship between the others. Let's eliminate \(a\) by manipulating the equations. Multiply equation (1) by 2: \[ 2(a + 2b + 3c) = 0 \] \[ 2a + 4b + 6c = 0 \quad \text{(1a)} \] Now subtract equation (2) from equation (1a): \[ (2a + 4b + 6c) - (2a + 3b + 4c) = 0 - 0 \] \[ (2a - 2a) + (4b - 3b) + (6c - 4c) = 0 \] \[ b + 2c = 0 \quad \text{(3)} \] **Step 2: Express one variable in terms of another** From equation (3), we can express \(b\) in terms of \(c\): \[ b = -2c \] **Step 3: Substitute back to find \(a\)** Substitute \(b = -2c\) into equation (1): \[ a + 2(-2c) + 3c = 0 \] \[ a - 4c + 3c = 0 \] \[ a - c = 0 \] \[ a = c \] **Step 4: Find the solution** Let \(c = t\), where \(t\) is any scalar (not necessarily zero). Then: \[ a = t \] \[ b = -2t \] \[ c = t \] **Step 5: Conclusion** Since we found nontrivial solutions (\(a = t\), \(b = -2t\), \(c = t\)) where \(t\) is any non-zero scalar, this means that the vectors are **linearly dependent**. **Answer:** *No; the vectors are linearly dependent because there exist scalars—\(a = t\), \(b = -2t\), \(c = t\) for any scalar \(t\)—not all zero, such that \(a[1, 2] + b[2, 3] + c[3, 4] = [0, 0]\).* |
False |
| GPT-4o | To determine if the vectors \(\mathbf{v}_1 = [1, 2]\), \(\mathbf{v}_2 = [2, 3]\), and \(\mathbf{v}_3 = [3, 4]\) are linearly independent, we need to check if a linear combination of these vectors equals the zero vector only when all the coefficients are zero. Mathematically, we need to solve the equation: \[ c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3 = \mathbf{0} \] Substituting the given vectors, this equation becomes: \[ c_1 [1, 2] + c_2 [2, 3] + c_3 [3, 4] = [0, 0] \] This can be written as a system of linear equations: \[ c_1 \cdot 1 + c_2 \cdot 2 + c_3 \cdot 3 = 0 \] \[ c_1 \cdot 2 + c_2 \cdot 3 + c_3 \cdot 4 = 0 \] We can write this system in matrix form as: \[ \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] We need to determine if the only solution to this equation is \(c_1 = c_2 = c_3 = 0\). One way to do this is to analyze the rank of the matrix: \[ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \end{bmatrix} \] A set of vectors is linearly independent if the rank of the coefficient matrix \(A\) is equal to the number of vectors. In this case, we have three vectors (\(n = 3\)) but only two equations (\(m = 2\)). For the vectors to be linearly independent, we would need \(m = n\). However, since \(m < n\), the maximum rank of \(A\) can be at most 2, which is less than 3. Therefore, the vectors cannot be linearly independent. To confirm this, we can perform row reduction on matrix \(A\): \[ \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \end{bmatrix} \] Subtract 2 times the first row from the second row: \[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & -1 & -2 \end{bmatrix} \] This matrix is in row-echelon form. The rank of the matrix is 2 (the number of non-zero rows). Since the rank is less than the number of vectors (3), the vectors \(\mathbf{v}_1\), \(\mathbf{v}_2\), and \(\mathbf{v}_3\) are linearly dependent. Thus, the vectors \([1, 2]\), \([2, 3]\), and \([3, 4]\) are not linearly independent. |
False |
| llama3.1-70B-Instruct | To determine if the vectors [1, 2], [2, 3], and [3, 4] are linearly independent, we need to check if any of the vectors can be expressed as a linear combination of the others. Let's assume that the vectors are linearly dependent, meaning that one of the vectors can be expressed as a linear combination of the others. We can write this as: [1, 2] = a[2, 3] + b[3, 4] where a and b are scalars. We can rewrite this equation as: [1, 2] = [2a + 3b, 3a + 4b] Equating the corresponding components, we get: 1 = 2a + 3b 2 = 3a + 4b We can solve this system of equations by substitution or elimination. Let's use substitution. Rearranging the first equation, we get: a = (1 - 3b) / 2 Substituting this expression for a into the second equation, we get: 2 = 3((1 - 3b) / 2) + 4b Simplifying, we get: 2 = 3/2 - 9b/2 + 4b Multiplying both sides by 2, we get: 4 = 3 - 9b + 8b Simplifying, we get: 4 = 3 - b Subtracting 3 from both sides, we get: 1 = -b Multiplying both sides by -1, we get: b = -1 Now that we have found b, we can find a: a = (1 - 3b) / 2 = (1 - 3(-1)) / 2 = (1 + 3) / 2 = 4/2 = 2 So, we have found that: [1, 2] = 2[2, 3] - [3, 4] This means that the vectors [1, 2], [2, 3], and [3, 4] are linearly dependent. Therefore, the answer is no, the vectors [1, 2], [2, 3], and [3, 4] are not linearly independent. |
False |
| llama3.1-405B-Instruct | To determine if the vectors [1, 2], [2, 3], and [3, 4] are linearly independent, we can use the following steps: 1. Write the vectors as columns of a matrix: | 1 2 3 | | 2 3 4 | 2. Calculate the determinant of the matrix: det(A) = 1*3 - 2*2 = 3 - 4 = -1 3. If the determinant is non-zero, the vectors are linearly independent. If the determinant is zero, the vectors are linearly dependent. In this case, the determinant is -1, which is non-zero. Therefore, the vectors [1, 2], [2, 3], and [3, 4] are linearly independent. |
False |
| DeepSeek-Coder-V2 | To determine if the vectors [1, 2], [2, 3], and [3, 4] are linearly independent, we need to check if the only solution to the equation \(a[1, 2] + b[2, 3] + c[3, 4] = [0, 0]\) is \(a = b = c = 0\). First, let's write the equation in matrix form: \[ \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] This can be expanded into a system of linear equations: 1. \(a + 2b + 3c = 0\) 2. \(2a + 3b + 4c = 0\) We can solve this system using the method of elimination or substitution. Let's use elimination: First, multiply the first equation by 2: \[ 2a + 4b + 6c = 0 \] Now subtract the second equation from this result: \[ (2a + 4b + 6c) - (2a + 3b + 4c) = 0 - 0 \] This simplifies to: \[ b + 2c = 0 \] So, \(b = -2c\). Now substitute \(b = -2c\) back into the first original equation: \[ a + 2(-2c) + 3c = 0 \] This simplifies to: \[ a - c = 0 \] So, \(a = c\). Now we have \(a = c\) and \(b = -2c\). If we set \(c = 0\), then \(a = 0\) and \(b = 0\). This means the only solution is \(a = b = c = 0\). Since the only solution to the equation \(a[1, 2] + b[2, 3] + c[3, 4] = [0, 0]\) is \(a = b = c = 0\), the vectors [1, 2], [2, 3], and [3, 4] are linearly independent. |
False |
| Qwen2-72B-Instruct | To determine if the vectors [1, 2], [2, 3], and [3, 4] are linearly independent, we need to see if they can be expressed as a linear combination of each other. In other words, we need to see if there exists a non-trivial solution (i.e., a solution where not all coefficients are zero) to the equation: c1 * [1, 2] + c2 * [2, 3] + c3 * [3, 4] = [0, 0] where c1, c2, and c3 are constants. We can write this equation in matrix form as: | 1 2 3 | | c1 | | 0 | | 2 3 4 | | c2 | = | 0 | We can solve this system of equations by finding the determinant of the matrix: | 1 2 3 | | 2 3 4 | The determinant of this matrix is: 1 * (3 * 4 - 4 * 3) - 2 * (2 * 4 - 4 * 2) + 3 * (2 * 3 - 3 * 2) = 0 Since the determinant is zero, the matrix is singular and the vectors are linearly dependent. Therefore, the vectors [1, 2], [2, 3], and [3, 4] are not linearly independent. |
False |
| baichuan4 | To determine if the vectors [1, 2], [2, 3], and [3, 4] are linearly independent, we need to see if there exists a unique solution to the equation c1[1, 2] + c2[2, 3] + c3[3, 4] = [0, 0], where c1, c2, and c3 are scalars. We can set up a system of equations: c1 + 2c2 + 3c3 = 0 2c1 + 3c2 + 4c3 = 0 Now, we can solve this system of equations. Multiplying the first equation by -2, we get: -2c1 - 4c2 - 6c3 = 0 Adding this equation to the second equation, we get: -c2 - 2c3 = 0 c2 = -2c3 Substituting c2 = -2c3 into the first equation, we get: c1 - 4c3 + 3c3 = 0 c1 - c3 = 0 c1 = c3 Now, substituting c1 = c3 and c2 = -2c3 into the second equation, we get: 2c3 - 6c3 + 4c3 = 0 0 = 0 This means that for any value of c3, there exists a solution for c1 and c2. Therefore, the vectors [1, 2], [2, 3], and [3, 4] are linearly dependent because they are not unique. |
False |
| doubao | To determine if the vectors are linearly independent, we can set up the equation: a[1, 2] + b[2, 3] + c[3, 4] = [0, 0] (where a, b, and c are scalars) This gives the system of equations: a + 2b + 3c = 0 2a + 3b + 4c = 0 Multiplying the first equation by 2 and subtracting the second equation gives: 2(a + 2b + 3c) - (2a + 3b + 4c) = 0 2a + 4b + 6c - 2a - 3b - 4c = 0 b + 2c = 0 b = -2c Substituting this into the first equation gives: a + 2(-2c) + 3c = 0 a - 4c + 3c = 0 a - c = 0 a = c So we can choose c = 1, then a = 1 and b = -2. This shows that there exist non-zero scalars a, b, and c such that the linear combination of the vectors is equal to the zero vector, so the vectors are linearly dependent. |
False |
| minimax | To determine if the vectors [1, 2], [2, 3], and [3, 4] are linearly independent, we need to check if there is a non-trivial solution to the equation: a[1, 2] + b[2, 3] + c[3, 4] = [0, 0] where a, b, and c are scalars, and [0, 0] is the zero vector. If the only solution is a = b = c = 0, then the vectors are linearly independent. Let's set up the system of equations based on the vectors given: a(1) + b(2) + c(3) = 0 a(2) + b(3) + c(4) = 0 This gives us two equations with three unknowns: 1a + 2b + 3c = 0 2a + 3b + 4c = 0 We can solve this system using various methods, such as substitution, elimination, or matrix methods. However, in this case, we can notice that the vectors are not linearly independent without solving the system because the second vector is a linear combination of the first and third vectors. If we subtract twice the first vector from the third vector, we get: [3, 4] - 2*[1, 2] = [3, 4] - [2, 4] = [1, 0] This shows that the third vector can be obtained by a linear combination of the first two vectors, which means that the vectors are not linearly independent. In other words, there exists a non-trivial solution to the equation with a, b, and c not all being zero. Therefore, the vectors [1, 2], [2, 3], and [3, 4] are linearly dependent. |
False |
| kimi | To determine if the vectors \(\mathbf{v}_1 = [1, 2]\), \(\mathbf{v}_2 = [2, 3]\), and \(\mathbf{v}_3 = [3, 4]\) are linearly independent, we can check if there exists a non-trivial linear combination of these vectors that equals the zero vector. In other words, we need to see if there are scalars \(a\), \(b\), and \(c\) (not all zero) such that: \[a\mathbf{v}_1 + b\mathbf{v}_2 + c\mathbf{v}_3 = \mathbf{0}\] This translates to the following system of linear equations: \[a(1) + b(2) + c(3) = 0\] \[a(2) + b(3) + c(4) = 0\] Which can be written in matrix form as: \[ \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] To solve for \(a\), \(b\), and \(c\), we can use the determinant of the matrix formed by the vectors. If the determinant is non-zero, the vectors are linearly independent. If the determinant is zero, the vectors are linearly dependent. The determinant of the matrix is: \[ \text{det} = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \end{vmatrix} \] Since the matrix is not a square matrix (it has more columns than rows), we can't directly compute the determinant. Instead, we can use the concept of row reduction to find the rank of the matrix, which will tell us if the vectors are linearly independent. Let's perform row reduction: \[ \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 2 & 3 \\ 0 & -1 & -2 \end{bmatrix} \] Subtract 2 times the first row from the second row: \[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & -1 & -2 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \end{bmatrix} \] Now, we can see that the second row is a scalar multiple of the first row (specifically, it is the negative of the first row). This means that the vectors are linearly dependent because there is a non-trivial solution to the system of equations (in this case, \(a = 3\), \(b = -1\), \(c = 0\) is one such solution). Therefore, the vectors \(\mathbf{v}_1 = [1, 2]\), \(\mathbf{v}_2 = [2, 3]\), and \(\mathbf{v}_3 = [3, 4]\) are linearly dependent. |
False |
| wenxinyiyan | To determine if the vectors [1, 2], [2, 3], and [3, 4] are linearly independent, we can use the concept of determinant. If the determinant of the matrix formed by these vectors is zero, then the vectors are linearly dependent. Otherwise, they are linearly independent. The determinant of a matrix A, denoted det(A), can be calculated using the formula: det(A) = ad - bc where a, b, c, and d are the elements of the 2x2 matrix A: A = [a, b] [c, d] For the given vectors, the matrix A would be: A = [1, 2] [2, 3] And the determinant would be: det(A) = (1 * 3) - (2 * 2) = 3 - 4 = -1 Since the determinant is non-zero, the vectors [1, 2] and [2, 3] are linearly independent. However, we need to check if the third vector [3, 4] is also linearly independent from the first two. We can do this by setting up a system of equations to see if there is a non-trivial solution (a solution where not all the coefficients are zero) to the equation: c1 * [1, 2] + c2 * [2, 3] + c3 * [3, 4] = [0, 0] This system of equations can be represented as: c1 + 2*c2 + 3*c3 = 0 2*c1 + 3*c2 + 4*c3 = 0 We can solve this system of equations using any suitable method (e.g., substitution, elimination, or matrix methods). However, for simplicity, we can observe that the third vector [3, 4] is a linear combination of the first two vectors: [3, 4] = (2 * [2, 3]) - ([1, 2]) This means that the vector [3, 4] can be expressed as a linear combination of the vectors [1, 2] and [2, 3], and therefore, the three vectors are linearly dependent. So, the vectors [1, 2], [2, 3], and [3, 4] are not linearly independent. |
False |
| zhipu | To determine if the vectors \(\mathbf{v}_1 = [1, 2]\), \(\mathbf{v}_2 = [2, 3]\), and \(\mathbf{v}_3 = [3, 4]\) are linearly independent, we need to check if the only solution to the equation \(c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3 = \mathbf{0}\) is the trivial solution \(c_1 = c_2 = c_3 = 0\). This can be written as: \[ c_1 \begin{bmatrix} 1 \\ 2 \end{bmatrix} + c_2 \begin{bmatrix} 2 \\ 3 \end{bmatrix} + c_3 \begin{bmatrix} 3 \\ 4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] Which translates to the system of linear equations: \[ \begin{cases} c_1 + 2c_2 + 3c_3 = 0 \\ 2c_1 + 3c_2 + 4c_3 = 0 \end{cases} \] We can represent this system in matrix form as: \[ \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] To determine if the vectors are linearly independent, we need to check if the coefficient matrix has full rank. Since we have three vectors in \(\mathbb{R}^2\), for them to be linearly independent, the matrix should be a \(2 \times 3\) matrix of rank 2. However, having three vectors in a two-dimensional space generally implies that they are linearly dependent. Let's row reduce the matrix to its echelon form to see if there are any free variables (which would imply linear dependence). The matrix is: \[ \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \end{bmatrix} \] We perform row operations to get it into row echelon form. 1. \( R_2 \rightarrow R_2 - 2R_1 \): \[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & -1 & -2 \end{bmatrix} \] 2. \( R_2 \rightarrow -R_2 \) (to make the leading coefficient positive): \[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \end{bmatrix} \] 3. \( R_1 \rightarrow R_1 - 2R_2 \): \[ \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \end{bmatrix} \] The row-reduced form shows that the matrix does not have full rank 2 because there is a column (the third column) that does not have a pivot. This indicates the presence of a free variable, meaning the system has non-trivial solutions. Therefore, the vectors \(\mathbf{v}_1\), \(\mathbf{v}_2\), and \(\mathbf{v}_3\) are linearly dependent. **Conclusion**: The vectors \([1, 2]\), \([2, 3]\), and \([3, 4]\) are not linearly independent. |
False |
| gpto1 | To determine whether the vectors \(\mathbf{v}_1 = [1, 2]\), \(\mathbf{v}_2 = [2, 3]\), and \(\mathbf{v}_3 = [3, 4]\) are linearly independent, we need to check if there exist scalars \(a\), \(b\), and \(c\), not all zero, such that: \[ a\mathbf{v}_1 + b\mathbf{v}_2 + c\mathbf{v}_3 = \mathbf{0} \] Let's set up the equation component-wise: For the first component: \[ a(1) + b(2) + c(3) = 0 \] \[ \Rightarrow a + 2b + 3c = 0 \quad \text{(1)} \] For the second component: \[ a(2) + b(3) + c(4) = 0 \] \[ \Rightarrow 2a + 3b + 4c = 0 \quad \text{(2)} \] Now, we have a system of two equations with three unknowns: \[ \begin{cases} a + 2b + 3c = 0 \quad \text{(1)} \\ 2a + 3b + 4c = 0 \quad \text{(2)} \end{cases} \] **Step 1: Eliminate one variable** First, we'll eliminate one variable to find a relationship between the others. Let's eliminate \(a\) by manipulating the equations. Multiply equation (1) by 2: \[ 2(a + 2b + 3c) = 0 \] \[ 2a + 4b + 6c = 0 \quad \text{(1a)} \] Now subtract equation (2) from equation (1a): \[ (2a + 4b + 6c) - (2a + 3b + 4c) = 0 - 0 \] \[ (2a - 2a) + (4b - 3b) + (6c - 4c) = 0 \] \[ b + 2c = 0 \quad \text{(3)} \] **Step 2: Express one variable in terms of another** From equation (3), we can express \(b\) in terms of \(c\): \[ b = -2c \] **Step 3: Substitute back to find \(a\)** Substitute \(b = -2c\) into equation (1): \[ a + 2(-2c) + 3c = 0 \] \[ a - 4c + 3c = 0 \] \[ a - c = 0 \] \[ a = c \] **Step 4: Find the solution** Let \(c = t\), where \(t\) is any scalar (not necessarily zero). Then: \[ a = t \] \[ b = -2t \] \[ c = t \] **Step 5: Conclusion** Since we found nontrivial solutions (\(a = t\), \(b = -2t\), \(c = t\)) where \(t\) is any non-zero scalar, this means that the vectors are **linearly dependent**. **Answer:** *No; the vectors are linearly dependent because there exist scalars—\(a = t\), \(b = -2t\), \(c = t\) for any scalar \(t\)—not all zero, such that \(a[1, 2] + b[2, 3] + c[3, 4] = [0, 0]\).* |
False |
| deepseekv3 | To determine whether the vectors \([1, 2]\), \([2, 3]\), and \([3, 4]\) are **linearly independent**, we can follow these steps: --- ### Step 1: Understand Linear Independence A set of vectors is **linearly independent** if no vector in the set can be written as a linear combination of the others. In other words, the only solution to the equation \[ c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \dots + c_n \mathbf{v}_n = \mathbf{0} \] is \(c_1 = c_2 = \dots = c_n = 0\). --- ### Step 2: Check the Number of Vectors vs. the Dimension The vectors \([1, 2]\), \([2, 3]\), and \([3, 4]\) are in \(\mathbb{R}^2\) (2-dimensional space). A set of more than 2 vectors in \(\mathbb{R}^2\) cannot be linearly independent, because at least one vector can always be written as a linear combination of the others. --- ### Step 3: Conclusion Since there are 3 vectors in \(\mathbb{R}^2\), they **cannot** be linearly independent. --- ### Final Answer: No, the vectors \([1, 2]\), \([2, 3]\), and \([3, 4]\) are **not linearly independent**. |
False |