| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the determinant of the given 4x4 matrix, we will expand it along the first row. The matrix is: \[ \left( \begin{array}{cccc} 0 & 4 & 3 & 0 \\ 0 & -1 & 1 & 4 \\ 4 & 0 & -1 & 1 \\ 5 & 1 & -2 & -1 \\ \end{array} \right) \] Expanding along the first row: \[ \text{det} = 0 \cdot C_{11} + 4 \cdot C_{12} + 3 \cdot C_{13} + 0 \cdot C_{14} \] Since the first and last elements of the first row are 0, their cofactors do not contribute to the determinant. We only need to calculate \(C_{12}\) and \(C_{13}\). ### Calculating \(C_{12}\): Remove the first row and second column to get the 3x3 submatrix: \[ \left( \begin{array}{ccc} 0 & 1 & 4 \\ 4 & -1 & 1 \\ 5 & -2 & -1 \\ \end{array} \right) \] The determinant of this 3x3 matrix is: \[ \text{det} = 0 \cdot \begin{vmatrix} -1 & 1 \\ -2 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 4 & 1 \\ 5 & -1 \end{vmatrix} + 4 \cdot \begin{vmatrix} 4 & -1 \\ 5 & -2 \end{vmatrix} \] Calculating the 2x2 determinants: \[ \begin{vmatrix} -1 & 1 \\ -2 & -1 \end{vmatrix} = (-1)(-1) - (1)(-2) = 1 + 2 = 3 \] \[ \begin{vmatrix} 4 & 1 \\ 5 & -1 \end{vmatrix} = 4(-1) - 1(5) = -4 - 5 = -9 \] \[ \begin{vmatrix} 4 & -1 \\ 5 & -2 \end{vmatrix} = 4(-2) - (-1)(5) = -8 + 5 = -3 \] Substituting back: \[ \text{det} = 0 \cdot 3 - 1 \cdot (-9) + 4 \cdot (-3) = 0 + 9 - 12 = -3 \] Since \(C_{12}\) is multiplied by \((-1)^{1+2}\), it is actually \(-(-3) = 3\). ### Calculating \(C_{13}\): Remove the first row and third column to get the 3x3 submatrix: \[ \left( \begin{array}{ccc} 0 & -1 & 4 \\ 4 & 0 & 1 \\ 5 & 1 & -1 \\ \end{array} \right) \] The determinant of this 3x3 matrix is: \[ \text{det} = 0 \cdot \begin{vmatrix} 0 & 1 \\ 1 & -1 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 4 & 1 \\ 5 & -1 \end{vmatrix} + 4 \cdot \begin{vmatrix} 4 & 0 \\ 5 & 1 \end{vmatrix} \] Calculating the 2x2 determinants: \[ \begin{vmatrix} 0 & 1 \\ 1 & -1 \end{vmatrix} = 0(-1) - 1(1) = 0 - 1 = -1 \] \[ \begin{vmatrix} 4 & 1 \\ 5 & -1 \end{vmatrix} = 4(-1) - 1(5) = -4 - 5 = -9 \] \[ \begin{vmatrix} 4 & 0 \\ 5 & 1 \end{vmatrix} = 4(1) - 0(5) = 4 - 0 = 4 \] Substituting back: \[ \text{det} = 0 \cdot (-1) + 1 \cdot (-9) + 4 \cdot 4 = 0 - 9 + 16 = 7 \] Since \(C_{13}\) is multiplied by \((-1)^{1+3}\), it remains 7. ### Final Determinant Calculation: \[ \text{det} = 4 \cdot 3 + 3 \cdot 7 = 12 + 21 = 33 \] The final answer is: \[ \boxed{33} \] |
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| deepseek-r1 | To find the determinant of the given 4x4 matrix: \[ \begin{pmatrix} 0 & 4 & 3 & 0 \\ 0 & -1 & 1 & 4 \\ 4 & 0 & -1 & 1 \\ 5 & 1 & -2 & -1 \\ \end{pmatrix} \] we use the cofactor expansion along the first column, which has two zeros, simplifying the calculation. The determinant is calculated as follows: 1. **Cofactor expansion along the first column**: - The non-zero entries in the first column are 4 (third row) and 5 (fourth row). - The determinant is calculated as \(4 \cdot C_{31} + 5 \cdot C_{41}\). 2. **Calculating the cofactors**: - **Minor \(M_{31}\)** (removing the third row and first column): \[ \begin{pmatrix} 4 & 3 & 0 \\ -1 & 1 & 4 \\ 1 & -2 & -1 \\ \end{pmatrix} \] The determinant of this matrix is calculated as: \[ 4 \cdot (1 \cdot (-1) - 4 \cdot (-2)) - 3 \cdot (-1 \cdot (-1) - 4 \cdot 1) + 0 = 4 \cdot 7 - 3 \cdot (-3) = 28 + 9 = 37 \] Cofactor \(C_{31} = (-1)^{3+1} \cdot 37 = 37\). - **Minor \(M_{41}\)** (removing the fourth row and first column): \[ \begin{pmatrix} 4 & 3 & 0 \\ -1 & 1 & 4 \\ 0 & -1 & 1 \\ \end{pmatrix} \] The determinant of this matrix is calculated as: \[ 4 \cdot (1 \cdot 1 - 4 \cdot (-1)) - 3 \cdot (-1 \cdot 1 - 4 \cdot 0) + 0 = 4 \cdot 5 - 3 \cdot (-1) = 20 + 3 = 23 \] Cofactor \(C_{41} = (-1)^{4+1} \cdot 23 = -23\). 3. **Summing the contributions**: - \(4 \cdot 37 = 148\) - \(5 \cdot (-23) = -115\) - Total determinant: \(148 - 115 = 33\) 4. **Verification using row operations**: - Swapping rows and converting the matrix to upper triangular form confirms the determinant calculation through the product of diagonal entries, resulting in the same value. Thus, the determinant of the matrix is \(\boxed{33}\). |
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| gpto1 | To find the determinant of the given matrix: \[ A = \begin{pmatrix} 0 & 4 & 3 & 0 \\ 0 & -1 & 1 & 4 \\ 4 & 0 & -1 & 1 \\ 5 & 1 & -2 & -1 \\ \end{pmatrix} \] First, we can swap Row 1 and Row 3 to bring the 4 to the upper-left corner: \[ A' = \begin{pmatrix} 4 & 0 & -1 & 1 \\ 0 & -1 & 1 & 4 \\ 0 & 4 & 3 & 0 \\ 5 & 1 & -2 & -1 \\ \end{pmatrix} \] Swapping rows changes the sign of the determinant. Now, we proceed to eliminate entries below the leading 4. Using row operations (which affect the determinant in known ways), we eliminate values below the pivot positions to create an upper triangular matrix. After performing appropriate row operations, we simplify the matrix and then compute the determinant. Next, we focus on computing the determinant of a 3x3 submatrix obtained after zeroing out elements below the leading positions. After simplifying, we find that the determinant of the matrix is: \[ \det(A) = -33 \] **Answer:** \(\boxed{-33}\) |
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| gpt-4o-2024-08-06 | To find the determinant of the given \(4 \times 4\) matrix, we can use cofactor expansion along any row or column. Here, we will expand along the first row for simplicity. The matrix is: \[ A = \begin{pmatrix} 0 & 4 & 3 & 0 \\ 0 & -1 & 1 & 4 \\ 4 & 0 & -1 & 1 \\ 5 & 1 & -2 & -1 \end{pmatrix} \] The determinant of \(A\) is given by: \[ \det(A) = 0 \cdot \det(M_{11}) - 4 \cdot \det(M_{12}) + 3 \cdot \det(M_{13}) - 0 \cdot \det(M_{14}) \] where \(M_{ij}\) is the \((i,j)\)-th minor of \(A\), obtained by deleting the \(i\)-th row and \(j\)-th column. Since the first and last terms are multiplied by zero, they do not contribute to the determinant. We only need to calculate the contributions from the second and third terms: ### Calculate \(\det(M_{12})\) The minor \(M_{12}\) is: \[ M_{12} = \begin{pmatrix} 0 & 1 & 4 \\ 4 & -1 & 1 \\ 5 & -2 & -1 \end{pmatrix} \] The determinant of this \(3 \times 3\) matrix is: \[ \det(M_{12}) = 0 \cdot \det\begin{pmatrix} -1 & 1 \\ -2 & -1 \end{pmatrix} - 1 \cdot \det\begin{pmatrix} 4 & 1 \\ 5 & -1 \end{pmatrix} + 4 \cdot \det\begin{pmatrix} 4 & -1 \\ 5 & -2 \end{pmatrix} \] Calculate each \(2 \times 2\) determinant: \[ \det\begin{pmatrix} -1 & 1 \\ -2 & -1 \end{pmatrix} = (-1)(-1) - (1)(-2) = 1 + 2 = 3 \] \[ \det\begin{pmatrix} 4 & 1 \\ 5 & -1 \end{pmatrix} = (4)(-1) - (1)(5) = -4 - 5 = -9 \] \[ \det\begin{pmatrix} 4 & -1 \\ 5 & -2 \end{pmatrix} = (4)(-2) - (-1)(5) = -8 + 5 = -3 \] Substitute these into the expression for \(\det(M_{12})\): \[ \det(M_{12}) = 0 - 1(-9) + 4(-3) = 9 - 12 = -3 \] ### Calculate \(\det(M_{13})\) The minor \(M_{13}\) is: \[ M_{13} = \begin{pmatrix} 0 & -1 & 4 \\ 4 & 0 & 1 \\ 5 & 1 & -1 \end{pmatrix} \] The determinant of this \(3 \times 3\) matrix is: \[ \det(M_{13}) = 0 \cdot \det\begin{pmatrix} 0 & 1 \\ 1 & -1 \end{pmatrix} + 1 \cdot \det\begin{pmatrix} 4 & 1 \\ 5 & -1 \end{pmatrix} - 4 \cdot \det\begin{pmatrix} 4 & 0 \\ 5 & 1 \end{pmatrix} \] We already calculated \(\det\begin{pmatrix} 4 & 1 \\ 5 & -1 \end{pmatrix} = -9\). Now calculate: \[ \det\begin{pmatrix} 4 & 0 \\ 5 & 1 \end{pmatrix} = (4)(1) - (0)(5) = 4 \] Substitute these into the expression for \(\det(M_{13})\): \[ \det(M_{13}) = 0 + 1(-9) - 4(4) = -9 - 16 = -25 \] ### Final Calculation Substitute \(\det(M_{12})\) and \(\det(M_{13})\) back into the expression for \(\det(A)\): \[ \det(A) = -4(-3) + 3(-25) = 12 - 75 = -63 \] Thus, the determinant of the matrix is \(\boxed{-63}\). |
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| DeepSeek-Coder-V2-Lite-Instruct | To find the determinant of the given matrix, we can use the method of cofactor expansion along the first row. The matrix is: \[ \left( \begin{array}{cccc} 0 & 4 & 3 & 0 \\ 0 & -1 & 1 & 4 \\ 4 & 0 & -1 & 1 \\ 5 & 1 & -2 & -1 \\ \end{array} \right) \] We will expand along the first row: \[ \text{det}(A) = 0 \cdot \text{det}(A_{11}) - 4 \cdot \text{det}(A_{12}) + 3 \cdot \text{det}(A_{13}) + 0 \cdot \text{det}(A_{14}) \] Since the first element in the first row is 0, the entire term involving \(\text{det}(A_{11})\) vanishes. Thus, we only need to compute the remaining terms: \[ \text{det}(A) = -4 \cdot \text{det}(A_{12}) + 3 \cdot \text{det}(A_{13}) \] Next, we need to compute the determinants of the 3x3 matrices \(A_{12}\), \(A_{13}\), and \(A_{14}\). 1. **Compute \(\text{det}(A_{12})\):** \[ A_{12} = \left( \begin{array}{ccc} 0 & -1 & 4 \\ 4 & 0 & 1 \\ 5 & 1 & -2 \\ \end{array} \right) \] Expanding along the first row: \[ \text{det}(A_{12}) = 0 \cdot \text{det}(A_{12,11}) - (-1) \cdot \text{det}(A_{12,12}) + 4 \cdot \text{det}(A_{12,13}) \] \[ \text{det}(A_{12,11}) = \left( \begin{array}{cc} 0 & 1 \\ 1 & -2 \\ \end{array} \right) = 0 \cdot 1 - 1 \cdot (-2) = 2 \] \[ \text{det}(A_{12,12}) = \left( \begin{array}{cc} 4 & 1 \\ 5 & -2 \\ \end{array} \right) = 4 \cdot (-2) - 1 \cdot 5 = -8 - 5 = -13 \] \[ \text{det}(A_{12,13}) = \left( \begin{array}{cc} 4 & 0 \\ 5 & 1 \\ \end{array} \right) = 4 \cdot 1 - 0 \cdot 5 = 4 \] \[ \text{det}(A_{12}) = 0 \cdot 2 - (-1) \cdot (-13) + 4 \cdot 4 = 0 - 13 + 16 = 3 \] 2. **Compute \(\text{det}(A_{13})\):** \[ A_{13} = \left( \begin{array}{ccc} 0 & -1 & 4 \\ 4 & -1 & 1 \\ 5 & 1 & -2 \\ \end{array} \right) \] Expanding along the first row: \[ \text{det}(A_{13}) = 0 \cdot \text{det}(A_{13,11}) - (-1) \cdot \text{det}(A_{13,12}) + 4 \cdot \text{det}(A_{13,13}) \] \[ \text{det}(A_{13,11}) = \left( \begin{array}{cc} -1 & 1 \\ 1 & -2 \\ \end{array} \right) = (-1) \cdot (-2) - 1 \cdot 1 = 2 - 1 = 1 \] \[ \text{det}(A_{13,12}) = \left( \begin{array}{cc} 4 & 1 \\ 5 & -2 \\ \end{array} \right) = 4 \cdot (-2) - 1 \cdot 5 = -8 - 5 = -13 \] \[ \text{det}(A_{13,13}) = \left( \begin{array}{cc} 4 & -1 \\ 5 & 1 \\ \end{array} \right) = 4 \cdot 1 - (-1) \cdot 5 = 4 + 5 = 9 \] \[ \text{det}(A_{13}) = 0 \cdot 1 - (-1) \cdot (-13) + 4 \cdot 9 = 0 - 13 + 36 = 23 \] Finally, substituting back into the original determinant expression: \[ \text{det}(A) = -4 \cdot 3 + 3 \cdot 23 = -12 + 69 = 57 \] Thus, the determinant of the matrix is: \[ \boxed{57} \] |
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| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: To find the determinant of the given 4x4 matrix, we can expand it along the first row. The determinant of a 4x4 matrix A can be found using the formula: \[|A| = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} + a_{14}C_{14}\] where \(C_{ij}\) is the cofactor of the element in the ith row and jth column. ## Step 2: Calculate the cofactors. The cofactor expansion along the first row gives us: \[|A| = 0*C_{11} + 4*C_{12} + 3*C_{13} + 0*C_{14}\] Since the first and last elements of the first row are 0, their cofactors do not contribute to the determinant. ## Step 3: Calculate the cofactors C12 and C13. Cofactor \(C_{12}\) is found by removing the row and column of the element \(a_{12}\), resulting in a 3x3 matrix, and multiplying it by \((-1)^{1+2}\). Cofactor \(C_{13}\) is found similarly, by removing the row and column of the element \(a_{13}\), resulting in a 3x3 matrix, and multiplying it by \((-1)^{1+3}\). ## Step 4: Calculate the determinant of the 3x3 matrices for C12 and C13. For \(C_{12}\), the 3x3 matrix is: \[ \begin{array}{ccc} 0 & 1 & 4 \\ 4 & -1 & 1 \\ 5 & -2 & -1 \\ \end{array} \] Its determinant is calculated as: \[0*(-1*(-1) - 1*(-2)) - 1*(4*(-1) - 1*5) + 4*(4*(-2) - (-1)*5)\] \[= 0 - 1*(-4 - 5) + 4*(-8 + 5)\] \[= 0 + 9 - 12\] \[= -3\] But since \(C_{12}\) is multiplied by \((-1)^{1+2}\), it is actually \(-(-3) = 3\). For \(C_{13}\), the 3x3 matrix is: \[ \begin{array}{ccc} 0 & -1 & 4 \\ 4 & 0 & 1 \\ 5 & 1 & -1 \\ \end{array} \] Its determinant is calculated as: \[0*(0*(-1) - 1*1) - (-1)*(4*(-1) - 1*5) + 4*(4*1 - 0*5)\] \[= 0 + 1*(-4 - 5) + 4*4\] \[= 0 - 9 + 16\] \[= 7\] Since \(C_{13}\) is multiplied by \((-1)^{1+3}\), it remains 7. ## Step 5: Substitute the values of C12 and C13 back into the determinant formula. \[|A| = 0 + 4*3 + 3*7 + 0\] \[|A| = 12 + 21\] \[|A| = 33\] The final answer is: $\boxed{33}$ |
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| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: To find the determinant of the given 4x4 matrix, we can expand it along the first row. The determinant of a 4x4 matrix A can be found using the formula: \[|A| = a(ei - fh) - b(di - fg) + c(dh - eg) - d(bi - cg)\] where e, f, g, i, h, and d are the 3x3 submatrices formed by removing the row and column of the respective element being multiplied. ## Step 2: Expanding along the first row, we get \[|A| = 0 \times \begin{vmatrix} -1 & 1 & 4 \\ 0 & -1 & 1 \\ 1 & -2 & -1 \end{vmatrix} - 4 \times \begin{vmatrix} 0 & 1 & 4 \\ 4 & -1 & 1 \\ 5 & -2 & -1 \end{vmatrix} + 3 \times \begin{vmatrix} 0 & -1 & 4 \\ 4 & 0 & 1 \\ 5 & 1 & -1 \end{vmatrix} - 0 \times \begin{vmatrix} 0 & -1 & 1 \\ 4 & 0 & -1 \\ 5 & 1 & -2 \end{vmatrix}\] ## Step 3: Calculating the first 3x3 determinant in the expansion. \[\begin{vmatrix} -1 & 1 & 4 \\ 0 & -1 & 1 \\ 1 & -2 & -1 \end{vmatrix} = (-1) \times \begin{vmatrix} -1 & 1 \\ -2 & -1 \end{vmatrix} - 1 \times \begin{vmatrix} 0 & 1 \\ 1 & -1 \end{vmatrix} + 4 \times \begin{vmatrix} 0 & -1 \\ 1 & -2 \end{vmatrix}\] \[= (-1) \times ((-1) \times (-1) - 1 \times (-2)) - 1 \times (0 \times (-1) - 1 \times 1) + 4 \times (0 \times (-2) - (-1) \times 1)\] \[= (-1) \times (1 + 2) - 1 \times (-1) + 4 \times (0 + 1)\] \[= (-1) \times 3 + 1 + 4\] \[= -3 + 1 + 4\] \[= 2\] ## Step 4: Calculating the second 3x3 determinant in the expansion. \[\begin{vmatrix} 0 & 1 & 4 \\ 4 & -1 & 1 \\ 5 & -2 & -1 \end{vmatrix} = 0 \times \begin{vmatrix} -1 & 1 \\ -2 & -1 \end{vmatrix} - 1 \times \begin{vmatrix} 4 & 1 \\ 5 & -1 \end{vmatrix} + 4 \times \begin{vmatrix} 4 & -1 \\ 5 & -2 \end{vmatrix}\] \[= 0 - 1 \times (4 \times (-1) - 1 \times 5) + 4 \times (4 \times (-2) - (-1) \times 5)\] \[= 0 - 1 \times (-4 - 5) + 4 \times (-8 + 5)\] \[= 0 + 9 + 4 \times (-3)\] \[= 9 - 12\] \[= -3\] ## Step 5: Calculating the third 3x3 determinant in the expansion. \[\begin{vmatrix} 0 & -1 & 4 \\ 4 & 0 & 1 \\ 5 & 1 & -1 \end{vmatrix} = 0 \times \begin{vmatrix} 0 & 1 \\ 1 & -1 \end{vmatrix} - (-1) \times \begin{vmatrix} 4 & 1 \\ 5 & -1 \end{vmatrix} + 4 \times \begin{vmatrix} 4 & 0 \\ 5 & 1 \end{vmatrix}\] \[= 0 + 1 \times (4 \times (-1) - 1 \times 5) + 4 \times (4 \times 1 - 0 \times 5)\] \[= 0 + 1 \times (-4 - 5) + 4 \times 4\] \[= 0 - 9 + 16\] \[= 7\] ## Step 6: Now, we can substitute the values of the 3x3 determinants back into the expansion. \[|A| = 0 \times 2 - 4 \times (-3) + 3 \times 7 - 0 \times 0\] \[= 0 + 12 + 21 - 0\] \[= 33\] The final answer is: $\boxed{33}$ |
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| Qwen2-72B-Instruct | To find the determinant of the given matrix, we can use the Laplace expansion (also known as cofactor expansion). The determinant of a 4x4 matrix can be calculated by expanding along any row or column. For simplicity, let's expand along the first column since it has two zeros which will simplify our calculations. Let's denote the matrix as $A$: $$ A = \left( \begin{array}{cccc} 0 & 4 & 3 & 0 \\ 0 & -1 & 1 & 4 \\ 4 & 0 & -1 & 1 \\ 5 & 1 & -2 & -1 \end{array} \right) $$ Expanding along the first column, we get: $$ \det(A) = 0 \cdot (-1)^{1+1} \cdot \det(M_{11}) + 0 \cdot (-1)^{2+1} \cdot \det(M_{21}) + 4 \cdot (-1)^{3+1} \cdot \det(M_{31}) + 5 \cdot (-1)^{4+1} \cdot \det(M_{41}) $$ where $M_{ij}$ is the minor obtained by deleting the $i$-th row and $j$-th column from $A$. Since the terms involving the first two elements of the first column are multiplied by zero, they will not contribute to the determinant. So we only need to calculate the determinants of $M_{31}$ and $M_{41}$. The minors $M_{31}$ and $M_{41}$ are: $$ M_{31} = \left( \begin{array}{ccc} 4 & 3 & 0 \\ -1 & 1 & 4 \\ 1 & -2 & -1 \end{array} \right), \quad M_{41} = \left( \begin{array}{ccc} 4 & 3 & 0 \\ -1 & 1 & 4 \\ 0 & -1 & 1 \end{array} \right) $$ Now we calculate their determinants using the Laplace expansion on the first row for each: $$ \det(M_{31}) = 4 \cdot \det \left( \begin{array}{cc} 1 & 4 \\ -2 & -1 \end{array} \right) - 3 \cdot \det \left( \begin{array}{cc} -1 & 4 \\ 1 & -1 \end{array} \right) = 4(-1 + 8) - 3(1 + 4) = 4 \cdot 7 - 3 \cdot 5 = 28 - 15 = 13 $$ $$ \det(M_{41}) = 4 \cdot \det \left( \begin{array}{cc} 1 & 4 \\ -1 & 1 \end{array} \right) - 3 \cdot \det \left( \begin{array}{cc} -1 & 4 \\ 0 & 1 \end{array} \right) = 4(1 + 4) - 3(-1) = 4 \cdot 5 + 3 = 20 + 3 = 23 $$ So the determinant of $A$ is: $$ \det(A) = 4 \cdot 13 - 5 \cdot 23 = 52 - 115 = -63 $$ Therefore, the determinant of the given matrix is $\boxed{-63}$. |
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| baichuan4 | To find the determinant of the given matrix, we can use the Laplace expansion method. However, since the first row contains two zeros, it would be more efficient to perform row operations to simplify the matrix before calculating the determinant. The matrix is: $$ A = \left( \begin{array}{cccc} 0 & 4 & 3 & 0 \\ 0 & -1 & 1 & 4 \\ 4 & 0 & -1 & 1 \\ 5 & 1 & -2 & -1 \\ \end{array} \right) $$ First, let's add the third row to the first row to replace the first row with non-zero elements: $$ R_1 \leftarrow R_1 + R_3 $$ This gives us: $$ \left( \begin{array}{cccc} 4 & 4 & 2 & 1 \\ 0 & -1 & 1 & 4 \\ 4 & 0 & -1 & 1 \\ 5 & 1 & -2 & -1 \\ \end{array} \right) $$ Next, let's subtract the first row from the third row to create a zero in the third row, first column: $$ R_3 \leftarrow R_3 - R_1 $$ This gives us: $$ \left( \begin{array}{cccc} 4 & 4 & 2 & 1 \\ 0 & -1 & 1 & 4 \\ 0 & -4 & -3 & 0 \\ 5 & 1 & -2 & -1 \\ \end{array} \right) $$ Now, let's subtract the first row multiplied by $\frac{5}{4}$ from the fourth row to create a zero in the fourth row, first column: $$ R_4 \leftarrow R_4 - \frac{5}{4}R_1 $$ This gives us: $$ \left( \begin{array}{cccc} 4 & 4 & 2 & 1 \\ 0 & -1 & 1 & 4 \\ 0 & -4 & -3 & 0 \\ 0 & -4 & -\frac{9}{2} & -\frac{9}{4} \\ \end{array} \right) $$ Next, let's add the second row to the third row to replace the third row with non-zero elements: $$ R_3 \leftarrow R_3 + R_2 $$ This gives us: $$ \left( \begin{array}{cccc} 4 & 4 & 2 & 1 \\ 0 & -1 & 1 & 4 \\ 0 & -5 & -2 & 4 \\ 0 & -4 & -\frac{9}{2} & -\frac{9}{4} \\ \end{array} \right) $$ Now, let's add the second row multiplied by $4$ to the fourth row to replace the fourth row with non-zero elements: $$ R_4 \leftarrow R_4 + 4R_2 $$ This gives us: $$ \left( \begin{array}{cccc} 4 & 4 & 2 & 1 \\ 0 & -1 & 1 & 4 \\ 0 & -5 & -2 & 4 \\ 0 & 0 & -1 & 3 \\ \end{array} \right) $$ At this point, we have a matrix in upper triangular form, and the determinant is simply the product of the diagonal elements. However, we need to account for the signs of the row operations we performed. We performed two row swaps initially (which does not change the determinant), and then we performed two row operations that each change the sign of the determinant. Therefore, the determinant is: $$ \text{det}(A) = 4 \cdot (-1) \cdot (-1) \cdot (-1) = -4 $$ Putting the final answer in a box, we have: $$ \boxed{-4} $$ |
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| doubao | Expanding along the first column: \[ \begin{align*} &\begin{vmatrix} 0 & 4 & 3 & 0 \\ 0 & -1 & 1 & 4 \\ 4 & 0 & -1 & 1 \\ 5 & 1 & -2 & -1 \end{vmatrix}\\ =&0\times(-1)^{1+1}\begin{vmatrix} -1 & 1 & 4 \\ 0 & -1 & 1 \\ 1 & -2 & -1 \end{vmatrix}+0\times(-1)^{2+1}\begin{vmatrix} 4 & 3 & 0 \\ 0 & -1 & 1 \\ 1 & -2 & -1 \end{vmatrix}+4\times(-1)^{3+1}\begin{vmatrix} 4 & 3 & 0 \\ -1 & 1 & 4 \\ 1 & -2 & -1 \end{vmatrix}+5\times(-1)^{4+1}\begin{vmatrix} 4 & 3 & 0 \\ -1 & 1 & 4 \\ 0 & -1 & 1 \end{vmatrix}\\ =&0+0+4\times\begin{vmatrix} 4 & 3 & 0 \\ -1 & 1 & 4 \\ 1 & -2 & -1 \end{vmatrix}-5\times\begin{vmatrix} 4 & 3 & 0 \\ -1 & 1 & 4 \\ 0 & -1 & 1 \end{vmatrix}\\ =&4\times((4\times((1\times(-1))-(-2)\times4))-3\times((-1)\times(-1)-4\times1)+0\times((-1)\times(-2)-1\times1))-5\times((4\times((1\times(-1))-(-2)\times4))-3\times((-1)\times(-1)-4\times1)+0\times((-1)\times(-2)-1\times1))\\ =&4\times((4\times(1+8)-3\times(1-4)+0)-5\times((4\times(1+8)-3\times(1-4)+0))\\ =&4\times((4\times9-3\times(-3)+0)-5\times((4\times9-3\times(-3)+0))\\ =&4\times((36+9+0)-5\times(36+9+0))\\ =&4\times(45-5\times45)\\ =&4\times(45-225)\\ =&4\times(-180)\\ =&-720 \end{align*} \] So the answer is $\boxed{-720}$. |
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| gpto1 | To find the determinant of the given matrix: \[ A = \begin{pmatrix} 0 & 4 & 3 & 0 \\ 0 & -1 & 1 & 4 \\ 4 & 0 & -1 & 1 \\ 5 & 1 & -2 & -1 \\ \end{pmatrix} \] First, we can swap Row 1 and Row 3 to bring the 4 to the upper-left corner: \[ A' = \begin{pmatrix} 4 & 0 & -1 & 1 \\ 0 & -1 & 1 & 4 \\ 0 & 4 & 3 & 0 \\ 5 & 1 & -2 & -1 \\ \end{pmatrix} \] Swapping rows changes the sign of the determinant. Now, we proceed to eliminate entries below the leading 4. Using row operations (which affect the determinant in known ways), we eliminate values below the pivot positions to create an upper triangular matrix. After performing appropriate row operations, we simplify the matrix and then compute the determinant. Next, we focus on computing the determinant of a 3x3 submatrix obtained after zeroing out elements below the leading positions. After simplifying, we find that the determinant of the matrix is: \[ \det(A) = -33 \] **Answer:** \(\boxed{-33}\) |
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| kimi | To find the determinant of the matrix \( A = \begin{pmatrix} 0 & 4 & 3 & 0 \\ 0 & -1 & 1 & 4 \\ 4 & 0 & -1 & 1 \\ 5 & 1 & -2 & -1 \end{pmatrix} \), we will use the cofactor expansion along the first row. The determinant of \( A \) is given by: \[ \det(A) = 0 \cdot C_{11} + 4 \cdot C_{12} + 3 \cdot C_{13} + 0 \cdot C_{14} \] where \( C_{ij} \) is the cofactor of the element \( a_{ij} \). Since the first element of the first row is 0, the determinant simplifies to: \[ \det(A) = 4 \cdot C_{12} \] Next, we need to find the cofactor \( C_{12} \). The cofactor \( C_{12} \) is given by: \[ C_{12} = (-1)^{1+2} \det\begin{pmatrix} 0 & 1 & 4 \\ 4 & -1 & 1 \\ 5 & -2 & -1 \end{pmatrix} \] This is because the cofactor is the determinant of the \( 3 \times 3 \) submatrix obtained by removing the first row and second column. Now, we calculate the determinant of the \( 3 \times 3 \) submatrix: \[ \det\begin{pmatrix} 0 & 1 & 4 \\ 4 & -1 & 1 \\ 5 & -2 & -1 \end{pmatrix} \] We expand this determinant along the first row: \[ \det\begin{pmatrix} 0 & 1 & 4 \\ 4 & -1 & 1 \\ 5 & -2 & -1 \end{pmatrix} = 0 \cdot \det\begin{pmatrix} -1 & 1 \\ -2 & -1 \end{pmatrix} - 1 \cdot \det\begin{pmatrix} 4 & 1 \\ 5 & -1 \end{pmatrix} + 4 \cdot \det\begin{pmatrix} 4 & -1 \\ 5 & -2 \end{pmatrix} \] Calculating each of these \( 2 \times 2 \) determinants: \[ \det\begin{pmatrix} -1 & 1 \\ -2 & -1 \end{pmatrix} = (-1)(-1) - (1)(-2) = 1 + 2 = 3 \] \[ \det\begin{pmatrix} 4 & 1 \\ 5 & -1 \end{pmatrix} = (4)(-1) - (1)(5) = -4 - 5 = -9 \] \[ \det\begin{pmatrix} 4 & -1 \\ 5 & -2 \end{pmatrix} = (4)(-2) - (-1)(5) = -8 + 5 = -3 \] Substituting these values back into the expansion: \[ \det\begin{pmatrix} 0 & 1 & 4 \\ 4 & -1 & 1 \\ 5 & -2 & -1 \end{pmatrix} = 0 \cdot 3 - 1 \cdot (-9) + 4 \cdot (-3) = 0 + 9 - 12 = -3 \] Now, we substitute this result back into the expression for \( C_{12} \): \[ C_{12} = (-1)^{1+2} \cdot (-3) = -1 \cdot (-3) = 3 \] Finally, we substitute \( C_{12} \) back into the expression for \( \det(A) \): \[ \det(A) = 4 \cdot 3 = 12 \] Thus, the determinant of the matrix is: \[ \boxed{12} \] |
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| minimax | To find the determinant of the given \(4 \times 4\) matrix, we can use the method of cofactor expansion. Let's expand along the first row, since it contains a zero, which will simplify our calculations. The matrix is: \[ \left(\begin{array}{cccc} 0 & 4 & 3 & 0 \\ 0 & -1 & 1 & 4 \\ 4 & 0 & -1 & 1 \\ 5 & 1 & -2 & -1 \\ \end{array}\right) \] The determinant, denoted as \( \text{det}(A) \), can be calculated as follows: \[ \text{det}(A) = 0 \cdot C_{11} + 4 \cdot C_{12} + 3 \cdot C_{13} + 0 \cdot C_{14} \] Where \( C_{ij} \) is the cofactor of the element in the \(i\)th row and \(j\)th column, which is calculated as \( (-1)^{i+j} \) times the determinant of the submatrix that remains after removing the \(i\)th row and \(j\)th column from \(A |
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| wenxinyiyan | To find the determinant of the given matrix, we will use the cofactor expansion method along the first row. The determinant can be calculated as follows: $$\begin{aligned} \text{det}(A) &= 0 \cdot \text{det}(A_{11}) + 4 \cdot \text{det}(A_{12}) + 3 \cdot \text{det}(A_{13}) + 0 \cdot \text{det}(A_{14}) \\ &= 4 \cdot \text{det} \left(\begin{array}{ccc} -1 & 1 & 4 \\ 0 & -1 & 1 \\ 1 & -2 & -1 \\ \end{array}\right) + 3 \cdot \text{det} \left(\begin{array}{ccc} 0 & 1 & 4 \\ 4 & -1 & 1 \\ 5 & -2 & -1 \\ \end{array}\right) \end{aligned}$$ Let's calculate the determinants of the two submatrices: 1. For the first submatrix: $$\begin{aligned} \text{det}(A_{12}) &= (-1) \cdot ((-1) \cdot (-1) - 1 \cdot (-2)) - 1 \cdot (0 \cdot (-1) - 1 \cdot 1) + 4 \cdot (0 \cdot (-2) - (-1) \cdot 1) \\ &= (-1) \cdot (-1 + 2) - 1 \cdot (0 + 1) + 4 \cdot (0 + 1) \\ &= (-1) \cdot 1 - 1 + 4 \\ &= -1 - 1 + 4 \\ &= 2 \end{aligned}$$ 2. For the second submatrix: $$\begin{aligned} \text{det}(A_{13}) &= 0 \cdot ((-1) \cdot (-1) - 1 \cdot (-2)) - 1 \cdot (4 \cdot (-1) - 1 \cdot 5) + 4 \cdot (4 \cdot (-2) - (-1) \cdot 5) \\ &= 0 \cdot (-1 + 2) - 1 \cdot (-4 - 5) + 4 \cdot (-8 + 5) \\ &= 0 - (-9) + 4 \cdot (-3) \\ &= 0 + 9 - 12 \\ &= -3 \end{aligned}$$ Now, let's multiply the determinants of the submatrices by the corresponding elements of the first row and sum them up: $$\begin{aligned} \text{det}(A) &= 4 \cdot 2 + 3 \cdot (-3) \\ &= 8 - 9 \\ &= -1 \end{aligned}$$ So, the determinant of the given matrix is $\boxed{-1}$. |
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| zhipu | To find the determinant of the given 4x4 matrix, we can use the method of cofactor expansion along any row or column. It is often advantageous to choose a row or column with the most zeros to simplify calculations. In this case, the first row has two zeros, so we will expand along the first row. The matrix is: \[ A = \begin{pmatrix} 0 & 4 & 3 & 0 \\ 0 & -1 & 1 & 4 \\ 4 & 0 & -1 & 1 \\ 5 & 1 & -2 & -1 \end{pmatrix} \] Expanding along the first row, we have: \[ \text{det}(A) = 0 \cdot \text{det}(M_{11}) - 4 \cdot \text{det}(M_{12}) + 3 \cdot \text{det}(M_{13}) + 0 \cdot \text{det}(M_{14}) \] where \(M_{ij}\) denotes the \((i,j)\)-minor, i.e., the matrix obtained by deleting the \(i\)-th row and \(j\)-th column. We need to compute the determinants of the 3x3 minors \(M_{12}\) and \(M_{13}\). 1. **Minor \(M_{12}\):** \[ M_{12} = \begin{pmatrix} 0 & 1 & 4 \\ 4 & -1 & 1 \\ 5 & -2 & -1 \end{pmatrix} \] \[ \text{det}(M_{12}) = 0 \cdot \begin{vmatrix} -1 & 1 \\ -2 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 4 & 1 \\ 5 & -1 \end{vmatrix} + 4 \cdot \begin{vmatrix} 4 & -1 \\ 5 & -2 \end{vmatrix} \] \[ = 0 - 1 \cdot (4 \cdot (-1) - 1 \cdot 5) + 4 \cdot (4 \cdot (-2) - (-1) \cdot 5) \] \[ = -1 \cdot (-4 - 5) + 4 \cdot (-8 + 5) \] \[ = -1 \cdot (-9) + 4 \cdot (-3) \] \[ = 9 - 12 = -3 \] 2. **Minor \(M_{13}\):** \[ M_{13} = \begin{pmatrix} 0 & -1 & 4 \\ 4 & 0 & 1 \\ 5 & 1 & -1 \end{pmatrix} \] \[ \text{det}(M_{13}) = 0 \cdot \begin{vmatrix} 0 & 1 \\ 1 & -1 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 4 & 1 \\ 5 & -1 \end{vmatrix} + 4 \cdot \begin{vmatrix} 4 & 0 \\ 5 & 1 \end{vmatrix} \] \[ = 0 + 1 \cdot (4 \cdot (-1) - 1 \cdot 5) + 4 \cdot (4 \cdot 1 - 0 \cdot 5) \] \[ = 1 \cdot (-4 - 5) + 4 \cdot 4 \] \[ = 1 \cdot (-9) + 16 \] \[ = -9 + 16 = 7 \] Now, substituting these results back into the expansion along the first row: \[ \text{det}(A) = 0 - 4 \cdot (-3) + 3 \cdot 7 + 0 \] \[ = 12 + 21 = 33 \] Thus, the determinant of the matrix is \(\boxed{33}\). |
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| deepseek-v3 | To find the determinant of the given \(4 \times 4\) matrix: \[ \left(\begin{array}{cccc} 0 & 4 & 3 & 0 \\ 0 & -1 & 1 & 4 \\ 4 & 0 & -1 & 1 \\ 5 & 1 & -2 & -1 \\ \end{array}\right) \] we will use the **cofactor expansion** method along the first row, since it contains two zeros, which simplifies calculations. ### Step 1: Apply Cofactor Expansion The determinant of a \(4 \times 4\) matrix \(A\) is computed as: \[ \det(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} + a_{14}C_{14} \] where \(a_{ij}\) are the elements of the first row and \(C_{ij}\) are the corresponding cofactors. Given the first row \([0, 4, 3, 0]\), the expression simplifies to: \[ \det(A) = 0 \cdot C_{11} + 4 \cdot C_{12} + 3 \cdot C_{13} + 0 \cdot C_{14} = 4C_{12} + 3C_{13} \] ### Step 2: Compute the Cofactors \(C_{12}\) and \(C_{13}\) **Cofactor \(C_{12}\):** Remove the first row and second column: \[ C_{12} = (-1)^{1+2} \det\left(\begin{array}{ccc} 0 & 1 & 4 \\ 4 & -1 & 1 \\ 5 & -2 & -1 \\ \end{array}\right) = -\det\left(\begin{array}{ccc} 0 & 1 & 4 \\ 4 & -1 & 1 \\ 5 & -2 & -1 \\ \end{array}\right) \] Compute the \(3 \times 3\) determinant: \[ \det\left(\begin{array}{ccc} 0 & 1 & 4 \\ 4 & -1 & 1 \\ 5 & -2 & -1 \\ \end{array}\right) = 0 \cdot \left((-1)(-1) - (1)(-2)\right) - 1 \cdot \left(4(-1) - (1)(5)\right) + 4 \cdot \left(4(-2) - (-1)(5)\right) \] \[ = 0 - 1 \cdot (-4 - 5) + 4 \cdot (-8 + 5) = 0 + 9 - 12 = -3 \] Thus, \[ C_{12} = -(-3) = 3 \] **Cofactor \(C_{13}\):** Remove the first row and third column: \[ C_{13} = (-1)^{1+3} \det\left(\begin{array}{ccc} 0 & -1 & 4 \\ 4 & 0 & 1 \\ 5 & 1 & -1 \\ \end{array}\right) = \det\left(\begin{array}{ccc} 0 & -1 & 4 \\ 4 & 0 & 1 \\ 5 & 1 & -1 \\ \end{array}\right) \] Compute the \(3 \times 3\) determinant: \[ \det\left(\begin{array}{ccc} 0 & -1 & 4 \\ 4 & 0 & 1 \\ 5 & 1 & -1 \\ \end{array}\right) = 0 \cdot \left(0(-1) - (1)(1)\right) - (-1) \cdot \left(4(-1) - (1)(5)\right) + 4 \cdot \left(4(1) - 0(5)\right) \] \[ = 0 + 1 \cdot (-4 - 5) + 4 \cdot (4 - 0) = 0 - 9 + 16 = 7 \] Thus, \[ C_{13} = 7 \] ### Step 3: Calculate the Determinant Substitute the cofactors back into the determinant expression: \[ \det(A) = 4 \cdot 3 + 3 \cdot 7 = 12 + 21 = 33 \] ### Final Answer \[ \boxed{33} \] |
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