| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the system of linear equations using Cramer's Rule, we first need to find the determinant of the coefficient matrix (D), and then the determinants of the matrices obtained by replacing the columns of the coefficient matrix with the constant terms (Dx, Dy, Dz). The solutions for x, y, and z will be Dx/D, Dy/D, and Dz/D respectively. Given the system: 1) 2x + 3y + z = 8 2) 4x + 4y + 4z = 12 3) x + y + 8z = 10 First, let's find the determinant of the coefficient matrix (D). The coefficient matrix is: | 2 3 1 | | 4 4 4 | | 1 1 8 | D = 2(4*8 - 4*1) - 3(4*8 - 4*1) + 1(4*1 - 4*1) = 2(32 - 4) - 3(32 - 4) + 1(4 - 4) = 2(28) - 3(28) + 0 = 56 - 84 = -28 Now, let's find Dx by replacing the first column with the constants: | 8 3 1 | |12 4 4 | |10 1 8 | Dx = 8(4*8 - 4*1) - 3(12*8 - 12*1) + 1(12*1 - 10*4) = 8(32 - 4) - 3(96 - 12) + 1(12 - 40) = 8(28) - 3(84) - 28 = 224 - 252 - 28 = -56 Now, let's find Dy by replacing the second column with the constants: | 2 8 1 | | 4 12 4 | | 1 10 8 | Dy = 2(12*8 - 4*10) - 8(4*8 - 4*1) + 1(4*10 - 12*1) = 2(96 - 40) - 8(32 - 4) + 1(40 - 12) = 2(56) - 8(28) + 28 = 112 - 224 + 28 = -84 Now, let's find Dz by replacing the third column with the constants: | 2 3 8 | | 4 4 12 | | 1 1 10 | Dz = 2(4*10 - 12*1) - 3(4*10 - 12*1) + 8(4*1 - 4*1) = 2(40 - 12) - 3(40 - 12) + 8(4 - 4) = 2(28) - 3(28) + 0 = 56 - 84 = -28 Now, we can find the solutions for x, y, and z: x = Dx / D = -56 / -28 = 2 y = Dy / D = -84 / -28 = 3 z = Dz / D = -28 / -28 = 1 So the solutions are x = 2, y = 3, and z = 1. |
[-1, 3, 1] |
| deepseekr1 | To solve the system of equations using Cramer's Rule, we follow these steps: **Given Equations:** 1. \(2x + 3y + z = 8\) 2. \(4x + 4y + 4z = 12\) 3. \(x + y + 8z = 10\) **Step 1: Compute the Determinant of the Coefficient Matrix (\(D\)):** \[ D = \begin{vmatrix} 2 & 3 & 1 \\ 4 & 4 & 4 \\ 1 & 1 & 8 \end{vmatrix} \] Expanding the determinant: \[ D = 2(4 \cdot 8 - 4 \cdot 1) - 3(4 \cdot 8 - 4 \cdot 1) + 1(4 \cdot 1 - 4 \cdot 1) = 2(28) - 3(28) + 0 = -28 \] **Step 2: Compute Determinants for Each Variable (\(D_x, D_y, D_z\)):** - **For \(x\) (\(D_x\)):** \[ D_x = \begin{vmatrix} 8 & 3 & 1 \\ 12 & 4 & 4 \\ 10 & 1 & 8 \end{vmatrix} \] \[ D_x = 8(32 - 4) - 3(96 - 40) + 1(12 - 40) = 28 \Rightarrow x = \frac{D_x}{D} = \frac{28}{-28} = -1 \] - **For \(y\) (\(D_y\)):** \[ D_y = \begin{vmatrix} 2 & 8 & 1 \\ 4 & 12 & 4 \\ 1 & 10 & 8 \end{vmatrix} \] \[ D_y = 2(96 - 40) - 8(32 - 4) + 1(40 - 12) = -84 \Rightarrow y = \frac{D_y}{D} = \frac{-84}{-28} = 3 \] - **For \(z\) (\(D_z\)):** \[ D_z = \begin{vmatrix} 2 & 3 & 8 \\ 4 & 4 & 12 \\ 1 & 1 & 10 \end{vmatrix} \] \[ D_z = 2(40 - 12) - 3(40 - 12) + 8(0) = -28 \Rightarrow z = \frac{D_z}{D} = \frac{-28}{-28} = 1 \] **Final Solution:** \[ \boxed{x = -1,\ y = 3,\ z = 1} \] Verification by substituting back into the original equations confirms the solution is correct. |
[-1, 3, 1] |
| gpto1 | To solve the system of linear equations using **Cramer's Rule**, we first need to set up the coefficient matrix and compute the necessary determinants. **Given equations:** 1. \(2x + 3y + z = 8\) 2. \(4x + 4y + 4z = 12\) 3. \(x + y + 8z = 10\) --- ### Step 1: Write the system in matrix form Let’s represent the system as \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the vector of variables, and \(B\) is the constants vector. **Coefficient matrix \(A\):** \[ A = \begin{pmatrix} 2 & 3 & 1 \\ 4 & 4 & 4 \\ 1 & 1 & 8 \\ \end{pmatrix} \] **Variables vector \(X\):** \[ X = \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} \] **Constants vector \(B\):** \[ B = \begin{pmatrix} 8 \\ 12 \\ 10 \\ \end{pmatrix} \] --- ### Step 2: Compute the determinant of matrix \(A\) The determinant \(\det(A)\) is computed as: \[ \det(A) = \begin{vmatrix} 2 & 3 & 1 \\ 4 & 4 & 4 \\ 1 & 1 & 8 \\ \end{vmatrix} \] We'll calculate the determinant using the method of **cofactor expansion** along the first row. **Cofactor expansion along the first row:** \[ \det(A) = a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13} \] Where: - \(M_{ij}\) is the minor of element \(a_{ij}\), calculated by deleting row \(i\) and column \(j\). **Compute minors:** 1. **Minor \(M_{11}\):** \[ M_{11} = \begin{vmatrix} 4 & 4 \\ 1 & 8 \\ \end{vmatrix} = (4)(8) - (4)(1) = 32 - 4 = 28 \] 2. **Minor \(M_{12}\):** \[ M_{12} = \begin{vmatrix} 4 & 4 \\ 1 & 8 \\ \end{vmatrix} = (4)(8) - (4)(1) = 32 - 4 = 28 \] _Note: The minors \(M_{11}\) and \(M_{12}\) are the same due to the symmetry in the matrix._ 3. **Minor \(M_{13}\):** \[ M_{13} = \begin{vmatrix} 4 & 4 \\ 1 & 1 \\ \end{vmatrix} = (4)(1) - (4)(1) = 4 - 4 = 0 \] **Calculate determinant:** \[ \det(A) = (2)(28) - (3)(28) + (1)(0) = 56 - 84 + 0 = -28 \] --- ### Step 3: Compute determinants for \(x\), \(y\), and \(z\) We replace the corresponding column in \(A\) with vector \(B\) to compute determinants \(D_x\), \(D_y\), and \(D_z\). #### 3.1 Determinant for \(x\): \(\det(A_x)\) Replace the first column of \(A\) with \(B\): \[ A_x = \begin{pmatrix} \color{blue}{8} & 3 & 1 \\ \color{blue}{12} & 4 & 4 \\ \color{blue}{10} & 1 & 8 \\ \end{pmatrix} \] Compute \(\det(A_x)\): \[ \det(A_x) = \begin{vmatrix} 8 & 3 & 1 \\ 12 & 4 & 4 \\ 10 & 1 & 8 \\ \end{vmatrix} \] **Cofactor expansion along the first row:** \[ \det(A_x) = (8)(M_{11}) - (3)(M_{12}) + (1)(M_{13}) \] **Compute minors:** 1. **Minor \(M_{11}\):** \[ M_{11} = \begin{vmatrix} 4 & 4 \\ 1 & 8 \\ \end{vmatrix} = (4)(8) - (4)(1) = 32 - 4 = 28 \] 2. **Minor \(M_{12}\):** \[ M_{12} = \begin{vmatrix} 12 & 4 \\ 10 & 8 \\ \end{vmatrix} = (12)(8) - (4)(10) = 96 - 40 = 56 \] 3. **Minor \(M_{13}\):** \[ M_{13} = \begin{vmatrix} 12 & 4 \\ 10 & 1 \\ \end{vmatrix} = (12)(1) - (4)(10) = 12 - 40 = -28 \] **Calculate determinant:** \[ \det(A_x) = (8)(28) - (3)(56) + (1)(-28) = 224 - 168 -28 = 28 \] --- #### 3.2 Determinant for \(y\): \(\det(A_y)\) Replace the second column of \(A\) with \(B\): \[ A_y = \begin{pmatrix} 2 & \color{blue}{8} & 1 \\ 4 & \color{blue}{12} & 4 \\ 1 & \color{blue}{10} & 8 \\ \end{pmatrix} \] Compute \(\det(A_y)\): \[ \det(A_y) = \begin{vmatrix} 2 & 8 & 1 \\ 4 & 12 & 4 \\ 1 & 10 & 8 \\ \end{vmatrix} \] **Cofactor expansion along the first row:** \[ \det(A_y) = (2)(M_{11}) - (8)(M_{12}) + (1)(M_{13}) \] **Compute minors:** 1. **Minor \(M_{11}\):** \[ M_{11} = \begin{vmatrix} 12 & 4 \\ 10 & 8 \\ \end{vmatrix} = (12)(8) - (4)(10) = 96 - 40 = 56 \] 2. **Minor \(M_{12}\):** \[ M_{12} = \begin{vmatrix} 4 & 4 \\ 1 & 8 \\ \end{vmatrix} = (4)(8) - (4)(1) = 32 - 4 = 28 \] 3. **Minor \(M_{13}\):** \[ M_{13} = \begin{vmatrix} 4 & 12 \\ 1 & 10 \\ \end{vmatrix} = (4)(10) - (12)(1) = 40 - 12 = 28 \] **Calculate determinant:** \[ \det(A_y) = (2)(56) - (8)(28) + (1)(28) = 112 - 224 + 28 = -84 \] --- #### 3.3 Determinant for \(z\): \(\det(A_z)\) Replace the third column of \(A\) with \(B\): \[ A_z = \begin{pmatrix} 2 & 3 & \color{blue}{8} \\ 4 & 4 & \color{blue}{12} \\ 1 & 1 & \color{blue}{10} \\ \end{pmatrix} \] Compute \(\det(A_z)\): \[ \det(A_z) = \begin{vmatrix} 2 & 3 & 8 \\ 4 & 4 & 12 \\ 1 & 1 & 10 \\ \end{vmatrix} \] **Cofactor expansion along the first row:** \[ \det(A_z) = (2)(M_{11}) - (3)(M_{12}) + (8)(M_{13}) \] **Compute minors:** 1. **Minor \(M_{11}\):** \[ M_{11} = \begin{vmatrix} 4 & 12 \\ 1 & 10 \\ \end{vmatrix} = (4)(10) - (12)(1) = 40 - 12 = 28 \] 2. **Minor \(M_{12}\):** \[ M_{12} = \begin{vmatrix} 4 & 12 \\ 1 & 10 \\ \end{vmatrix} = (4)(10) - (12)(1) = 40 - 12 = 28 \] 3. **Minor \(M_{13}\):** \[ M_{13} = \begin{vmatrix} 4 & 4 \\ 1 & 1 \\ \end{vmatrix} = (4)(1) - (4)(1) = 4 - 4 = 0 \] **Calculate determinant:** \[ \det(A_z) = (2)(28) - (3)(28) + (8)(0) = 56 - 84 + 0 = -28 \] --- ### Step 4: Compute the values of \(x\), \(y\), and \(z\) Using Cramer's Rule: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{28}{-28} = -1 \] \[ y = \frac{\det(A_y)}{\det(A)} = \frac{-84}{-28} = 3 \] \[ z = \frac{\det(A_z)}{\det(A)} = \frac{-28}{-28} = 1 \] --- ### **Solution:** \[ \boxed{ \begin{cases} x = -1 \\ y = 3 \\ z = 1 \\ \end{cases} } \] --- ### Verification To ensure the solution is correct, substitute \(x\), \(y\), and \(z\) back into the original equations. 1. **First equation:** \[ 2x + 3y + z = 2(-1) + 3(3) + (1) = -2 + 9 + 1 = 8 \quad \checkmark \] 2. **Second equation:** \[ 4x + 4y + 4z = 4(-1) + 4(3) + 4(1) = -4 + 12 + 4 = 12 \quad \checkmark \] 3. **Third equation:** \[ x + y + 8z = (-1) + 3 + 8(1) = -1 + 3 + 8 = 10 \quad \checkmark \] All original equations are satisfied, confirming that the solution is correct. |
[-1, 3, 1] |
| GPT-4o | Cramer's rule is a mathematical theorem used to solve a system of linear equations with as many equations as unknowns, provided that the determinant of the coefficient matrix is non-zero. Given the system of linear equations: \[ 2x + 3y + z = 8 \] \[ 4x + 4y + 4z = 12 \] \[ x + y + 8z = 10 \] We can express this system in matrix form as \( A\mathbf{x} = \mathbf{b} \), where: \[ A = \begin{pmatrix} 2 & 3 & 1 \\ 4 & 4 & 4 \\ 1 & 1 & 8 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 8 \\ 12 \\ 10 \end{pmatrix} \] To use Cramer's rule, we need to compute the determinant of the coefficient matrix \( A \) and the determinants of matrices formed by replacing each column of \( A \) with the vector \( \mathbf{b} \). First, let's calculate \( \det(A) \): \[ A = \begin{pmatrix} 2 & 3 & 1 \\ 4 & 4 & 4 \\ 1 & 1 & 8 \end{pmatrix} \] To find \( \det(A) \), we can use the rule of Sarrus (for a 3x3 matrix): \[ \det(A) = 2 \cdot (4 \cdot 8 - 4 \cdot 1) - 3 \cdot (4 \cdot 8 - 4 \cdot 1) + 1 \cdot (4 \cdot 1 - 4 \cdot 1) \] \[ = 2 \cdot (32 - 4) - 3 \cdot (32 - 4) + 1 \cdot (4 - 4) \] \[ = 2 \cdot 28 - 3 \cdot 28 + 0 \] \[ = 56 - 84 \] \[ = -28 \] Since \(\det(A) \neq 0\), we can proceed with Cramer's rule. Next, we calculate the determinants of the matrices formed by replacing each column of \( A \) with \( \mathbf{b} \): 1. \( A_x \) (replace the first column of \( A \) with \( \mathbf{b} \)): \[ A_x = \begin{pmatrix} 8 & 3 & 1 \\ 12 & 4 & 4 \\ 10 & 1 & 8 \end{pmatrix} \] \[ \det(A_x) = 8 \cdot (4 \cdot 8 - 4 \cdot 1) - 3 \cdot (12 \cdot 8 - 4 \cdot 10) + 1 \cdot (12 \cdot 1 - 4 \cdot 10) \] \[ = 8 \cdot (32 - 4) - 3 \cdot (96 - 40) + 1 \cdot (12 - 40) \] \[ = 8 \cdot 28 - 3 \cdot 56 + 1 \cdot (-28) \] \[ = 224 - 168 - 28 \] \[ = 28 \] 2. \( A_y \) (replace the second column of \( A \) with \( \mathbf{b} \)): \[ A_y = \begin{pmatrix} 2 & 8 & 1 \\ 4 & 12 & 4 \\ 1 & 10 & 8 \end{pmatrix} \] \[ \det(A_y) = 2 \cdot (12 \cdot 8 - 4 \cdot 10) - 8 \cdot (4 \cdot 8 - 4 \cdot 1) + 1 \cdot (4 \cdot 10 - 12 \cdot 1) \] \[ = 2 \cdot (96 - 40) - 8 \cdot (32 - 4) + 1 \cdot (40 - 12) \] \[ = 2 \cdot 56 - 8 \cdot 28 + 1 \cdot 28 \] \[ = 112 - 224 + 28 \] \[ = -84 \] 3. \( A_z \) (replace the third column of \( A \) with \( \mathbf{b} \)): \[ A_z = |
[-1, 3, 1] |
| llama3.1-70B-Instruct | To solve the system of linear equations using Cramer's rule, we first need to find the determinant of the coefficient matrix (D) and the determinants of the matrices obtained by replacing each column of the coefficient matrix with the constant terms (Dx, Dy, Dz). The given equations are: 1) 2x + 3y + z = 8 2) 4x + 4y + 4z = 12 3) x + y + 8z = 10 The coefficient matrix (A) is: | 2 3 1 | | 4 4 4 | | 1 1 8 | The constant matrix (B) is: | 8 | | 12| | 10| Now, let's calculate the determinants: D = | 2 3 1 | | 4 4 4 | | 1 1 8 | D = 2(32 - 4) - 3(32 - 4) + 1(4 - 4) D = 2(28) - 3(28) + 1(0) D = 56 - 84 + 0 D = -28 Dx = | 8 3 1 | | 12 4 4 | | 10 1 8 | Dx = 8(32 - 4) - 3(96 - 40) + 1(12 - 40) Dx = 8(28) - 3(56) + 1(-28) Dx = 224 - 168 - 28 Dx = 28 Dy = | 2 8 1 | | 4 12 4 | | 1 10 8 | Dy = 2(96 - 40) - 8(32 - 4) + 1(40 - 12) Dy = 2(56) - 8(28) + 1(28) Dy = 112 - 224 + 28 Dy = -84 Dz = | 2 3 8 | | 4 4 12 | | 1 1 10 | Dz = 2(40 - 12) - 3(40 - 12) + 8(4 - 4) Dz = 2(28) - 3(28) + 8(0) Dz = 56 - 84 + 0 Dz = -28 Now, we can find the values of x, y, and z using Cramer's rule: x = Dx / D = 28 / -28 = -1 y = Dy / D = -84 / -28 = 3 z = Dz / D = -28 / -28 = 1 Therefore, the solution to the system of linear equations is x = -1, y = 3, and z = 1. |
[-1, 3, 1] |
| llama3.1-405B-Instruct | To solve the system of linear equations using Cramer's rule, we need to find the determinants of the coefficient matrix and the matrices obtained by replacing each column of the coefficient matrix with the constant terms. The coefficient matrix is: | 2 3 1 | | 4 4 4 | | 1 1 8 | The determinant of the coefficient matrix is: | 2 3 1 | | 4 4 4 | | 1 1 8 | = 2(32-4) - 3(32-4) + 1(4-4) = 2(28) - 3(28) + 1(0) = 56 - 84 + 0 = -28 Now, we need to find the determinants of the matrices obtained by replacing each column of the coefficient matrix with the constant terms. For x: | 8 3 1 | | 12 4 4 | | 10 1 8 | = 8(32-4) - 3(96-40) + 1(12-40) = 8(28) - 3(56) + 1(-28) = 224 - 168 - 28 = 28 For y: | 2 8 1 | | 4 12 4 | | 1 10 8 | = 2(96-40) - 8(32-4) + 1(40-12) = 2(56) - 8(28) + 1(28) = 112 - 224 + 28 = -84 For z: | 2 3 8 | | 4 4 12 | | 1 1 10 | = 2(40-12) - 3(40-12) + 8(4-4) = 2(28) - 3(28) + 8(0) = 56 - 84 + 0 = -28 Now, we can use Cramer's rule to find the values of x, y, and z: x = | 8 3 1 | | 12 4 4 | | 10 1 8 | / | 2 3 1 | | 4 4 4 | | 1 1 8 | = 28 / -28 = -1 y = | 2 8 1 | | 4 12 4 | | 1 10 8 | / | 2 3 1 | | 4 4 4 | | 1 1 8 | = -84 / -28 = 3 z = | 2 3 8 | | 4 4 12 | | 1 1 10 | / | 2 3 1 | | 4 4 4 | | 1 1 8 | = -28 / -28 = 1 Therefore, the values of x, y, and z are -1, 3, and 1, respectively. |
[-1, 3, 1] |
| DeepSeek-Coder-V2 | To solve the given system of linear equations using Cramer's Rule, we first need to write the system in matrix form. The system of equations is: 1. \(2x + 3y + z = 8\) 2. \(4x + 4y + 4z = 12\) 3. \(x + y + 8z = 10\) We can represent this system as \(AX = B\), where: \[ A = \begin{pmatrix} 2 & 3 & 1 \\ 4 & 4 & 4 \\ 1 & 1 & 8 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 8 \\ 12 \\ 10 \end{pmatrix} \] Cramer's Rule states that the solution for each variable \(x, y, z\) can be found by replacing the corresponding column of \(A\) with \(B\) and then dividing the determinant of this new matrix by the determinant of \(A\). First, we need to calculate the determinant of \(A\): \[ \text{det}(A) = \begin{vmatrix} 2 & 3 & 1 \\ 4 & 4 & 4 \\ 1 & 1 & 8 \end{vmatrix} \] Using the cofactor expansion along the first row: \[ \text{det}(A) = 2 \begin{vmatrix} 4 & 4 \\ 1 & 8 \end{vmatrix} - 3 \begin{vmatrix} 4 & 4 \\ 1 & 8 \end{vmatrix} + 1 \begin{vmatrix} 4 & 4 \\ 1 & 1 \end{vmatrix} \] Calculating the 2x2 determinants: \[ \begin{vmatrix} 4 & 4 \\ 1 & 8 \end{vmatrix} = 4 \cdot 8 - 4 \cdot 1 = 32 - 4 = 28 \] \[ \begin{vmatrix} 4 & 4 \\ 1 & 1 \end{vmatrix} = 4 \cdot 1 - 4 \cdot 1 = 4 - 4 = 0 \] So, \[ \text{det}(A) = 2 \cdot 28 - 3 \cdot 28 + 1 \cdot 0 = 56 - 84 + 0 = -28 \] Since \(\text{det}(A) \neq 0\), we can proceed with Cramer's Rule. Next, we calculate the determinants of the matrices formed by replacing each column of \(A\) with \(B\): 1. For \(x\): \[ A_1 = \begin{pmatrix} 8 & 3 & 1 \\ 12 & 4 & 4 \\ 10 & 1 & 8 \end{pmatrix} \] \[ \text{det}(A_1) = 8 \begin{vmatrix} 4 & 4 \\ 1 & 8 \end{vmatrix} - 3 \begin{vmatrix} 12 & 4 \\ 10 & 8 \end{vmatrix} + 1 \begin{vmatrix} 12 & 4 \\ 10 & 1 \end{vmatrix} \] \[ \text{det}(A_1) = 8 \cdot 28 - 3 \cdot (12 \cdot 8 - 4 \cdot 10) + 1 \cdot (12 \cdot 1 - 4 \cdot 10) \] \[ \text{det}(A_1) = 224 - 3 \cdot (96 - 40) + 1 \cdot (12 - 40) \] \[ \text{det}(A_1) = 224 - 3 \cdot 56 - 28 \] \[ \text{det}(A_1) = 224 - 168 - 28 = 28 \] \[ x = \frac{\text{det}(A_1)}{\text{det}(A)} = \frac{28}{-28} = -1 \] 2. For \(y\): \[ A_2 = \begin{pmatrix} 2 & 8 & 1 \\ 4 & 12 & 4 \\ 1 & 10 & 8 \end{pmatrix} \] \[ \text{det}(A_2) = 2 \begin{vmatrix} 12 & 4 \\ 10 & 8 \end{vmatrix} - 8 \begin{vmatrix} 4 & 4 \\ 1 & 8 \end{vmatrix} + 1 \begin{vmatrix} 4 & 12 \\ 1 & 10 \end{vmatrix} \] \[ \text{det}(A_2) = 2 \cdot (12 \cdot 8 - 4 \cdot 10) - 8 \cdot 28 + 1 \cdot (4 \cdot 10 - 12 \cdot 1) \] \[ \text{det}(A_2) = 2 \cdot (96 - 40) - 224 + 1 \cdot (40 - 12) \] \[ \text{det}(A_2) = 2 \cdot 56 - 224 + 28 \] \[ \text{det}(A_2) = 112 - 224 + 28 = -84 \] \[ y = \frac{\text{det}(A_2)}{\text{det}(A)} = \frac{-84}{-28} = 3 \] 3. For \(z\): \[ A_3 = \begin{pmatrix} 2 & 3 & 8 \\ 4 & 4 & 12 \\ 1 & 1 & 10 \end{pmatrix} \] \[ \text{det}(A_3) = 2 \begin{vmatrix} 4 & 12 \\ 1 & 10 \end{vmatrix} - 3 \begin{vmatrix} 4 & 12 \\ 1 & 10 \end{vmatrix} + 8 \begin{vmatrix} 4 & 4 \\ 1 & 1 \end{vmatrix} \] \[ \text{det}(A_3) = 2 \cdot (4 \cdot 10 - 12 \cdot 1) - 3 \cdot (4 \cdot 10 - 12 \cdot 1) + 8 \cdot 0 \] \[ \text{det}(A_3) = 2 \cdot (40 - 12) - 3 \cdot (40 - 12) + 0 \] \[ \text{det}(A_3) = 2 \cdot 28 - 3 \cdot 28 \] \[ \text{det}(A_3) = 56 - 84 = -28 \] \[ z = \frac{\text{det}(A_3)}{\text{det}(A)} = \frac{-28}{-28} = 1 \] Thus, the solution to the system of equations is: \[ x = -1, \quad y = 3, \quad z = 1 \] |
[-1, 3, 1] |
| Qwen2-72B-Instruct | To solve the system of linear equations using Cramer's Rule, we first need to find the determinant of the coefficient matrix (D), and then the determinants of the matrices obtained by replacing the columns of the coefficient matrix with the constant terms (Dx, Dy, Dz). The solutions for x, y, and z will be Dx/D, Dy/D, and Dz/D respectively. Given the system: 1) 2x + 3y + z = 8 2) 4x + 4y + 4z = 12 3) x + y + 8z = 10 First, let's find the determinant of the coefficient matrix (D). The coefficient matrix is: | 2 3 1 | | 4 4 4 | | 1 1 8 | D = 2(4*8 - 4*1) - 3(4*8 - 4*1) + 1(4*1 - 4*1) = 2(32 - 4) - 3(32 - 4) + 1(4 - 4) = 2(28) - 3(28) + 0 = 56 - 84 = -28 Now, let's find Dx by replacing the first column with the constants: | 8 3 1 | |12 4 4 | |10 1 8 | Dx = 8(4*8 - 4*1) - 3(12*8 - 12*1) + 1(12*1 - 10*4) = 8(32 - 4) - 3(96 - 12) + 1(12 - 40) = 8(28) - 3(84) - 28 = 224 - 252 - 28 = -56 Now, let's find Dy by replacing the second column with the constants: | 2 8 1 | | 4 12 4 | | 1 10 8 | Dy = 2(12*8 - 4*10) - 8(4*8 - 4*1) + 1(4*10 - 12*1) = 2(96 - 40) - 8(32 - 4) + 1(40 - 12) = 2(56) - 8(28) + 28 = 112 - 224 + 28 = -84 Now, let's find Dz by replacing the third column with the constants: | 2 3 8 | | 4 4 12 | | 1 1 10 | Dz = 2(4*10 - 12*1) - 3(4*10 - 12*1) + 8(4*1 - 4*1) = 2(40 - 12) - 3(40 - 12) + 8(4 - 4) = 2(28) - 3(28) + 0 = 56 - 84 = -28 Now, we can find the solutions for x, y, and z: x = Dx / D = -56 / -28 = 2 y = Dy / D = -84 / -28 = 3 z = Dz / D = -28 / -28 = 1 So the solutions are x = 2, y = 3, and z = 1. |
[-1, 3, 1] |
| baichuan4 | To apply Cramer's Rule to solve the system of linear equations: $$2x + 3y + z = 8 \quad (1)$$ $$4x + 4y + 4z = 12 \quad (2)$$ $$x + y + 8z = 10 \quad (3)$$ We first need to write the coefficient matrix $A$, the right-hand side vector $b$, and then calculate the determinants of $A$, $A_x$, $A_y$, and $A_z$ where $A_x$, $A_y$, and $A_z$ are matrices formed by replacing the respective columns of $A$ with the vector $b$. The coefficient matrix $A$ and the right-hand side vector $b$ are: $$A = \begin{bmatrix} 2 & 3 & 1 \\ 4 & 4 & 4 \\ 1 & 1 & 8 \end{bmatrix}, \quad b = \begin{bmatrix} 8 \\ 12 \\ 10 \end{bmatrix}$$ Now, we calculate the determinant of $A$: $$|A| = 2(4*8 - 4*1) - 3(4*8 - 4*1) + 1(4*1 - 4*1) = 2(32 - 4) - 3(32 - 4) + 0 = 56 - 88 = -32$$ Next, we calculate the determinants of $A_x$, $A_y$, and $A_z$: $$A_x = \begin{bmatrix} 8 & 3 & 1 \\ 12 & 4 & 4 \\ 10 & 1 & 8 \end{bmatrix}, \quad |A_x| = 8(4*8 - 4*1) - 3(12*8 - 4*10) + 1(12*1 - 4*10) = 32(8 - 1) - 3(96 - 40) - 32 = 224 - 144 - 32 = 48$$ $$A_y = \begin{bmatrix} 2 & 8 & 1 \\ 4 & 12 & 4 \\ 1 & 10 & 8 \end{bmatrix}, \quad |A_y| = 2(12*8 - 4*10) - 8(4*8 - 4*1) + 1(4*10 - 12*1) = 2(96 - 40) - 8(32 - 4) + 32 - 12 = 112 - 224 + 20 = -92$$ $$A_z = \begin{bmatrix} 2 & 3 & 8 \\ 4 & 4 & 12 \\ 1 & 1 & 10 \end{bmatrix}, \quad |A_z| = 2(4*10 - 12*1) - 3(4*10 - 4*1) + 8(4*1 - 4*1) = 2(40 - 12) - 3(40 - 4) + 0 = 56 - 108 = -52$$ Now, we can use Cramer's Rule to find the values of $x$, $y$, and $z$: $$x = \frac{|A_x|}{|A|} = \frac{48}{-32} = -1.5$$ $$y = \frac{|A_y|}{|A|} = \frac{-92}{-32} = 2.875$$ $$z = \frac{|A_z|}{|A|} = \frac{-52}{-32} = 1.625$$ So, the solution to the system of equations is $x = -1.5$, $y = 2.875$, and $z = 1.625$. |
[-1, 3, 1] |
| doubao | First find the determinant of the coefficient matrix: \[ \begin{vmatrix} 2 & 3 & 1 \\ 4 & 4 & 4 \\ 1 & 1 & 8 \end{vmatrix} = 2\begin{vmatrix} 4 & 4 \\ 1 & 8 \end{vmatrix} - 3\begin{vmatrix} 4 & 4 \\ 1 & 8 \end{vmatrix} + \begin{vmatrix} 4 & 4 \\ 1 & 1 \end{vmatrix} = 2(32-4) - 3(32-4) + (4-4) = 2(28) - 3(28) = 56 - 84 = -28 \] Now find the determinants for x, y, and z: For x: \[ \begin{vmatrix} 8 & 3 & 1 \\ 12 & 4 & 4 \\ 10 & 1 & 8 \end{vmatrix} = 8\begin{vmatrix} 4 & 4 \\ 1 & 8 \end{vmatrix} - 3\begin{vmatrix} 12 & 4 \\ 10 & 8 \end{vmatrix} + \begin{vmatrix} 12 & 4 \\ 10 & 1 \end{vmatrix} = 8(32-4) - 3(96-40) + (12-40) = 8(28) - 3(56) - 28 = 224 - 168 - 28 = 28 \] For y: \[ \begin{vmatrix} 2 & 8 & 1 \\ 4 & 12 & 4 \\ 1 & 10 & 8 \end{vmatrix} = 2\begin{vmatrix} 12 & 4 \\ 10 & 8 \end{vmatrix} - 8\begin{vmatrix} 4 & 4 \\ 1 & 8 \end{vmatrix} + \begin{vmatrix} 4 & 12 \\ 1 & 10 \end{vmatrix} = 2(96-40) - 8(32-4) + (40-12) = 2(56) - 8(28) + 28 = 112 - 224 + 28 = -84 \] For z: \[ \begin{vmatrix} 2 & 3 & 8 \\ 4 & 4 & 12 \\ 1 & 1 & 10 \end{vmatrix} = 2\begin{vmatrix} 4 & 12 \\ 1 & 10 \end{vmatrix} - 3\begin{vmatrix} 4 & 12 \\ 1 & 10 \end{vmatrix} + 8\begin{vmatrix} 4 & 4 \\ 1 & 1 \end{vmatrix} = 2(40-12) - 3(40-12) + 8(4-4) = 2(28) - 3(28) + 0 = 56 - 84 = -28 \] So $x = \frac{28}{-28} = -1$, $y = \frac{-84}{-28} = 3$, $z = \frac{-28}{-28} = 1$. |
[-1, 3, 1] |
| minimax | Cramer's Rule is a method for solving systems of linear equations with as many equations as unknowns, using determinants. For a system of three linear equations: \[ \begin{align*} a_1x + b_1y + c_1z &= d_1, \\ a_2x + b_2y + c_2z &= d_2, \\ a_3x + b_3y + c_3z &= d_3, \end{align*} \] the solution can be found using the following formulas: \[ x = \frac{\Delta_x}{\Delta}, \quad y = \frac{\Delta_y}{\Delta}, \quad z = \frac{\Delta_z}{\Delta}, \] where $\Delta$ is the determinant of the coefficient matrix, and $\Delta_x$, $\Delta_y$, and $\Delta_z$ are the determinants of matrices formed by replacing the x, y, and z columns of the coefficient matrix with the constants from the right-hand side of the equations, respectively. First, we need to find the determinant $\Delta$ of the coefficient matrix: \[ \Delta = \begin{vmatrix} 2 & 3 & 1 \\ 4 & 4 & 4 \\ 1 & 1 & 8 \\ \end{vmatrix}. \] Calculating the determinant: \[ \Delta = 2(4 \cdot 8 - 4 \cdot 1) - 3(4 \cdot 8 - 4 \cdot 1) + 1(4 \cdot 1 - 4 \cdot 1) = 2(32 - 4) - 3(32 - 4) + 1(4 - 4) = 2(28) - 3(28) + 0 = 56 - 84 = -28. \] Next, we find $\Delta_x$, $\Delta_y$, and $\Delta_z$ by replacing the respective columns with the constants from the right-hand side of the equations: \[ \Delta_x = \begin{vmatrix} 8 & 3 & 1 \\ 12 & 4 & 4 \\ 10 & 1 & 8 \\ \end{vmatrix}, \quad \Delta_y = \begin{vmatrix} 2 & 8 & 1 \\ 4 & 12 & 4 \\ 1 & 10 & 8 \\ \end{vmatrix}, \quad \Delta_z = \begin{vmatrix} 2 & 3 & 8 \\ 4 & 4 & 12 \\ 1 & 1 & 10 \\ \end{vmatrix}. \] Calculating $\Delta_x$: \[ \Delta_x = 8(4 \cdot 8 - 4 \cdot 1) - 3(12 \cdot 8 - 4 \cdot 10) + 1(12 \cdot 1 - 4 \cdot 10) = 8(32 - 4) - 3(96 - 40) + 1(12 - 40) = 8(28) - 3(56) - 28 = 224 - 168 - 28 = 28. \] Calculating $\Delta_y$: \[ \Delta_y = 2(12 \cdot 8 - 12 \cdot 10) - 8(4 \cdot 8 - 4 \cdot 10) + 1(4 \cdot 10 - 12 \cdot 1) = 2(96 - 120) - 8(32 - 40) + 1(40 - 12) = 2(-24) - 8(-8) + 28 = -48 + 64 + 28 = 44. \] Calculating $\Delta_z$: \[ \Delta_z = 2(4 \cdot 10 - 4 \cdot 1) - 3(4 \cdot 10 - 12 \cdot 1) + 8(4 \cdot 1 - 12 \cdot 1) = 2(40 - 4) - 3(40 - 12) + 8(4 - 12) = 2(36) - 3(28) - 8(8) = 72 - 84 - 64 = -76. \] Now we can find the values of x, y, and z using Cramer's Rule: \[ x = \frac{\Delta_x}{\Delta} = \frac{28}{-28} = -1, \quad y = \frac{\Delta_y}{\Delta} = \frac{44}{-28} = -\frac{11}{7}, \quad z = \frac{\Delta_z}{\Delta} = \frac{-76}{-28} = \frac{19}{7}. \] Therefore, the solution to the system of equations is: \[ x = -1, \quad y = -\frac{11}{7}, \quad z = \frac{19}{7}. \] |
[-1, 3, 1] |
| kimi | To solve the system of linear equations using Cramer's Rule, we first need to write the system in matrix form: \[ \begin{align*} 2x + 3y + z &= 8 \\ 4x + 4y + 4z &= 12 \\ x + y + 8z &= 10 \end{align*} \] This can be represented as: \[ \begin{bmatrix} 2 & 3 & 1 \\ 4 & 4 & 4 \\ 1 & 1 & 8 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 8 \\ 12 \\ 10 \end{bmatrix} \] Cramer's Rule states that for a system of linear equations \(Ax = b\), the solution is given by: \[ x_i = \frac{\det(A_i)}{\det(A)} \] where \(A_i\) is the matrix obtained by replacing the \(i\)-th column of \(A\) with the column vector \(b\), and \(\det(A)\) is the determinant of matrix \(A\). Let's calculate the determinant of \(A\): \[ \det(A) = \begin{vmatrix} 2 & 3 & 1 \\ 4 & 4 & 4 \\ 1 & 1 & 8 \end{vmatrix} \] \[ \det(A) = 2 \begin{vmatrix} 4 & 4 \\ 1 & 8 \end{vmatrix} - 3 \begin{vmatrix} 4 & 4 \\ 1 & 8 \end{vmatrix} + 1 \begin{vmatrix} 4 & 4 \\ 1 & 1 \end{vmatrix} \] \[ \det(A) = 2(4*8 - 4*1) - 3(4*8 - 4*1) + (4*1 - 4*1) \] \[ \det(A) = 2(32 - 4) - 3(32 - 4) + 0 \] \[ \det(A) = 2(28) - 3(28) + 0 \] \[ \det(A) = 56 - 84 \] \[ \det(A) = -28 \] Now, let's find \(\det(A_x)\) by replacing the first column of \(A\) with the column vector \(b\): \[ \det(A_x) = \begin{vmatrix} 8 & 3 & 1 \\ 12 & 4 & 4 \\ 10 & 1 & 8 \end{vmatrix} \] \[ \det(A_x) = 8 \begin{vmatrix} 4 & 4 \\ 1 & 8 \end{vmatrix} - 3 \begin{vmatrix} 12 & 4 \\ 10 & 8 \end{vmatrix} + 1 \begin{vmatrix} 12 & 4 \\ 10 & 1 \end{vmatrix} \] \[ \det(A_x) = 8(4*8 - 4*1) - 3(12*8 - 4*10) + (12*1 - 4*10) \] \[ \det(A_x) = 8(32 - 4) - 3(96 - 40) + (12 - 40) \] \[ \det(A_x) = 8(28) - 3(56) - 28 \] \[ \det(A_x) = 224 - 168 - 28 \] \[ \det(A_x) = 28 \] So, \(x = \frac{\det(A_x)}{\det(A)} = \frac{28}{-28} = -1\). Next, let's find \(\det(A_y)\) by replacing the second column of \(A\) with the column vector \(b\): \[ \det(A_y) = \begin{vmatrix} 2 & 8 & 1 \\ 4 & 12 & 4 \\ 1 & 10 & 8 \end{vmatrix} \] \[ \det(A_y) = 2 \begin{vmatrix} 12 & 4 \\ 10 & 8 \end{vmatrix} - 8 \begin{vmatrix} 4 & 4 \\ 1 & 8 \end{vmatrix} + 1 \begin{vmatrix} 4 & |
[-1, 3, 1] |
| wenxinyiyan | To solve the given system of linear equations using Cramer's rule, we first need to find the determinant of the coefficient matrix (D), and then the determinants of the matrices obtained by replacing the columns with the constant terms (Dx, Dy, and Dz). The system of equations is: 1. $2x + 3y + z = 8$ 2. $4x + 4y + 4z = 12$ 3. $x + y + 8z = 10$ First, let's write the coefficient matrix (A) and the constant vector (B): A = $\begin{vmatrix} 2 & 3 & 1 \\ 4 & 4 & 4 \\ 1 & 1 & 8 \end{vmatrix}$ B = $\begin{vmatrix} 8 \\ 12 \\ 10 \end{vmatrix}$ Now, we calculate the determinant of A (D): D = $\begin{vmatrix} 2 & 3 & 1 \\ 4 & 4 & 4 \\ 1 & 1 & 8 \end{vmatrix}$ = $2\begin{vmatrix} 4 & 4 \\ 1 & 8 \end{vmatrix} - 3\begin{vmatrix} 4 & 4 \\ 1 & 8 \end{vmatrix} + 1\begin{vmatrix} 4 & 4 \\ 1 & 1 \end{vmatrix}$ = $2(32 - 4) - 3(32 - 4) + 1(4 - 4)$ = $2(28) - 3(28) + 0$ = $56 - 84 + 0$ = $-28$ Next, we calculate Dx by replacing the first column of A with B and finding the determinant: Dx = $\begin{vmatrix} 8 & 3 & 1 \\ 12 & 4 & 4 \\ 10 & 1 & 8 \end{vmatrix}$ = $8\begin{vmatrix} 4 & 4 \\ 1 & 8 \end{vmatrix} - 3\begin{vmatrix} 12 & 4 \\ 10 & 8 \end{vmatrix} + 1\begin{vmatrix} 12 & 4 \\ 10 & 1 \end{vmatrix}$ = $8(32 - 4) - 3(96 - 40) + 1(12 - 40)$ = $8(28) - 3(56) + 1(-28)$ = $224 - 168 - 28$ = $28$ Now, we calculate Dy by replacing the second column of A with B and finding the determinant: Dy = $\begin{vmatrix} 2 & 8 & 1 \\ 4 & 12 & 4 \\ 1 & 10 & 8 \end{vmatrix}$ = $2\begin{vmatrix} 12 & 4 \\ 10 & 8 \end{vmatrix} - 8\begin{vmatrix} 4 & 4 \\ 1 & 8 \end{vmatrix} + 1\begin{vmatrix} 4 & 12 \\ 1 & 10 \end{vmatrix}$ = $2(96 - 40) - 8(32 - 4) + 1(40 - 12)$ = $2(56) - 8(28) + 1(28)$ = $112 - 224 + 28$ = $-84$ Finally, we calculate Dz by replacing the third column of A with B and finding the determinant: Dz = $\begin{vmatrix} 2 & 3 & 8 \\ 4 & 4 & 12 \\ 1 & 1 & 10 \end{vmatrix}$ = $2\begin{vmatrix} 4 & 12 \\ 1 & 10 \end{vmatrix} - 3\begin{vmatrix} 4 & 12 \\ 1 & 10 \end{vmatrix} + 8\begin{vmatrix} 4 & 4 \\ 1 & 1 \end{vmatrix}$ = $2(40 - 12) - 3(40 - 12) + 8(4 |
[-1, 3, 1] |
| zhipu | To solve the system of linear equations using Cramer's Rule, we first need to arrange the equations in the standard form Ax + By + Cz = D. The given system is: 1) \( 2x + 3y + z = 8 \) 2) \( 4x + 4y + 4z = 12 \) 3) \( x + y + 8z = 10 \) Let's denote the coefficient matrix as A, the variable matrix as X, and the constant matrix as B: \[ A = \begin{pmatrix} 2 & 3 & 1 \\ 4 & 4 & 4 \\ 1 & 1 & 8 \end{pmatrix} \] \[ X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \] \[ B = \begin{pmatrix} 8 \\ 12 \\ 10 \end{pmatrix} \] The system can be written in matrix form as AX = B. Cramer's Rule states that for a system of linear equations, the solution for each variable xi is given by: \[ x_i = \frac{\det(A_i)}{\det(A)} \] where \( A_i \) is the matrix formed by replacing the i-th column of A with the matrix B. First, we need to find the determinant of matrix A, \( \det(A) \): \[ \det(A) = \begin{vmatrix} 2 & 3 & 1 \\ 4 & 4 & 4 \\ 1 & 1 & 8 \end{vmatrix} \] \[ \det(A) = 2(4*8 - 4*1) - 3(4*8 - 4*1) + 1(4*1 - 4*1) \] \[ \det(A) = 2(32 - 4) - 3(32 - 4) + 1(4 - 4) \] \[ \det(A) = 2(28) - 3(28) + 1(0) \] \[ \det(A) = 56 - 84 + 0 \] \[ \det(A) = -28 \] Now we need to find the determinants of the matrices \( A_x \), \( A_y \), and \( A_z \), which are obtained by replacing the x, y, and z columns of A with the constant matrix B, respectively. \[ A_x = \begin{pmatrix} 8 & 3 & 1 \\ 12 & 4 & 4 \\ 10 & 1 & 8 \end{pmatrix} \] \[ \det(A_x) = \begin{vmatrix} 8 & 3 & 1 \\ 12 & 4 & 4 \\ 10 & 1 & 8 \end{vmatrix} \] \[ \det(A_x) = 8(4*8 - 4*1) - 3(12*8 - 4*10) + 1(12*1 - 4*10) \] \[ \det(A_x) = 8(32 - 4) - 3(96 - 40) + 1(12 - 40) \] \[ \det(A_x) = 8(28) - 3(56) + 1(-28) \] \[ \det(A_x) = 224 - 168 - 28 \] \[ \det(A_x) = 28 \] \[ A_y = \begin{pmatrix} 2 & 8 & 1 \\ 4 & 12 & 4 \\ 1 & 10 & 8 \end{pmatrix} \] \[ \det(A_y) = \begin{vmatrix} 2 & 8 & 1 \\ 4 & 12 & 4 \\ 1 & 10 & 8 \end{vmatrix} \] \[ \det(A_y) = 2(12*8 - 4*10) - 8(4*8 - 4*1) + 1(4*10 - 12*1) \] \[ \det(A_y) = 2(96 - 40) - 8(32 - 4) + 1(40 - 12) \] \[ \det(A_y) = 2(56) - 8(28) + 1(28) \] \[ \det(A_y) = 112 - 224 + 28 \] \[ \det(A_y) = -84 \] \[ A_z = \begin{pmatrix} 2 & 3 & 8 \\ 4 & 4 & 12 \\ 1 & 1 & 10 \end{pmatrix} \] \[ \det(A_z) = \begin{vmatrix} 2 & 3 & 8 \\ 4 & 4 & 12 \\ 1 & 1 & 10 \end{vmatrix} \] \[ \det(A_z) = 2(4*10 - 12*1) - 3(4*10 - 12*1) + 8(4*1 - 4*1) \] \[ \det(A_z) = 2(40 - 12) - 3(40 - 12) + 8(4 - 4) \] \[ \det(A_z) = 2(28) - 3(28) + 8(0) \] \[ \det(A_z) = 56 - 84 + 0 \] \[ \det(A_z) = -28 \] Now we can find the values of x, y, and z using Cramer's Rule: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{28}{-28} = -1 \] \[ y = \frac{\det(A_y)}{\det(A)} = \frac{-84}{-28} = 3 \] \[ z = \frac{\det(A_z)}{\det(A)} = \frac{-28}{-28} = 1 \] So the solution to the system of equations is: \[ x = -1 \] \[ y = 3 \] \[ z = 1 \] |
[-1, 3, 1] |
| gpto1 | To solve the system of linear equations using **Cramer's Rule**, we first need to set up the coefficient matrix and compute the necessary determinants. **Given equations:** 1. \(2x + 3y + z = 8\) 2. \(4x + 4y + 4z = 12\) 3. \(x + y + 8z = 10\) --- ### Step 1: Write the system in matrix form Let’s represent the system as \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the vector of variables, and \(B\) is the constants vector. **Coefficient matrix \(A\):** \[ A = \begin{pmatrix} 2 & 3 & 1 \\ 4 & 4 & 4 \\ 1 & 1 & 8 \\ \end{pmatrix} \] **Variables vector \(X\):** \[ X = \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} \] **Constants vector \(B\):** \[ B = \begin{pmatrix} 8 \\ 12 \\ 10 \\ \end{pmatrix} \] --- ### Step 2: Compute the determinant of matrix \(A\) The determinant \(\det(A)\) is computed as: \[ \det(A) = \begin{vmatrix} 2 & 3 & 1 \\ 4 & 4 & 4 \\ 1 & 1 & 8 \\ \end{vmatrix} \] We'll calculate the determinant using the method of **cofactor expansion** along the first row. **Cofactor expansion along the first row:** \[ \det(A) = a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13} \] Where: - \(M_{ij}\) is the minor of element \(a_{ij}\), calculated by deleting row \(i\) and column \(j\). **Compute minors:** 1. **Minor \(M_{11}\):** \[ M_{11} = \begin{vmatrix} 4 & 4 \\ 1 & 8 \\ \end{vmatrix} = (4)(8) - (4)(1) = 32 - 4 = 28 \] 2. **Minor \(M_{12}\):** \[ M_{12} = \begin{vmatrix} 4 & 4 \\ 1 & 8 \\ \end{vmatrix} = (4)(8) - (4)(1) = 32 - 4 = 28 \] _Note: The minors \(M_{11}\) and \(M_{12}\) are the same due to the symmetry in the matrix._ 3. **Minor \(M_{13}\):** \[ M_{13} = \begin{vmatrix} 4 & 4 \\ 1 & 1 \\ \end{vmatrix} = (4)(1) - (4)(1) = 4 - 4 = 0 \] **Calculate determinant:** \[ \det(A) = (2)(28) - (3)(28) + (1)(0) = 56 - 84 + 0 = -28 \] --- ### Step 3: Compute determinants for \(x\), \(y\), and \(z\) We replace the corresponding column in \(A\) with vector \(B\) to compute determinants \(D_x\), \(D_y\), and \(D_z\). #### 3.1 Determinant for \(x\): \(\det(A_x)\) Replace the first column of \(A\) with \(B\): \[ A_x = \begin{pmatrix} \color{blue}{8} & 3 & 1 \\ \color{blue}{12} & 4 & 4 \\ \color{blue}{10} & 1 & 8 \\ \end{pmatrix} \] Compute \(\det(A_x)\): \[ \det(A_x) = \begin{vmatrix} 8 & 3 & 1 \\ 12 & 4 & 4 \\ 10 & 1 & 8 \\ \end{vmatrix} \] **Cofactor expansion along the first row:** \[ \det(A_x) = (8)(M_{11}) - (3)(M_{12}) + (1)(M_{13}) \] **Compute minors:** 1. **Minor \(M_{11}\):** \[ M_{11} = \begin{vmatrix} 4 & 4 \\ 1 & 8 \\ \end{vmatrix} = (4)(8) - (4)(1) = 32 - 4 = 28 \] 2. **Minor \(M_{12}\):** \[ M_{12} = \begin{vmatrix} 12 & 4 \\ 10 & 8 \\ \end{vmatrix} = (12)(8) - (4)(10) = 96 - 40 = 56 \] 3. **Minor \(M_{13}\):** \[ M_{13} = \begin{vmatrix} 12 & 4 \\ 10 & 1 \\ \end{vmatrix} = (12)(1) - (4)(10) = 12 - 40 = -28 \] **Calculate determinant:** \[ \det(A_x) = (8)(28) - (3)(56) + (1)(-28) = 224 - 168 -28 = 28 \] --- #### 3.2 Determinant for \(y\): \(\det(A_y)\) Replace the second column of \(A\) with \(B\): \[ A_y = \begin{pmatrix} 2 & \color{blue}{8} & 1 \\ 4 & \color{blue}{12} & 4 \\ 1 & \color{blue}{10} & 8 \\ \end{pmatrix} \] Compute \(\det(A_y)\): \[ \det(A_y) = \begin{vmatrix} 2 & 8 & 1 \\ 4 & 12 & 4 \\ 1 & 10 & 8 \\ \end{vmatrix} \] **Cofactor expansion along the first row:** \[ \det(A_y) = (2)(M_{11}) - (8)(M_{12}) + (1)(M_{13}) \] **Compute minors:** 1. **Minor \(M_{11}\):** \[ M_{11} = \begin{vmatrix} 12 & 4 \\ 10 & 8 \\ \end{vmatrix} = (12)(8) - (4)(10) = 96 - 40 = 56 \] 2. **Minor \(M_{12}\):** \[ M_{12} = \begin{vmatrix} 4 & 4 \\ 1 & 8 \\ \end{vmatrix} = (4)(8) - (4)(1) = 32 - 4 = 28 \] 3. **Minor \(M_{13}\):** \[ M_{13} = \begin{vmatrix} 4 & 12 \\ 1 & 10 \\ \end{vmatrix} = (4)(10) - (12)(1) = 40 - 12 = 28 \] **Calculate determinant:** \[ \det(A_y) = (2)(56) - (8)(28) + (1)(28) = 112 - 224 + 28 = -84 \] --- #### 3.3 Determinant for \(z\): \(\det(A_z)\) Replace the third column of \(A\) with \(B\): \[ A_z = \begin{pmatrix} 2 & 3 & \color{blue}{8} \\ 4 & 4 & \color{blue}{12} \\ 1 & 1 & \color{blue}{10} \\ \end{pmatrix} \] Compute \(\det(A_z)\): \[ \det(A_z) = \begin{vmatrix} 2 & 3 & 8 \\ 4 & 4 & 12 \\ 1 & 1 & 10 \\ \end{vmatrix} \] **Cofactor expansion along the first row:** \[ \det(A_z) = (2)(M_{11}) - (3)(M_{12}) + (8)(M_{13}) \] **Compute minors:** 1. **Minor \(M_{11}\):** \[ M_{11} = \begin{vmatrix} 4 & 12 \\ 1 & 10 \\ \end{vmatrix} = (4)(10) - (12)(1) = 40 - 12 = 28 \] 2. **Minor \(M_{12}\):** \[ M_{12} = \begin{vmatrix} 4 & 12 \\ 1 & 10 \\ \end{vmatrix} = (4)(10) - (12)(1) = 40 - 12 = 28 \] 3. **Minor \(M_{13}\):** \[ M_{13} = \begin{vmatrix} 4 & 4 \\ 1 & 1 \\ \end{vmatrix} = (4)(1) - (4)(1) = 4 - 4 = 0 \] **Calculate determinant:** \[ \det(A_z) = (2)(28) - (3)(28) + (8)(0) = 56 - 84 + 0 = -28 \] --- ### Step 4: Compute the values of \(x\), \(y\), and \(z\) Using Cramer's Rule: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{28}{-28} = -1 \] \[ y = \frac{\det(A_y)}{\det(A)} = \frac{-84}{-28} = 3 \] \[ z = \frac{\det(A_z)}{\det(A)} = \frac{-28}{-28} = 1 \] --- ### **Solution:** \[ \boxed{ \begin{cases} x = -1 \\ y = 3 \\ z = 1 \\ \end{cases} } \] --- ### Verification To ensure the solution is correct, substitute \(x\), \(y\), and \(z\) back into the original equations. 1. **First equation:** \[ 2x + 3y + z = 2(-1) + 3(3) + (1) = -2 + 9 + 1 = 8 \quad \checkmark \] 2. **Second equation:** \[ 4x + 4y + 4z = 4(-1) + 4(3) + 4(1) = -4 + 12 + 4 = 12 \quad \checkmark \] 3. **Third equation:** \[ x + y + 8z = (-1) + 3 + 8(1) = -1 + 3 + 8 = 10 \quad \checkmark \] All original equations are satisfied, confirming that the solution is correct. |
[-1, 3, 1] |
| deepseekv3 | To solve the system of linear equations using **Cramer's Rule**, we first need to represent the system in matrix form: \[ \begin{cases} 2x + 3y + z = 8 \\ 4x + 4y + 4z = 12 \\ x + y + 8z = 10 \end{cases} \] The system can be written as \( AX = B \), where: \[ A = \begin{bmatrix} 2 & 3 & 1 \\ 4 & 4 & 4 \\ 1 & 1 & 8 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 8 \\ 12 \\ 10 \end{bmatrix} \] ### Step 1: Compute the determinant of \( A \) (\( \det(A) \)) The determinant of \( A \) is: \[ \det(A) = 2(4 \cdot 8 - 4 \cdot 1) - 3(4 \cdot 8 - 4 \cdot 1) + 1(4 \cdot 1 - 4 \cdot 1) \] \[ \det(A) = 2(32 - 4) - 3(32 - 4) + 1(4 - 4) \] \[ \det(A) = 2(28) - 3(28) + 1(0) \] \[ \det(A) = 56 - 84 + 0 = -28 \] Since \( \det(A) \neq 0 \), the system has a unique solution. ### Step 2: Compute \( \det(A_x) \) Replace the first column of \( A \) with \( B \): \[ A_x = \begin{bmatrix} 8 & 3 & 1 \\ 12 & 4 & 4 \\ 10 & 1 & 8 \end{bmatrix} \] The determinant of \( A_x \) is: \[ \det(A_x) = 8(4 \cdot 8 - 4 \cdot 1) - 3(12 \cdot 8 - 4 \cdot 10) + 1(12 \cdot 1 - 4 \cdot 10) \] \[ \det(A_x) = 8(32 - 4) - 3(96 - 40) + 1(12 - 40) \] \[ \det(A_x) = 8(28) - 3(56) + 1(-28) \] \[ \det(A_x) = 224 - 168 - 28 = 28 \] \[ x = \frac{\det(A_x)}{\det(A)} = \frac{28}{-28} = -1 \] ### Step 3: Compute \( \det(A_y) \) Replace the second column of \( A \) with \( B \): \[ A_y = \begin{bmatrix} 2 & 8 & 1 \\ 4 & 12 & 4 \\ 1 & 10 & 8 \end{bmatrix} \] The determinant of \( A_y \) is: \[ \det(A_y) = 2(12 \cdot 8 - 4 \cdot 10) - 8(4 \cdot 8 - 4 \cdot 1) + 1(4 \cdot 10 - 12 \cdot 1) \] \[ \det(A_y) = 2(96 - 40) - 8(32 - 4) + 1(40 - 12) \] \[ \det(A_y) = 2(56) - 8(28) + 1(28) \] \[ \det(A_y) = 112 - 224 + 28 = -84 \] \[ y = \frac{\det(A_y)}{\det(A)} = \frac{-84}{-28} = 3 \] ### Step 4: Compute \( \det(A_z) \) Replace the third column of \( A \) with \( B \): \[ A_z = \begin{bmatrix} 2 & 3 & 8 \\ 4 & 4 & 12 \\ 1 & 1 & 10 \end{bmatrix} \] The determinant of \( A_z \) is: \[ \det(A_z) = 2(4 \cdot 10 - 12 \cdot 1) - 3(4 \cdot 10 - 12 \cdot 1) + 8(4 \cdot 1 - 4 \cdot 1) \] \[ \det(A_z) = 2(40 - 12) - 3(40 - 12) + 8(4 - 4) \] \[ \det(A_z) = 2(28) - 3(28) + 8(0) \] \[ \det(A_z) = 56 - 84 + 0 = -28 \] \[ z = \frac{\det(A_z)}{\det(A)} = \frac{-28}{-28} = 1 \] ### Final Solution: \[ x = -1, \quad y = 3, \quad z = 1 \] |
[-1, 3, 1] |