| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | The given group element is ((3, 5), (4, 6)) in Z_7. This is a matrix group, where the operation is matrix multiplication modulo 7. To find the inverse of a 2x2 matrix in Z_7, we use the formula: M^(-1) = (1/det(M)) * adj(M) where det(M) is the determinant of M, and adj(M) is the adjugate of M. First, we compute the determinant of M: det(M) = (3 * 6) - (5 * 4) = 18 - 20 = -2 Since we are working in Z_7, we need to find the modular inverse of -2 modulo 7. The modular inverse of a number a in Z_n is a number b such that (a * b) % n = 1. In this case, we need to find a number b such that (-2 * b) % 7 = 1. By trial and error, we find that b = 4, since (-2 * 4) % 7 = 1. Next, we compute the adjugate of M. The adjugate of a 2x2 matrix is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements: adj(M) = ((6, -5), (-4, 3)) Now, we can compute the inverse of M: M^(-1) = (1/det(M)) * adj(M) = (4) * ((6, -5), (-4, 3)) Multiplying each element of the adjugate matrix by 4 and taking the result modulo 7, we get: M^(-1) = ((24 % 7, -20 % 7), (-16 % 7, 12 % 7)) = ((3, 2), (5, 5)) Finally, we can find M[0][1], which is the element in the first row and second column of M^(-1): M[0][1] = 2 Therefore, the answer is 2. |
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| deepseekr1 | To find the inverse of the group element \(\begin{pmatrix} 3 & 5 \\ 4 & 6 \end{pmatrix}\) in \(\mathbb{Z}_7\) and determine the entry \(M[0][1]\), we follow these steps: 1. **Compute the determinant**: \[ \text{det} = (3 \cdot 6 - 5 \cdot 4) \mod 7 = (18 - 20) \mod 7 = -2 \mod 7 = 5 \] 2. **Find the inverse of the determinant modulo 7**: \[ 5^{-1} \mod 7 = 3 \quad \text{(since } 5 \cdot 3 = 15 \equiv 1 \mod 7\text{)} \] 3. **Compute the adjugate matrix**: \[ \text{Adjugate} = \begin{pmatrix} 6 & -5 \\ -4 & 3 \end{pmatrix} \] 4. **Multiply each entry of the adjugate matrix by the inverse determinant modulo 7**: - First row entries: \[ 6 \cdot 3 \mod 7 = 18 \mod 7 = 4 \] \[ -5 \cdot 3 \mod 7 = -15 \mod 7 = 6 \] - Second row entries: \[ -4 \cdot 3 \mod 7 = -12 \mod 7 = 2 \] \[ 3 \cdot 3 \mod 7 = 9 \mod 7 = 2 \] 5. **Construct the inverse matrix**: \[ \begin{pmatrix} 4 & 6 \\ 2 & 2 \end{pmatrix} \] 6. **Verify the inverse matrix by multiplication with the original matrix**: - The product of the original matrix and the inverse matrix yields the identity matrix in \(\mathbb{Z}_7\). Thus, the entry \(M[0][1]\) in the inverse matrix is the element in the first row and second column, which is \(6\). \[ \boxed{6} \] |
6 |
| gpto1 | To solve the problem, we need to find the inverse \( M \) of the matrix \( A = \begin{pmatrix} 3 & 5 \\ 4 & 6 \end{pmatrix} \) over the field \( \mathbb{Z}_7 \), the integers modulo 7. Then, we will find the element \( M[0][1] \), which is the element in the first row and second column of \( M \). **Step 1: Compute the Determinant of \( A \) Modulo 7** The determinant \( D \) of \( A \) is calculated as: \[ D = (3 \times 6) - (5 \times 4) = 18 - 20 = -2 \] Calculating modulo 7: \[ D \equiv -2 \mod 7 \\ \Rightarrow D \equiv 5 \mod 7 \] So, \( D = 5 \) in \( \mathbb{Z}_7 \). **Step 2: Find the Multiplicative Inverse of the Determinant Modulo 7** We need to find an integer \( D^{-1} \) such that: \[ 5 \times D^{-1} \equiv 1 \mod 7 \] Testing integers from 1 upwards: - \( 5 \times 1 = 5 \mod 7 \neq 1 \) - \( 5 \times 2 = 10 \mod 7 = 3 \) - \( 5 \times 3 = 15 \mod 7 = 1 \) So, \( D^{-1} = 3 \). **Step 3: Compute the Adjugate Matrix of \( A \) Modulo 7** The adjugate \( \text{Adj}(A) \) of a 2x2 matrix \( A \) is: \[ \text{Adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \] Applying this to \( A \) and reducing modulo 7: \[ \text{Adj}(A) = \begin{pmatrix} 6 & -5 \\ -4 & 3 \end{pmatrix} \mod 7 \\ \Rightarrow \begin{pmatrix} 6 & 2 \\ 3 & 3 \end{pmatrix} \] (Note: \( -5 \mod 7 = 2 \) and \( -4 \mod 7 = 3 \)). **Step 4: Compute the Inverse Matrix \( M \)** Multiply each element of the adjugate matrix by \( D^{-1} \) and reduce modulo 7: \[ M = D^{-1} \times \text{Adj}(A) \mod 7 \] \[ M = 3 \times \begin{pmatrix} 6 & 2 \\ 3 & 3 \end{pmatrix} \mod 7 \\ = \begin{pmatrix} 3 \times 6 & 3 \times 2 \\ 3 \times 3 & 3 \times 3 \end{pmatrix} \mod 7 \\ = \begin{pmatrix} 18 & 6 \\ 9 & 9 \end{pmatrix} \mod 7 \\ = \begin{pmatrix} 4 & 6 \\ 2 & 2 \end{pmatrix} \] (Notes: \( 18 \mod 7 = 4 \) and \( 9 \mod 7 = 2 \)). **Step 5: Extract \( M[0][1] \)** \( M[0][1] \) is the element in the first row, second column: \[ M[0][1] = 6 \] **Answer:** 6 |
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| GPT-4o | To solve for \( M[0][1] \), we first need to find the inverse of the given matrix \(\mathbf{A} = \begin{pmatrix} 3 & 5 \\ 4 & 6 \end{pmatrix}\) in the group \(\mathbb{Z}_7\). The first step is to calculate the determinant of \(\mathbf{A}\) modulo 7. The determinant \(\det(\mathbf{A})\) is given by: \[ \det(\mathbf{A}) = 3 \cdot 6 - 5 \cdot 4 = 18 - 20 = -2. \] Now, we take \(-2\) modulo 7: \[ -2 \equiv 5 \pmod{7}. \] So, the determinant of \(\mathbf{A}\) modulo 7 is 5. Next, we need to find the multiplicative inverse of the determinant modulo 7. We need a number \(x\) such that: \[ 5x \equiv 1 \pmod{7}. \] By testing values or using the extended Euclidean algorithm, we find that the multiplicative inverse of 5 modulo 7 is 3, because: \[ 5 \cdot 3 = 15 \equiv 1 \pmod{7}. \] Now, we can find the inverse of matrix \(\mathbf{A}\) using the formula for the inverse of a \(2 \times 2\) matrix: \[ \mathbf{A}^{-1} = \frac{1}{\det(\mathbf{A})} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}. \] For \(\mathbf{A} = \begin{pmatrix} 3 & 5 \\ 4 & 6 \end{pmatrix}\): \[ \mathbf{A}^{-1} = 3 \cdot \begin{pmatrix} 6 & -5 \\ -4 & 3 \end{pmatrix} \mod 7. \] We now compute each entry modulo 7: \[ \begin{pmatrix} 6 & -5 \\ -4 & 3 \end{pmatrix} \equiv \begin{pmatrix} 6 & 2 \\ 3 & 3 \end{pmatrix} \pmod{7}, \] where \(-5 \equiv 2 \pmod{7}\) and \(-4 \equiv 3 \pmod{7}\). Now, multiply each entry by 3: \[ 3 \cdot \begin{pmatrix} 6 & 2 \\ 3 & 3 \end{pmatrix} = \begin{pmatrix} 18 & 6 \\ 9 & 9 \end{pmatrix} \equiv \begin{pmatrix} 4 & 6 \\ 2 & 2 \end{pmatrix} \pmod{7}, \] where \(18 \equiv 4 \pmod{7}\), \(6 \equiv 6 \pmod{7}\), \(9 \equiv 2 \pmod{7}\), and \(9 \equiv 2 \pmod{7}\). Thus, the inverse matrix \(M\) is: \[ M = \begin{pmatrix} 4 & 6 \\ 2 & 2 \end{pmatrix}. \] The element \(M[0][1]\) is 6. Therefore, \(M[0][1] = 6\). |
6 |
| llama3.1-70B-Instruct | To find the inverse of the group element ((3, 5), (4, 6)) in Z_7, we need to find the matrix M such that: ((3, 5), (4, 6)) × M = I where I is the identity matrix. To find the inverse, we can use the formula: M = (adj(A)) / det(A) where A is the matrix ((3, 5), (4, 6)). First, we need to find the determinant of A: det(A) = 3 × 6 - 5 × 4 = 18 - 20 = -2 Since we are working in Z_7, we need to find the multiplicative inverse of -2 modulo 7. We can do this by finding the number x such that: -2 × x ≡ 1 (mod 7) We can find x by trying out different values: -2 × 1 ≡ -2 (mod 7) -2 × 2 ≡ -4 (mod 7) -2 × 3 ≡ -6 (mod 7) -2 × 4 ≡ -8 (mod 7) -2 × 5 ≡ -10 (mod 7) -2 × 6 ≡ -12 (mod 7) We can see that -2 × 4 ≡ 1 (mod 7), so x = 4. Now, we can find the adjugate of A: adj(A) = (6, -5) (-4, 3) Finally, we can find the inverse of A: M = (adj(A)) / det(A) = (6, -5) (-4, 3) / -2 M = (3, 2) (2, 4) Therefore, M[0][1] = 2. |
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| llama3.1-405B-Instruct | To find the inverse of the group element ((3, 5), (4, 6)) in Z_7, we need to find the modular multiplicative inverse of each element. The modular multiplicative inverse of an element a in Z_n is an element b such that a*b ≡ 1 (mod n). In this case, we need to find the modular multiplicative inverse of 3, 5, 4, and 6 in Z_7. The modular multiplicative inverse of 3 in Z_7 is 5, since 3*5 ≡ 1 (mod 7). The modular multiplicative inverse of 5 in Z_7 is 3, since 5*3 ≡ 1 (mod 7). The modular multiplicative inverse of 4 in Z_7 is 2, since 4*2 ≡ 1 (mod 7). The modular multiplicative inverse of 6 in Z_7 is 6, since 6*6 ≡ 1 (mod 7). Therefore, the inverse of the group element ((3, 5), (4, 6)) in Z_7 is ((5, 3), (2, 6)). M[0][1] refers to the second element of the first row of the matrix M, which is 3. So, M[0][1] = 3. |
6 |
| DeepSeek-Coder-V2 | To solve the problem, we need to find the inverse of the group element ((3, 5), (4, 6)) in the group \( Z_7 \). Here, \( Z_7 \) refers to the set of integers modulo 7, which is a group under addition modulo 7. First, let's understand the given matrix: \[ M = \begin{pmatrix} 3 & 5 \\ 4 & 6 \end{pmatrix} \] To find the inverse of this matrix in \( Z_7 \), we need to use the formula for the inverse of a 2x2 matrix: \[ M^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \] For our matrix \( M \): \[ a = 3, \quad b = 5, \quad c = 4, \quad d = 6 \] First, calculate the determinant: \[ \text{det}(M) = ad - bc = 3 \cdot 6 - 5 \cdot 4 = 18 - 20 = -2 \] Since we are working in \( Z_7 \), we need to convert -2 to a positive number modulo 7: \[ -2 \equiv 5 \pmod{7} \] So, the determinant of \( M \) in \( Z_7 \) is 5. Now, we need to find the modular inverse of 5 in \( Z_7 \). The modular inverse of 5 modulo 7 is a number \( x \) such that: \[ 5x \equiv 1 \pmod{7} \] By testing values, we find: \[ 5 \cdot 3 = 15 \equiv 1 \pmod{7} \] Thus, the modular inverse of 5 in \( Z_7 \) is 3. Now, we can find the inverse matrix: \[ M^{-1} = 3 \begin{pmatrix} 6 & -5 \\ -4 & 3 \end{pmatrix} \] Next, we need to convert the elements of the matrix to be within the range of \( Z_7 \): \[ M^{-1} = 3 \begin{pmatrix} 6 & 2 \\ 3 & 3 \end{pmatrix} \] Now, multiply each element by 3: \[ M^{-1} = \begin{pmatrix} 18 & 6 \\ 9 & 9 \end{pmatrix} \] Convert each element to modulo 7: \[ M^{-1} = \begin{pmatrix} 4 & 6 \\ 2 & 2 \end{pmatrix} \] Finally, we need to find \( M[0][1] \): \[ M[0][1] = 6 \] Thus, the answer is: \[ \boxed{6} \] |
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| Qwen2-72B-Instruct | The given group element is ((3, 5), (4, 6)) in Z_7. This is a matrix group, where the operation is matrix multiplication modulo 7. To find the inverse of a 2x2 matrix in Z_7, we use the formula: M^(-1) = (1/det(M)) * adj(M) where det(M) is the determinant of M, and adj(M) is the adjugate of M. First, we compute the determinant of M: det(M) = (3 * 6) - (5 * 4) = 18 - 20 = -2 Since we are working in Z_7, we need to find the modular inverse of -2 modulo 7. The modular inverse of a number a in Z_n is a number b such that (a * b) % n = 1. In this case, we need to find a number b such that (-2 * b) % 7 = 1. By trial and error, we find that b = 4, since (-2 * 4) % 7 = 1. Next, we compute the adjugate of M. The adjugate of a 2x2 matrix is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements: adj(M) = ((6, -5), (-4, 3)) Now, we can compute the inverse of M: M^(-1) = (1/det(M)) * adj(M) = (4) * ((6, -5), (-4, 3)) Multiplying each element of the adjugate matrix by 4 and taking the result modulo 7, we get: M^(-1) = ((24 % 7, -20 % 7), (-16 % 7, 12 % 7)) = ((3, 2), (5, 5)) Finally, we can find M[0][1], which is the element in the first row and second column of M^(-1): M[0][1] = 2 Therefore, the answer is 2. |
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| baichuan4 | To find the inverse of the group element ((3, 5), (4, 6)) in Z_7, we need to solve the equation: (3a mod 7, 5a mod 7) + (4b mod 7, 6b mod 7) = (1, 0) This gives us two equations: 3a + 4b ≡ 1 (mod 7) 5a + 6b ≡ 0 (mod 7) We can solve this system of equations using the Euclidean algorithm for modular inverses. First, we find the inverse of 3 modulo 7. We need to find a number a such that 3a ≡ 1 (mod 7). By trial and error, we find that a = 5 satisfies this equation: 3*5 ≡ 15 ≡ 1 (mod 7). Now we can substitute a = 5 into the second equation: 5*5 + 6b ≡ 0 (mod 7) 25 + 6b ≡ 0 (mod 7) 6b ≡ -25 (mod 7) 6b ≡ 2 (mod 7) To find b, we need to find the inverse of 6 modulo 7. By trial and error, we find that b = 4 satisfies this equation: 6*4 ≡ 24 ≡ 2 (mod 7). So, the inverse of ((3, 5), (4, 6)) in Z_7 is ((5, 4), (?, ?)). To find the missing elements, we need to solve the equation: 4c + 6d ≡ 0 (mod 7) We already know that d = 4, so: 4c + 6*4 ≡ 0 (mod 7) 4c + 24 ≡ 0 (mod 7) 4c ≡ -24 (mod 7) 4c ≡ 3 (mod 7) By trial and error, we find that c = 5 satisfies this equation: 4*5 ≡ 20 ≡ 3 (mod 7). Therefore, the inverse of ((3, 5), (4, 6)) in Z_7 is ((5, 4), (5, 4)), and M[0][1] = 4. |
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| doubao | First find the determinant of the given matrix: det = (3*6 - 4*5) mod 7 = (18 - 20) mod 7 = -2 mod 7 = 5. The adjoint of the matrix is ((6, -5), (-4, 3)). To find the inverse, divide each element of the adjoint by the determinant. So the inverse M is ((6/5, -5/5), (-4/5, 3/5)) = ((6/5, -1), (-4/5, 3/5)). So M[0][1] = -1. |
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| minimax | To solve this problem, we need to understand that we are working within the group \( \mathbb{Z}_7 \), which is the set of integers modulo 7. The group operation is addition modulo 7. The element ((3, 5), (4, 6)) is a permutation of the elements of \( \mathbb{Z}_7 \), and we can interpret it as a 2x2 matrix over \( \mathbb{Z}_7 \), where the entries are the images of the elements under the permutation. The permutation ((3, 5), (4, 6)) can be represented as a matrix: \[ \begin{pmatrix} 3 & 5 \\ 4 & 6 \\ \end{pmatrix} \] However, to find the inverse of this matrix in \( \mathbb{Z}_7 \), we need to ensure that the matrix is invertible. A matrix is invertible if its determinant is non-zero in the field over which it is defined. In \( \mathbb{Z}_7 \), we calculate the determinant modulo 7. The determinant of the matrix is: \[ \text{det} = (3 \times 6) - (5 \times 4) = 18 - 20 = -2 \] Since we are working in \( \mathbb{Z}_7 \), we need to reduce this modulo 7: \[ \text{det} \equiv -2 \equiv 5 \pmod{7} \] Since 5 is non-zero in \( \mathbb{Z}_7 \), the matrix is invertible. Now, to find the inverse of the matrix, we use the formula for the inverse of a 2x2 matrix: \[ M^{-1} = \frac{1}{\text{det}} \begin{pmatrix} d & -b \\ -c & a \\ \end{pmatrix} \] where \( a, b, c, d \) are the entries of the original matrix: \[ \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} = \begin{pmatrix} 3 & 5 \\ 4 & 6 \\ \end{pmatrix} \] So the inverse matrix \( M^{-1} \) is: \[ M^{-1} = \frac{1}{5} \begin{pmatrix} 6 & -5 \\ -4 & 3 \\ \end{pmatrix} \] We need to find the multiplicative inverse of 5 modulo 7, which is the number that when multiplied by 5 gives 1 modulo 7. The multiplicative inverse of 5 in \( \mathbb{Z}_7 \) is 3 because: \[ 5 \times 3 \equiv 15 \equiv 1 \pmod{7} \] Now we can multiply each entry of the matrix by 3 modulo 7 to get the inverse matrix: \[ M^{-1} = 3 \begin{pmatrix} 6 & -5 \\ -4 & 3 \\ \end{pmatrix} = \begin{pmatrix} 18 & -15 \\ -12 & 9 \\ \end{pmatrix} \] Reducing each entry modulo 7: \[ M^{-1} = \begin{pmatrix} 4 & 2 \\ 5 & 2 \\ \end{pmatrix} \] Therefore, \( M[0][1] \) is the entry in the first row and second column of the inverse matrix, which is 2. |
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| kimi | To find the inverse of a permutation in the group Z_n (where Z_n is the set of integers modulo n), we need to find a permutation that, when composed with the original permutation, results in the identity permutation. Given the permutation (3, 5), (4, 6) in Z_7, let's first write it out in cycle notation: (3 5)(4 6) means: - 3 is mapped to 5 - 5 is mapped to 3 - 4 is mapped to 6 - 6 is mapped to 4 - 1 is mapped to 1 (since it's not in the cycle) - 2 is mapped to 2 (since it's not in the cycle) - 7 is mapped to 7 (since it's not in the cycle) The identity permutation for Z_7 is (1)(2)(3)(4)(5)(6)(7), which means each element maps to itself. To find the inverse, we need to reverse the effect of each cycle. So the inverse permutation would be: - 3 is mapped to 5 - 5 is mapped to 3 - 4 is mapped to 6 - 6 is mapped to 4 - 1 is mapped to 1 - 2 is mapped to 2 - 7 is mapped to 7 In cycle notation, this is (3 5)(4 6). However, we need to consider the entire group Z_7, which means we need to include the elements that were not in the original cycles. The inverse permutation in cycle notation for Z_7 would be (3 5)(4 6)(1)(2)(7), which is the same as the original permutation since it is a set of disjoint cycles. Now, we can write the inverse permutation in matrix form: ``` M = | 0 0 1 0 1 0 0 | | 0 0 0 0 0 1 0 | | 0 0 0 1 0 0 0 | | 0 0 0 0 1 0 0 | | 1 0 0 0 0 0 0 | | 0 1 0 0 0 0 0 | | 0 0 0 0 0 0 1 | ``` Here, M[0][1] is the element in the first row and second column of the matrix. From the matrix above, we can see that M[0][1] = 0. |
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| wenxinyiyan | To find the inverse of the group element ((3, 5), (4, 6)) in Z_7, we first need to understand that we are working with elements of a special group called the General Linear Group GL(2, Z_7), which consists of 2x2 matrices with entries in Z_7 (integers modulo 7) that have an inverse. The inverse of a matrix in this context is another matrix such that when the two matrices are multiplied (using matrix multiplication), the result is the identity matrix. The identity matrix in GL(2, Z_7) is the 2x2 matrix with 1s on the diagonal and 0s elsewhere, i.e., ((1, 0), (0, 1)). To find the inverse, we can use the formula for the inverse of a 2x2 matrix, which is: If A = ((a, b), (c, d)), then A^(-1) = (1/det(A)) * ((d, -b), (-c, a)), where det(A) = ad - bc is the determinant of A, and the multiplication and inversion are done modulo 7. For our matrix ((3, 5), (4, 6)), the determinant is (3*6) - (5*4) = 18 - 20 = -2, which is equivalent to 5 modulo 7 (since -2 + 7 = 5). So, the inverse of our matrix is: (1/5) * ((6, -5), (-4, 3)) in Z_7. However, we need to be careful with the division. In Z_7, "1/5" actually means the multiplicative inverse of 5, which is a number x such that 5x ≡ 1 (mod 7). It turns out that 3 is the multiplicative inverse of 5 in Z_7 because 5*3 = 15 ≡ 1 (mod 7). Therefore, the inverse of our matrix is: 3 * ((6, -5), (-4, 3)) = ((18, -15), (-12, 9)) = ((4, 2), (2, 2)) in Z_7, since 18 ≡ 4, -15 ≡ 2, -12 ≡ 2, and 9 ≡ 2 (mod 7). Finally, M[0][1] refers to the element in the first row and second column of the inverse matrix, which is 2. |
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| zhipu | To find the inverse of the group element ((3, 5), (4, 6)) in \( \mathbb{Z}_7 \), we need to understand that we are dealing with a 2x2 matrix and looking for its inverse within the context of the group of invertible 2x2 matrices under modulo 7 arithmetic. Let's denote the matrix as \( A \): \[ A = \begin{pmatrix} 3 & 5 \\ 4 & 6 \end{pmatrix} \] The inverse of a 2x2 matrix \( A \) is given by the formula: \[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \] where \( a, b, c, \) and \( d \) are the elements of \( A \), and \( \text{det}(A) \) is the determinant of \( A \). First, we calculate the determinant of \( A \): \[ \text{det}(A) = (3 \cdot 6) - (5 \cdot 4) \] \[ \text{det}(A) = 18 - 20 \] \[ \text{det}(A) = -2 \] In modulo 7 arithmetic, we convert \(-2\) to a positive equivalent by adding 7 until we get a non-negative result: \[ -2 + 7 = 5 \] So, \( \text{det}(A) \equiv 5 \mod 7 \). Next, we need to find the multiplicative inverse of the determinant modulo 7. The multiplicative inverse of 5 modulo 7 is the number \( x \) such that: \[ 5x \equiv 1 \mod 7 \] By testing values, we find: \[ 5 \cdot 3 = 15 \equiv 1 \mod 7 \] So, the multiplicative inverse of 5 modulo 7 is 3. Now we can compute the inverse matrix \( A^{-1} \): \[ A^{-1} = 3 \begin{pmatrix} 6 & -5 \\ -4 & 3 \end{pmatrix} \] We perform the matrix multiplication by 3 modulo 7: \[ A^{-1} = \begin{pmatrix} 3 \cdot 6 & 3 \cdot (-5) \\ 3 \cdot (-4) & 3 \cdot 3 \end{pmatrix} \mod 7 \] \[ A^{-1} = \begin{pmatrix} 18 & -15 \\ -12 & 9 \end{pmatrix} \mod 7 \] \[ A^{-1} = \begin{pmatrix} 4 & 2 \\ 2 & 2 \end{pmatrix} \] Here, we have converted each element modulo 7: - \( 18 \mod 7 = 4 \) - \( -15 \mod 7 = 2 \) (since \(-15 + 14 = -1\) and \(-1 + 7 = 6\)) - \( -12 \mod 7 = 2 \) (since \(-12 + 14 = 2\)) - \( 9 \mod 7 = 2 \) Finally, we need to find \( M[0][1] \), which corresponds to the element in the first row and second column of the inverse matrix \( A^{-1} \): \[ M[0][1] = 2 \] So, the value of \( M[0][1] \) is \( 2 \). |
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| gpto1 | To solve the problem, we need to find the inverse \( M \) of the matrix \( A = \begin{pmatrix} 3 & 5 \\ 4 & 6 \end{pmatrix} \) over the field \( \mathbb{Z}_7 \), the integers modulo 7. Then, we will find the element \( M[0][1] \), which is the element in the first row and second column of \( M \). **Step 1: Compute the Determinant of \( A \) Modulo 7** The determinant \( D \) of \( A \) is calculated as: \[ D = (3 \times 6) - (5 \times 4) = 18 - 20 = -2 \] Calculating modulo 7: \[ D \equiv -2 \mod 7 \\ \Rightarrow D \equiv 5 \mod 7 \] So, \( D = 5 \) in \( \mathbb{Z}_7 \). **Step 2: Find the Multiplicative Inverse of the Determinant Modulo 7** We need to find an integer \( D^{-1} \) such that: \[ 5 \times D^{-1} \equiv 1 \mod 7 \] Testing integers from 1 upwards: - \( 5 \times 1 = 5 \mod 7 \neq 1 \) - \( 5 \times 2 = 10 \mod 7 = 3 \) - \( 5 \times 3 = 15 \mod 7 = 1 \) So, \( D^{-1} = 3 \). **Step 3: Compute the Adjugate Matrix of \( A \) Modulo 7** The adjugate \( \text{Adj}(A) \) of a 2x2 matrix \( A \) is: \[ \text{Adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \] Applying this to \( A \) and reducing modulo 7: \[ \text{Adj}(A) = \begin{pmatrix} 6 & -5 \\ -4 & 3 \end{pmatrix} \mod 7 \\ \Rightarrow \begin{pmatrix} 6 & 2 \\ 3 & 3 \end{pmatrix} \] (Note: \( -5 \mod 7 = 2 \) and \( -4 \mod 7 = 3 \)). **Step 4: Compute the Inverse Matrix \( M \)** Multiply each element of the adjugate matrix by \( D^{-1} \) and reduce modulo 7: \[ M = D^{-1} \times \text{Adj}(A) \mod 7 \] \[ M = 3 \times \begin{pmatrix} 6 & 2 \\ 3 & 3 \end{pmatrix} \mod 7 \\ = \begin{pmatrix} 3 \times 6 & 3 \times 2 \\ 3 \times 3 & 3 \times 3 \end{pmatrix} \mod 7 \\ = \begin{pmatrix} 18 & 6 \\ 9 & 9 \end{pmatrix} \mod 7 \\ = \begin{pmatrix} 4 & 6 \\ 2 & 2 \end{pmatrix} \] (Notes: \( 18 \mod 7 = 4 \) and \( 9 \mod 7 = 2 \)). **Step 5: Extract \( M[0][1] \)** \( M[0][1] \) is the element in the first row, second column: \[ M[0][1] = 6 \] **Answer:** 6 |
6 |
| deepseekv3 | To find the inverse \( M \) of the matrix \( \begin{pmatrix} 3 & 5 \\ 4 & 6 \end{pmatrix} \) in \( \mathbb{Z}_7 \), we follow these steps: ### Step 1: Compute the determinant The determinant \( \det(A) \) of matrix \( A = \begin{pmatrix} 3 & 5 \\ 4 & 6 \end{pmatrix} \) is calculated as: \[ \det(A) = (3 \times 6) - (5 \times 4) = 18 - 20 = -2 \] In \( \mathbb{Z}_7 \), \( -2 \) is equivalent to \( 5 \) (since \( -2 \equiv 5 \mod 7 \)). So, \( \det(A) = 5 \). ### Step 2: Find the modular inverse of the determinant We need to find the modular inverse of \( 5 \) in \( \mathbb{Z}_7 \). A number \( x \) is the modular inverse of \( 5 \) if \( 5x \equiv 1 \mod 7 \). Testing values: \[ 5 \times 3 = 15 \equiv 1 \mod 7 \] Thus, the modular inverse of \( 5 \) is \( 3 \). ### Step 3: Compute the adjugate matrix The adjugate matrix \( \text{adj}(A) \) is obtained by swapping the diagonal elements and changing the sign of the off-diagonal elements: \[ \text{adj}(A) = \begin{pmatrix} 6 & -5 \\ -4 & 3 \end{pmatrix} \] In \( \mathbb{Z}_7 \), \( -5 \equiv 2 \) and \( -4 \equiv 3 \), so: \[ \text{adj}(A) = \begin{pmatrix} 6 & 2 \\ 3 & 3 \end{pmatrix} \] ### Step 4: Compute the inverse matrix The inverse matrix \( M \) is given by: \[ M = \det(A)^{-1} \times \text{adj}(A) \] Substituting the values: \[ M = 3 \times \begin{pmatrix} 6 & 2 \\ 3 & 3 \end{pmatrix} = \begin{pmatrix} 18 & 6 \\ 9 & 9 \end{pmatrix} \] Reducing modulo 7: \[ M = \begin{pmatrix} 18 \mod 7 & 6 \mod 7 \\ 9 \mod 7 & 9 \mod 7 \end{pmatrix} = \begin{pmatrix} 4 & 6 \\ 2 & 2 \end{pmatrix} \] ### Step 5: Extract \( M[0][1] \) The element \( M[0][1] \) is the entry in the first row and second column of \( M \), which is \( 6 \). Thus, the final answer is: \[ \boxed{6} \] |
6 |