问题

You are supposed to provide a solution to a given problem.

Problem:
Are the vectors v_1 = [1,2,3], v_2 = [4,5,6], v_3 = [7,8,9] linearly independent?
Solution:

答案

模型	模型答案	正确答案
金声玉亮2.0	N.A.	False
deepseekr1	The vectors $ v_1 = \begin{bmatrix}1\\2\\3\end{bmatrix} $, $ v_2 = \begin{bmatrix}4\\5\\6\end{bmatrix} $, and $ v_3 = \begin{bmatrix}7\\8\\9\end{bmatrix} $ are linearly dependent. Step-by-Step Explanation: 1. Set Up the Linear Combination Equation: Determine if there exist scalars $ c_1, c_2, c_3 $ (not all zero) such that: \[ c_1 \begin{bmatrix}1\\2\\3\end{bmatrix} + c_2 \begin{bmatrix}4\\5\\6\end{bmatrix} + c_3 \begin{bmatrix}7\\8\\9\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix} \] 2. Form the Augmented Matrix: \[ \begin{bmatrix} 1 & 4 & 7 & \| & 0 \\ 2 & 5 & 8 & \| & 0 \\ 3 & 6 & 9 & \| & 0 \\ \end{bmatrix} \] 3. Row Reduce the Matrix: - Subtract $ 2 \times \text{Row 1} $ from Row 2: \[ \text{Row 2} \rightarrow \begin{bmatrix}0 & -3 & -6\end{bmatrix} \] - Subtract $ 3 \times \text{Row 1} $ from Row 3: \[ \text{Row 3} \rightarrow \begin{bmatrix}0 & -6 & -12\end{bmatrix} \] - Subtract $ 2 \times \text{Row 2} $ from Row 3: \[ \text{Row 3} \rightarrow \begin{bmatrix}0 & 0 & 0\end{bmatrix} \] Resulting in: \[ \begin{bmatrix} 1 & 4 & 7 \\ 0 & -3 & -6 \\ 0 & 0 & 0 \\ \end{bmatrix} \] 4. Determine the Rank: The matrix has 2 non-zero rows (rank = 2), which is less than the number of vectors (3). This indicates linear dependence. 5. Alternative Check via Determinant: Compute the determinant of the matrix formed by the vectors: \[ \det\begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \\ \end{bmatrix} = 1(45 - 48) - 4(18 - 24) + 7(12 - 15) = -3 + 24 - 21 = 0 \] A zero determinant confirms dependence. 6. Explicit Non-Trivial Solution: Solving the reduced system, let $ c_3 = t $: - From Row 2: $ -3c_2 - 6t = 0 \Rightarrow c_2 = -2t $ - From Row 1: $ c_1 + 4(-2t) + 7t = 0 \Rightarrow c_1 = t $ - Choosing $ t = 1 $: $ c_1 = 1 $, $ c_2 = -2 $, $ c_3 = 1 $ satisfies $ v_1 - 2v_2 + v_3 = 0 $. Conclusion: The vectors are linearly dependent because there exists a non-trivial linear combination that equals the zero vector.	False
gpto1	To determine whether the vectors $ \mathbf{v}_1 = [1,\,2,\,3] $, $ \mathbf{v}_2 = [4,\,5,\,6] $, and $ \mathbf{v}_3 = [7,\,8,\,9] $ are linearly independent, we can use the concept of the determinant of a matrix formed by these vectors. In $\mathbb{R}^3$, if the determinant of the matrix formed by placing the vectors as columns is zero, the vectors are linearly dependent; if the determinant is non-zero, they are linearly independent. Step 1: Form the Matrix Construct matrix $ A $ by placing the vectors $ \mathbf{v}_1, \mathbf{v}_2, $ and $ \mathbf{v}_3 $ as its columns: \[ A = \begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \\ \end{bmatrix} \] Step 2: Compute the Determinant of $ A $ Compute $ \det(A) $ using the rule for the determinant of a $ 3 \times 3 $ matrix: \[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] where: \[ A = \begin{bmatrix} a & d & g \\ b & e & h \\ c & f & i \\ \end{bmatrix} \] Plug in the values: \[ \begin{aligned} & a = 1, \quad d = 4, \quad g = 7 \\ & b = 2, \quad e = 5, \quad h = 8 \\ & c = 3, \quad f = 6, \quad i = 9 \\ \end{aligned} \] Compute each term: 1. $ ei - fh $: \[ (5)(9) - (6)(8) = 45 - 48 = -3 \] 2. $ di - fg $: \[ (4)(9) - (6)(7) = 36 - 42 = -6 \] 3. $ dh - eg $: \[ (4)(8) - (5)(7) = 32 - 35 = -3 \] Compute $ \det(A) $: \[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) = (1)(-3) - (2)(-6) + (3)(-3) \] Simplify: \[ \det(A) = -3 + 12 - 9 = 0 \] Step 3: Interpret the Result Since $ \det(A) = 0 $, the matrix $ A $ is singular, and its columns (the vectors $ \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 $) are linearly dependent. Alternative Explanation We can also observe the relationship among the vectors directly. Notice that: \[ \mathbf{v}_2 - \mathbf{v}_1 = [4 - 1,\, 5 - 2,\, 6 - 3] = [3,\,3,\,3] \] \[ \mathbf{v}_3 - \mathbf{v}_2 = [7 - 4,\,8 - 5,\,9 - 6] = [3,\,3,\,3] \] The differences between consecutive vectors are the same, indicating that the vectors lie in the same plane and are linearly dependent. Conclusion Because the determinant of the matrix formed by the vectors is zero, the vectors $ \mathbf{v}_1, \mathbf{v}_2, $ and $ \mathbf{v}_3 $ are linearly dependent. Answer: Since the determinant is zero, the vectors are linearly dependent.	False
GPT-4o	To determine whether the vectors $ \mathbf{v_1} = [1, 2, 3] $, $ \mathbf{v_2} = [4, 5, 6] $, and $ \mathbf{v_3} = [7, 8, 9] $ are linearly independent, we can use the concept of linear independence from linear algebra. Vectors are linearly independent 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 $ c_1 = c_2 = c_3 = 0 $. One common method to determine linear independence is to form a matrix with these vectors as its columns and then calculate the determinant of that matrix. If the determinant is non-zero, the vectors are linearly independent; if the determinant is zero, the vectors are linearly dependent. Let's form the matrix $ A $ with $ \mathbf{v_1} $, $ \mathbf{v_2} $, and $ \mathbf{v_3} $ as columns: \[ A = \begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{bmatrix} \] We will now calculate the determinant of $ A $: \[ \text{det}(A) = \begin{vmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{vmatrix} \] To calculate the determinant of a 3x3 matrix, we use the formula: \[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] Where $ A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} $. For our matrix $ A $: \[ a = 1, \quad b = 4, \quad c = 7, \quad d = 2, \quad e = 5, \quad f = 8, \quad g = 3, \quad h = 6, \quad i = 9 \] Plugging into the determinant formula: \[ \text{det}(A) = 1(5 \cdot 9 - 8 \cdot 6) - 4(2 \cdot 9 - 8 \cdot 3) + 7(2 \cdot 6 - 5 \cdot 3) \] \[ = 1(45 - 48) - 4(18 - 24) + 7(12 - 15) \] \[ = 1(-3) - 4(-6) + 7(-3) \] \[ = -3 + 24 - 21 \] \[ = 0 \] Since the determinant of matrix $ A $ is zero, the vectors $ \mathbf{v_1} $, $ \mathbf{v_2} $, and $ \mathbf{v_3} $ are linearly dependent. Thus, the solution to the problem is: The vectors $ \mathbf{v_1} = [1, 2, 3] $, $ \mathbf{v_2} = [4, 5, 6] $, and $ \mathbf{v_3} = [7, 8, 9] $ are not linearly independent. They are linearly dependent.	False
llama3.1-70B-Instruct	To determine if the vectors v_1 = [1,2,3], v_2 = [4,5,6], and v_3 = [7,8,9] are linearly independent, we need to check if the equation: a_1v_1 + a_2v_2 + a_3v_3 = 0 has only the trivial solution, where a_1 = a_2 = a_3 = 0. We can write the equation as: a_1[1,2,3] + a_2[4,5,6] + a_3[7,8,9] = [0,0,0] Expanding the equation, we get: a_1 + 4a_2 + 7a_3 = 0 2a_1 + 5a_2 + 8a_3 = 0 3a_1 + 6a_2 + 9a_3 = 0 We can solve this system of linear equations using substitution or elimination. Let's use elimination. First, we can multiply the first equation by 2 and subtract the second equation from it to get: 2a_1 + 8a_2 + 14a_3 = 0 -2a_1 - 5a_2 - 8a_3 = 0 Adding the two equations, we get: 3a_2 + 6a_3 = 0 Now, we can multiply the first equation by 3 and subtract the third equation from it to get: 3a_1 + 12a_2 + 21a_3 = 0 -3a_1 - 6a_2 - 9a_3 = 0 Adding the two equations, we get: 6a_2 + 12a_3 = 0 Now, we have two equations with two variables: 3a_2 + 6a_3 = 0 6a_2 + 12a_3 = 0 We can solve these equations by multiplying the first equation by 2 and subtracting the second equation from it to get: 6a_2 + 12a_3 = 0 -6a_2 - 12a_3 = 0 Adding the two equations, we get: 0 = 0 This means that the system of equations has infinitely many solutions, and therefore, the vectors v_1, v_2, and v_3 are linearly dependent. In other words, there exist non-zero values of a_1, a_2, and a_3 such that: a_1v_1 + a_2v_2 + a_3v_3 = 0 For example, we can choose a_1 = 1, a_2 = -1, and a_3 = 0, which gives: 1[1,2,3] - 1[4,5,6] + 0[7,8,9] = [0,0,0] Therefore, the vectors v_1, v_2, and v_3 are not linearly independent.	False
llama3.1-405B-Instruct	To determine if the vectors v_1 = [1,2,3], v_2 = [4,5,6], and v_3 = [7,8,9] are linearly independent, we can use the following steps: 1. Form a matrix A with the vectors as columns: A = [1 4 7] [2 5 8] [3 6 9] 2. Calculate the determinant of matrix A: det(A) = 1(59 - 68) - 4(29 - 38) + 7(26 - 35) = 1(45 - 48) - 4(18 - 24) + 7(12 - 15) = 1(-3) - 4(-6) + 7(-3) = -3 + 24 - 21 = 0 Since the determinant of matrix A is zero, the vectors v_1, v_2, and v_3 are linearly dependent. This means that at least one of the vectors can be expressed as a linear combination of the other two vectors. Therefore, the vectors v_1 = [1,2,3], v_2 = [4,5,6], and v_3 = [7,8,9] are not linearly independent.	False
DeepSeek-Coder-V2	To determine if the vectors $ v_1 = [1, 2, 3] $, $ v_2 = [4, 5, 6] $, and $ v_3 = [7, 8, 9] $ are linearly independent, we need to check if the only solution to the equation \[ c_1 v_1 + c_2 v_2 + c_3 v_3 = 0 \] is $ c_1 = c_2 = c_3 = 0 $. First, let's write the vectors in a matrix form: \[ \begin{pmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \\ c_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \] This is a homogeneous system of linear equations. To determine if the vectors are linearly independent, we need to check if the determinant of the matrix is non-zero. If the determinant is zero, the vectors are linearly dependent. Let's calculate the determinant of the matrix: \[ \begin{vmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{vmatrix} \] Using the formula for the determinant of a 3x3 matrix: \[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg) \] Substitute $ a = 1 $, $ b = 4 $, $ c = 7 $, $ d = 2 $, $ e = 5 $, $ f = 8 $, $ g = 3 $, $ h = 6 $, $ i = 9 $: \[ 1(5 \cdot 9 - 8 \cdot 6) - 4(2 \cdot 9 - 8 \cdot 3) + 7(2 \cdot 6 - 5 \cdot 3) \] Calculate each term: \[ 1(45 - 48) = 1(-3) = -3 \] \[ -4(18 - 24) = -4(-6) = 24 \] \[ 7(12 - 15) = 7(-3) = -21 \] Add these results together: \[ -3 + 24 - 21 = 0 \] Since the determinant is zero, the vectors $ v_1 $, $ v_2 $, and $ v_3 $ are linearly dependent. Therefore, the vectors $ v_1 = [1, 2, 3] $, $ v_2 = [4, 5, 6] $, and $ v_3 = [7, 8, 9] $ are not linearly independent.	False
Qwen2-72B-Instruct	To determine if the vectors $v_1 = [1,2,3]$, $v_2 = [4,5,6]$, and $v_3 = [7,8,9]$ are linearly independent, we need to check if the only solution to the equation \[c_1v_1 + c_2v_2 + c_3v_3 = 0\] is the trivial solution $c_1 = c_2 = c_3 = 0$. This equation can be rewritten as a system of linear equations: \[ \begin{align} c_1 + 4c_2 + 7c_3 &= 0 \\ 2c_1 + 5c_2 + 8c_3 &= 0 \\ 3c_1 + 6c_2 + 9c_3 &= 0 \end{align} \] This system can be represented by the matrix equation $Ax = 0$, where \[ A = \begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \\ \end{bmatrix} \] and $x = [c_1, c_2, c_3]^T$. To solve this system, we can find the determinant of $A$. If the determinant is non-zero, the vectors are linearly independent. If the determinant is zero, the vectors are linearly dependent. The determinant of $A$ can be found as follows: \[ \begin{align} \det(A) &= 1(5\cdot9 - 6\cdot8) - 4(2\cdot9 - 3\cdot8) + 7(2\cdot6 - 3\cdot5) \\ &= 1(45 - 48) - 4(18 - 24) + 7(12 - 15) \\ &= 1(-3) + 4(6) - 7(3) \\ &= -3 + 24 - 21 \\ &= 0 \end{align} \] Since $\det(A) = 0$, the matrix $A$ is singular, and the vectors are linearly dependent. Therefore, the vectors $v_1$, $v_2$, and $v_3$ are not linearly independent.	False
baichuan4	To determine if the vectors $v_1 = [1,2,3]$, $v_2 = [4,5,6]$, and $v_3 = [7,8,9]$ are linearly independent, we need to see if there exists a nontrivial linear combination of these vectors that equals the zero vector. We can set up a matrix with these vectors as rows and perform row reduction to find if the system has a nontrivial solution. The matrix is: $$ \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{bmatrix} $$ Performing row reduction, we get: 1. Subtract 4 times the first row from the second row: $$ \begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 7 & 8 & 9 \\ \end{bmatrix} $$ 2. Subtract 7 times the first row from the third row: $$ \begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & -6 & -12 \\ \end{bmatrix} $$ 3. Add 2 times the second row to the third row: $$ \begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & 0 & 0 \\ \end{bmatrix} $$ Since we have obtained a row of zeros, this means that the system has a nontrivial solution, and therefore the vectors $v_1$, $v_2$, and $v_3$ are linearly dependent.	False
doubao	To determine if the vectors are linearly independent, we can set up the equation c1v1 + c2v2 + c3v3 = 0, where c1, c2, and c3 are scalars. So we have: c1[1,2,3] + c2[4,5,6] + c3[7,8,9] = [0,0,0] This gives the system of equations: c1 + 4c2 + 7c3 = 0 2c1 + 5c2 + 8c3 = 0 3c1 + 6c2 + 9c3 = 0 We can solve this system of equations. If the only solution is c1 = c2 = c3 = 0, then the vectors are linearly independent. Using elimination or other methods to solve the system, we find that the only solution is indeed c1 = c2 = c3 = 0. So the vectors v1 = [1,2,3], v2 = [4,5,6], and v3 = [7,8,9] are linearly independent.	False
minimax	To determine if the vectors $ \mathbf{v}_1 = [1,2,3] $, $ \mathbf{v}_2 = [4,5,6] $, and $ \mathbf{v}_3 = [7,8,9] $ are linearly independent, we need to check if the equation \[ c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3 = \mathbf{0} \] has only the trivial solution $ c_1 = c_2 = c_3 = 0 $, where $ \mathbf{0} $ is the zero vector. If there is a non-trivial solution, the vectors are linearly dependent. Let's write the equation in terms of components: \[ c_1[1,2,3] + c_2[4,5,6] + c_3[7,8,9] = [0,0,0] \] This gives us a system of linear equations: \[ \begin{align} c_1 + 4c_2 + 7c_3 &= 0 \\ 2c_1 + 5c_2 + 8c_3 &= 0 \\ 3c_1 + 6c_2 + 9c_3 &= 0 \end{align} \] We can represent this system as a matrix and perform row reduction to find the solutions: \[ \begin{bmatrix} 1 & 4 & 7 & \| & 0 \\ 2 & 5 & 8 & \| & 0 \\ 3 & 6 & 9 & \| & 0 \end{bmatrix} \] Performing row reduction (Gaussian elimination), we get: \[ \begin{bmatrix} 1 & 4 & 7 & \| & 0 \\ 0 & -3 & -6 & \| & 0 \\ 0 & -6 & -12 & \| & 0 \end{bmatrix} \] Further simplifying the second and third rows by dividing the second row by -3 and subtracting appropriate multiples of the second row from the third row: \[ \begin{bmatrix} 1 & 4 & 7 & \| & 0 \\ 0 & 1 & 2 & \| & 0 \\ 0 & 0 & 0 & \| & 0 \end{bmatrix} \] The third row now represents the equation $ 0 = 0 $, which is always true and does not provide any information about the variables $ c_1 $, $ c_2 $, and $ c_3 $. The second row represents the equation $ c_2 + 2c_3 = 0 $, which means that $ c_2 $ can be expressed in terms of $ c_3 $. The first row represents the equation $ c_1 + 4c_2 + 7c_3 = 0 $, and since $ c_2 $ is dependent on $ c_3 $, $ c_1 $ can also be expressed in terms of $ c_3 $. Since we have one free variable, $ c_3 $, and the other two variables $ c_1 $ and $ c_2 $ can be expressed in terms of $ c_3 $, the vectors are linearly dependent. There is a non-trivial solution to the equation, and thus the vectors $ \mathbf{v}_1 $, $ \mathbf{v}_2 $, and $ \mathbf{v}_3 $ are not linearly independent.	False
kimi	To determine if the vectors $\mathbf{v}_1 = [1,2,3]$, $\mathbf{v}_2 = [4,5,6]$, and $\mathbf{v}_3 = [7,8,9]$ are linearly independent, we can 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, where $c_1 = c_2 = c_3 = 0$. This equation translates to the following system of linear equations: \[ \begin{align} c_1 + 4c_2 + 7c_3 &= 0, \\ 2c_1 + 5c_2 + 8c_3 &= 0, \\ 3c_1 + 6c_2 + 9c_3 &= 0. \end{align} \] We can write this system in matrix form as: \[ \begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}. \] To determine if the vectors are linearly independent, we can calculate the determinant of the matrix. If the determinant is non-zero, the vectors are linearly independent. If the determinant is zero, the vectors are linearly dependent. Let's calculate the determinant of the matrix: \[ \text{det} = \begin{vmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{vmatrix}. \] Using the cofactor expansion along the first row: \[ \text{det} = 1 \cdot \begin{vmatrix} 5 & 8 \\ 6 & 9 \end{vmatrix} - 4 \cdot \begin{vmatrix} 2 & 8 \\ 3 & 9 \end{vmatrix} + 7 \cdot \begin{vmatrix} 2 & 5 \\ 3 & 6 \end{vmatrix}. \] Calculating the 2x2 determinants: \[ \begin{vmatrix} 5 & 8 \\ 6 & 9 \end{vmatrix} = (5 \cdot 9) - (8 \cdot 6) = 45 - 48 = -3, \] \[ \begin{vmatrix} 2 & 8 \\ 3 & 9 \end{vmatrix} = (2 \cdot 9) - (8 \cdot 3) = 18 - 24 = -6, \] \[ \begin{vmatrix} 2 & 5 \\ 3 & 6 \end{vmatrix} = (2 \cdot 6) - (5 \cdot 3) = 12 - 15 = -3. \] Substituting these values back into the determinant equation: \[ \text{det} = 1 \cdot (-3) - 4 \cdot (-6) + 7 \cdot (-3) = -3 + 24 - 21 = 0. \] Since the determinant is zero, the vectors $\mathbf{v}_1$, $\mathbf{v}_2$, and $\mathbf{v}_3$ are linearly dependent.	False
wenxinyiyan	To determine if the vectors v_1 = [1,2,3], v_2 = [4,5,6], v_3 = [7,8,9] are linearly independent, we can use the concept of determinants. We first form a matrix A with the vectors as its columns: A = [1 2 3; 4 5 6; 7 8 9] The determinant of a 3x3 matrix A, denoted as \|A\|, can be calculated using the formula: \|A\| = a11(a22a33 - a23a32) - a12(a21a33 - a23a31) + a13(a21a32 - a22a31) where aij denotes the element in the ith row and jth column of matrix A. For our matrix A, the determinant would be: \|A\| = 1(59 - 68) - 2(49 - 67) + 3(48 - 57) = 1(45 - 48) - 2(36 - 42) + 3(32 - 35) = -3 + 12 - 9 = 0 Since the determinant of matrix A is 0, the vectors v_1, v_2, and v_3 are linearly dependent. This means that one of the vectors can be expressed as a linear combination of the other two vectors.	False
zhipu	To determine if the vectors $ v_1 = [1, 2, 3] $, $ v_2 = [4, 5, 6] $, and $ v_3 = [7, 8, 9] $ are linearly independent, we need to check if the only solution to the equation $ c_1 v_1 + c_2 v_2 + c_3 v_3 = 0 $ is $ c_1 = c_2 = c_3 = 0 $. This can be written as: \[ c_1 \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} + c_2 \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix} + c_3 \begin{bmatrix} 7 \\ 8 \\ 9 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \] This leads to the system of linear equations: 1. $ c_1 + 4c_2 + 7c_3 = 0 $ 2. $ 2c_1 + 5c_2 + 8c_3 = 0 $ 3. $ 3c_1 + 6c_2 + 9c_3 = 0 $ We can write this system in matrix form as: \[ \begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \] To determine if the vectors are linearly independent, we need to check the determinant of the coefficient matrix: \[ \text{det} \begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{bmatrix} \] Calculating the determinant: \[ \text{det} \begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{bmatrix} = 1 \cdot (5 \cdot 9 - 8 \cdot 6) - 4 \cdot (2 \cdot 9 - 8 \cdot 3) + 7 \cdot (2 \cdot 6 - 5 \cdot 3) \] \[ = 1 \cdot (45 - 48) - 4 \cdot (18 - 24) + 7 \cdot (12 - 15) \] \[ = 1 \cdot (-3) - 4 \cdot (-6) + 7 \cdot (-3) \] \[ = -3 + 24 - 21 \] \[ = 0 \] Since the determinant of the matrix is 0, the vectors $ v_1 $, $ v_2 $, and $ v_3 $ are linearly dependent. Therefore, the vectors $ v_1 = [1, 2, 3] $, $ v_2 = [4, 5, 6] $, and $ v_3 = [7, 8, 9] $ are not linearly independent.	False
gpto1	To determine whether the vectors $ \mathbf{v}_1 = [1,\,2,\,3] $, $ \mathbf{v}_2 = [4,\,5,\,6] $, and $ \mathbf{v}_3 = [7,\,8,\,9] $ are linearly independent, we can use the concept of the determinant of a matrix formed by these vectors. In $\mathbb{R}^3$, if the determinant of the matrix formed by placing the vectors as columns is zero, the vectors are linearly dependent; if the determinant is non-zero, they are linearly independent. Step 1: Form the Matrix Construct matrix $ A $ by placing the vectors $ \mathbf{v}_1, \mathbf{v}_2, $ and $ \mathbf{v}_3 $ as its columns: \[ A = \begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \\ \end{bmatrix} \] Step 2: Compute the Determinant of $ A $ Compute $ \det(A) $ using the rule for the determinant of a $ 3 \times 3 $ matrix: \[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] where: \[ A = \begin{bmatrix} a & d & g \\ b & e & h \\ c & f & i \\ \end{bmatrix} \] Plug in the values: \[ \begin{aligned} & a = 1, \quad d = 4, \quad g = 7 \\ & b = 2, \quad e = 5, \quad h = 8 \\ & c = 3, \quad f = 6, \quad i = 9 \\ \end{aligned} \] Compute each term: 1. $ ei - fh $: \[ (5)(9) - (6)(8) = 45 - 48 = -3 \] 2. $ di - fg $: \[ (4)(9) - (6)(7) = 36 - 42 = -6 \] 3. $ dh - eg $: \[ (4)(8) - (5)(7) = 32 - 35 = -3 \] Compute $ \det(A) $: \[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) = (1)(-3) - (2)(-6) + (3)(-3) \] Simplify: \[ \det(A) = -3 + 12 - 9 = 0 \] Step 3: Interpret the Result Since $ \det(A) = 0 $, the matrix $ A $ is singular, and its columns (the vectors $ \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 $) are linearly dependent. Alternative Explanation We can also observe the relationship among the vectors directly. Notice that: \[ \mathbf{v}_2 - \mathbf{v}_1 = [4 - 1,\, 5 - 2,\, 6 - 3] = [3,\,3,\,3] \] \[ \mathbf{v}_3 - \mathbf{v}_2 = [7 - 4,\,8 - 5,\,9 - 6] = [3,\,3,\,3] \] The differences between consecutive vectors are the same, indicating that the vectors lie in the same plane and are linearly dependent. Conclusion Because the determinant of the matrix formed by the vectors is zero, the vectors $ \mathbf{v}_1, \mathbf{v}_2, $ and $ \mathbf{v}_3 $ are linearly dependent. Answer: Since the determinant is zero, the vectors are linearly dependent.	False
deepseekv3	To determine whether the vectors $ v_1 = [1, 2, 3] $, $ v_2 = [4, 5, 6] $, and $ v_3 = [7, 8, 9] $ are linearly independent, we can use the concept of linear independence in vector spaces. Specifically, we will check if the only solution to the equation: \[ c_1 v_1 + c_2 v_2 + c_3 v_3 = 0 \] is $ c_1 = c_2 = c_3 = 0 $. If this is the case, the vectors are linearly independent; otherwise, they are linearly dependent. ### Step 1: Write the vectors as a matrix Arrange the vectors as columns of a matrix: \[ A = \begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{bmatrix} \] ### Step 2: Compute the determinant of the matrix The vectors are linearly independent if and only if the determinant of $ A $ is nonzero. Compute the determinant: \[ \det(A) = 1 \cdot (5 \cdot 9 - 6 \cdot 8) - 4 \cdot (2 \cdot 9 - 3 \cdot 8) + 7 \cdot (2 \cdot 6 - 3 \cdot 5) \] Simplify each term: \[ \det(A) = 1 \cdot (45 - 48) - 4 \cdot (18 - 24) + 7 \cdot (12 - 15) \] \[ \det(A) = 1 \cdot (-3) - 4 \cdot (-6) + 7 \cdot (-3) \] \[ \det(A) = -3 + 24 - 21 \] \[ \det(A) = 0 \] ### Step 3: Analyze the determinant Since $ \det(A) = 0 $, the vectors $ v_1, v_2, v_3 $ are linearly dependent. ### Step 4: Verify dependency To confirm, observe that $ v_3 = 2v_2 - v_1 $: \[ [7, 8, 9] = 2 \cdot [4, 5, 6] - [1, 2, 3] = [8, 10, 12] - [1, 2, 3] \] This shows that $ v_3 $ is a linear combination of $ v_1 $ and $ v_2 $, confirming linear dependence. ### Final Answer: The vectors $ v_1 = [1, 2, 3] $, $ v_2 = [4, 5, 6] $, and $ v_3 = [7, 8, 9] $ are linearly dependent.	False