| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | 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 of the coefficient matrix with the constants from the right-hand side of the equations (Dx and Dy). The given system of linear equations is: \[2x + 3y = 10\] \[4x + 4y = 12\] The coefficient matrix (A) is: \[A = \begin{vmatrix} 2 & 3 \\ 4 & 4 \end{vmatrix}\] To find the determinant of A (D): \[D = (2 * 4) - (3 * 4) = 8 - 12 = -4\] Next, we find the determinant Dx by replacing the first column of A with the constants from the right-hand side of the equations (10 and 12): \[Dx = \begin{vmatrix} 10 & 3 \\ 12 & 4 \end{vmatrix}\] \[Dx = (10 * 4) - (3 * 12) = 40 - 36 = 4\] Then, we find the determinant Dy by replacing the second column of A with the constants from the right-hand side of the equations: \[Dy = \begin{vmatrix} 2 & 10 \\ 4 & 12 \end{vmatrix}\] \[Dy = (2 * 12) - (10 * 4) = 24 - 40 = -16\] Now, we can find the values of x and y using Cramer's Rule: \[x = \frac{Dx}{D} = \frac{4}{-4} = -1\] \[y = \frac{Dy}{D} = \frac{-16}{-4} = 4\] Therefore, the solution to the given system of linear equations is: \[x = -1, y = 4\] |
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| deepseekr1 | To solve the system of equations using Cramer's Rule: **Given Equations:** 1. \(2x + 3y = 10\) 2. \(4x + 4y = 12\) **Step 1: Write in Matrix Form \(AX = B\)** \[ A = \begin{bmatrix} 2 & 3 \\ 4 & 4 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \end{bmatrix}, \quad B = \begin{bmatrix} 10 \\ 12 \end{bmatrix} \] **Step 2: Compute Determinant of \(A\)** \[ \det(A) = (2)(4) - (3)(4) = 8 - 12 = -4 \] **Step 3: Solve for \(x\) (Replace 1st Column with \(B\))** \[ A_x = \begin{bmatrix} 10 & 3 \\ 12 & 4 \end{bmatrix} \] \[ \det(A_x) = (10)(4) - (3)(12) = 40 - 36 = 4 \] \[ x = \frac{\det(A_x)}{\det(A)} = \frac{4}{-4} = -1 \] **Step 4: Solve for \(y\) (Replace 2nd Column with \(B\))** \[ A_y = \begin{bmatrix} 2 & 10 \\ 4 & 12 \end{bmatrix} \] \[ \det(A_y) = (2)(12) - (10)(4) = 24 - 40 = -16 \] \[ y = \frac{\det(A_y)}{\det(A)} = \frac{-16}{-4} = 4 \] **Final Solution:** \[ x = -1, \quad y = 4 \] **Verification:** Substitute \(x = -1\) and \(y = 4\) into both equations: 1. \(2(-1) + 3(4) = -2 + 12 = 10\) ✓ 2. \(4(-1) + 4(4) = -4 + 16 = 12\) ✓ **Answer:** \(x = -1\), \(y = 4\) |
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| gpto1 | To solve the system of equations using **Cramer's Rule**, we'll follow these steps: **Given equations:** \[ \begin{cases} 2x + 3y = 10 \quad \text{(1)} \\ 4x + 4y = 12 \quad \text{(2)} \end{cases} \] **Cramer's Rule Overview:** For a system of two linear equations: \[ \begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases} \] The solutions are: \[ x = \frac{D_x}{D}, \quad y = \frac{D_y}{D} \] where: - \( D \) is the determinant of the coefficients: \[ D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = a_1 b_2 - a_2 b_1 \] - \( D_x \) is the determinant when the constants replace the \( x \)-coefficients: \[ D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = c_1 b_2 - c_2 b_1 \] - \( D_y \) is the determinant when the constants replace the \( y \)-coefficients: \[ D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = a_1 c_2 - a_2 c_1 \] --- ### **Step 1: Compute the Determinant \( D \)** Extract coefficients from the equations: - From Equation (1): \( a_1 = 2 \), \( b_1 = 3 \) - From Equation (2): \( a_2 = 4 \), \( b_2 = 4 \) Calculate \( D \): \[ D = \begin{vmatrix} 2 & 3 \\ 4 & 4 \end{vmatrix} = (2)(4) - (4)(3) = 8 - 12 = \mathbf{-4} \] --- ### **Step 2: Compute the Determinant \( D_x \)** Replace \( x \)-coefficients with constants: - Constants: \( c_1 = 10 \), \( c_2 = 12 \) Calculate \( D_x \): \[ D_x = \begin{vmatrix} 10 & 3 \\ 12 & 4 \end{vmatrix} = (10)(4) - (12)(3) = 40 - 36 = \mathbf{4} \] --- ### **Step 3: Compute the Determinant \( D_y \)** Replace \( y \)-coefficients with constants: Calculate \( D_y \): \[ D_y = \begin{vmatrix} 2 & 10 \\ 4 & 12 \end{vmatrix} = (2)(12) - (4)(10) = 24 - 40 = \mathbf{-16} \] --- ### **Step 4: Solve for \( x \) and \( y \)** Compute \( x \): \[ x = \frac{D_x}{D} = \frac{4}{-4} = \mathbf{-1} \] Compute \( y \): \[ y = \frac{D_y}{D} = \frac{-16}{-4} = \mathbf{4} \] --- ### **Solution** The solution to the system is: \[ \boxed{\begin{cases} x = -1 \\ y = 4 \end{cases}} \] **Verification:** Plug \( x = -1 \) and \( y = 4 \) back into the original equations to verify. **Equation (1):** \[ 2(-1) + 3(4) = -2 + 12 = 10 \quad \checkmark \] **Equation (2):** \[ 4(-1) + 4(4) = -4 + 16 = 12 \quad \checkmark \] Both equations are satisfied, confirming the solution is correct. --- **Answer:** By using Cramer's Rule, the solution to the system is \( x = -1 \) and \( y = 4 \). |
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| GPT-4o | To solve the given system of linear equations using Cramer's Rule, we follow these steps: Given the linear equations: 1. \( 2x + 3y = 10 \) 2. \( 4x + 4y = 12 \) First, write the equations in the standard form: \[ \begin{cases} 2x + 3y = 10 \\ 4x + 4y = 12 \end{cases} \] 1. **Form the coefficient matrix \( A \):** \[ A = \begin{pmatrix} 2 & 3 \\ 4 & 4 \end{pmatrix} \] 2. **Calculate the determinant of the coefficient matrix \( \Delta \):** \[ \Delta = \det(A) = \begin{vmatrix} 2 & 3 \\ 4 & 4 \end{vmatrix} = (2 \times 4) - (3 \times 4) = 8 - 12 = -4 \] Since \(\Delta \neq 0\), the system has a unique solution. 3. **Form the matrices \( A_x \) and \( A_y \) by replacing the respective columns with the constants on the right-hand side of the equations:** \[ A_x = \begin{pmatrix} 10 & 3 \\ 12 & 4 \end{pmatrix} \] \[ A_y = \begin{pmatrix} 2 & 10 \\ 4 & 12 \end{pmatrix} \] 4. **Calculate the determinants \( \Delta_x \) and \( \Delta_y \):** \[ \Delta_x = \det(A_x) = \begin{vmatrix} 10 & 3 \\ 12 & 4 \end{vmatrix} = (10 \times 4) - (3 \times 12) = 40 - 36 = 4 \] \[ \Delta_y = \det(A_y) = \begin{vmatrix} 2 & 10 \\ 4 & 12 \end{vmatrix} = (2 \times 12) - (10 \times 4) = 24 - 40 = -16 \] 5. **Calculate the solutions for \( x \) and \( y \) using Cramer's Rule:** \[ x = \frac{\Delta_x}{\Delta} = \frac{4}{-4} = -1 \] \[ y = \frac{\Delta_y}{\Delta} = \frac{-16}{-4} = 4 \] Therefore, the solution to the system of equations is: \[ x = -1 \] \[ y = 4 \] |
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| llama3.1-70B-Instruct | A nice linear algebra problem! Cramer's rule is a method for solving systems of linear equations using determinants. Here's how to apply it to the given problem: **Step 1: Write the system of equations in matrix form** We have two linear equations with two variables, x and y. We can write them in matrix form as: 2x + 3y = 10 ... (1) 4x + 4y = 12 ... (2) The matrix form is: | 2 3 | | x | | 10 | | 4 4 | | y | = | 12 | **Step 2: Find the determinant of the coefficient matrix (D)** The coefficient matrix is: | 2 3 | | 4 4 | The determinant of this matrix is: D = (2)(4) - (3)(4) = 8 - 12 = -4 **Step 3: Find the determinants of the matrices obtained by replacing each column of the coefficient matrix with the constant matrix (Dx and Dy)** Replace the first column (coefficients of x) with the constant matrix: | 10 3 | | 12 4 | The determinant of this matrix is: Dx = (10)(4) - (3)(12) = 40 - 36 = 4 Replace the second column (coefficients of y) with the constant matrix: | 2 10 | | 4 12 | The determinant of this matrix is: Dy = (2)(12) - (10)(4) = 24 - 40 = -16 **Step 4: Apply Cramer's rule to find the values of x and y** Cramer's rule states that: x = Dx / D y = Dy / D Substitute the values: x = 4 / (-4) = -1 y = -16 / (-4) = 4 Therefore, the solution to the system of linear equations is: x = -1 y = 4 Great! We have solved the system of linear equations using Cramer's rule. |
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| 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 constant matrix. The coefficient matrix is: | 2 3 | | 4 4 | The determinant of the coefficient matrix is: det(A) = 2*4 - 3*4 = 8 - 12 = -4 The constant matrix is: | 10 3 | | 12 4 | The determinant of the constant matrix is: det(B) = 10*4 - 3*12 = 40 - 36 = 4 Now, we can use Cramer's rule to find the values of x and y: x = det(B) / det(A) = 4 / -4 = -1 y = det(C) / det(A) where det(C) is the determinant of the matrix obtained by replacing the second column of the coefficient matrix with the constant matrix: | 2 10 | | 4 12 | det(C) = 2*12 - 10*4 = 24 - 40 = -16 y = -16 / -4 = 4 Therefore, the solution to the system of linear equations is x = -1 and y = 4. |
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| DeepSeek-Coder-V2 | To solve the given system of linear equations using Cramer's rule, we first need to write the equations in standard form and identify the coefficients and constants. The given equations are: 1. \(2x + 3y = 10\) 2. \(4x + 4y = 12\) ### Step 1: Identify the Coefficient Matrix (A), the Variable Matrix (X), and the Constant Matrix (B) The coefficient matrix \(A\) is: \[ A = \begin{pmatrix} 2 & 3 \\ 4 & 4 \end{pmatrix} \] The variable matrix \(X\) is: \[ X = \begin{pmatrix} x \\ y \end{pmatrix} \] The constant matrix \(B\) is: \[ B = \begin{pmatrix} 10 \\ 12 \end{pmatrix} \] ### Step 2: Calculate the Determinant of the Coefficient Matrix (\(|A|\)) The determinant of \(A\) is calculated as follows: \[ |A| = \begin{vmatrix} 2 & 3 \\ 4 & 4 \end{vmatrix} = (2 \cdot 4) - (3 \cdot 4) = 8 - 12 = -4 \] ### Step 3: Calculate the Determinants of the Modified Matrices (\(|A_x|\) and \(|A_y|\)) To find \(|A_x|\), replace the first column of \(A\) with the constants from \(B\): \[ A_x = \begin{pmatrix} 10 & 3 \\ 12 & 4 \end{pmatrix} \] \[ |A_x| = \begin{vmatrix} 10 & 3 \\ 12 & 4 \end{vmatrix} = (10 \cdot 4) - (3 \cdot 12) = 40 - 36 = 4 \] To find \(|A_y|\), replace the second column of \(A\) with the constants from \(B\): \[ A_y = \begin{pmatrix} 2 & 10 \\ 4 & 12 \end{pmatrix} \] \[ |A_y| = \begin{vmatrix} 2 & 10 \\ 4 & 12 \end{vmatrix} = (2 \cdot 12) - (10 \cdot 4) = 24 - 40 = -16 \] ### Step 4: Solve for \(x\) and \(y\) Using Cramer's Rule Cramer's rule states that: \[ x = \frac{|A_x|}{|A|} = \frac{4}{-4} = -1 \] \[ y = \frac{|A_y|}{|A|} = \frac{-16}{-4} = 4 \] ### Final Solution The solution to the system of equations is: \[ x = -1 \] \[ y = 4 \] Thus, the solution is: \[ \boxed{(-1, 4)} \] |
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| Qwen2-72B-Instruct | 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 of the coefficient matrix with the constants from the right-hand side of the equations (Dx and Dy). The given system of linear equations is: \[2x + 3y = 10\] \[4x + 4y = 12\] The coefficient matrix (A) is: \[A = \begin{vmatrix} 2 & 3 \\ 4 & 4 \end{vmatrix}\] To find the determinant of A (D): \[D = (2 * 4) - (3 * 4) = 8 - 12 = -4\] Next, we find the determinant Dx by replacing the first column of A with the constants from the right-hand side of the equations (10 and 12): \[Dx = \begin{vmatrix} 10 & 3 \\ 12 & 4 \end{vmatrix}\] \[Dx = (10 * 4) - (3 * 12) = 40 - 36 = 4\] Then, we find the determinant Dy by replacing the second column of A with the constants from the right-hand side of the equations: \[Dy = \begin{vmatrix} 2 & 10 \\ 4 & 12 \end{vmatrix}\] \[Dy = (2 * 12) - (10 * 4) = 24 - 40 = -16\] Now, we can find the values of x and y using Cramer's Rule: \[x = \frac{Dx}{D} = \frac{4}{-4} = -1\] \[y = \frac{Dy}{D} = \frac{-16}{-4} = 4\] Therefore, the solution to the given system of linear equations is: \[x = -1, y = 4\] |
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| baichuan4 | To use Cramer's Rule to solve the system of linear equations, we first need to set up the coefficient matrix (A), the constant matrix (B), and the determinant of A (|A|). Given the equations: 1. $2x + 3y = 10$ 2. $4x + 4y = 12$ The coefficient matrix A is: $$ A = \begin{bmatrix} 2 & 3 \\ 4 & 4 \end{bmatrix} $$ The constant matrix B is: $$ B = \begin{bmatrix} 10 \\ 12 \end{bmatrix} $$ First, we need to calculate the determinant of A: $$ |A| = 2*4 - 3*4 = 8 - 12 = -4 $$ Now, to find $x$, we replace the first column of A with the constants from B to create matrix $A_x$ and calculate its determinant: $$ A_x = \begin{bmatrix} 10 & 3 \\ 12 & 4 \end{bmatrix} $$ $$ |A_x| = 10*4 - 3*12 = 40 - 36 = 4 $$ To find $y$, we replace the second column of A with the constants from B to create matrix $A_y$ and calculate its determinant: $$ A_y = \begin{bmatrix} 2 & 10 \\ 4 & 12 \end{bmatrix} $$ $$ |A_y| = 2*12 - 10*4 = 24 - 40 = -16 $$ Now, using Cramer's Rule, we can find the values of $x$ and $y$: $$ x = \frac{|A_x|}{|A|} = \frac{4}{-4} = -1 $$ $$ y = \frac{|A_y|}{|A|} = \frac{-16}{-4} = 4 $$ Therefore, the solution to the system of equations is $x = -1$ and $y = 4$. |
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| doubao | First find the determinant of the coefficient matrix $D$: \[ D = \begin{vmatrix} 2 & 3 \\ 4 & 4 \end{vmatrix} = 2\times4 - 4\times3 = 8 - 12 = -4. \] Then find $D_x$ by replacing the first column of the coefficient matrix with the constants on the right side of the equations: \[ D_x = \begin{vmatrix} 10 & 3 \\ 12 & 4 \end{vmatrix} = 10\times4 - 12\times3 = 40 - 36 = 4. \] And find $D_y$ by replacing the second column of the coefficient matrix with the constants on the right side of the equations: \[ D_y = \begin{vmatrix} 2 & 10 \\ 4 & 12 \end{vmatrix} = 2\times12 - 4\times10 = 24 - 40 = -16. \] Finally, $x = \frac{D_x}{D} = \frac{4}{-4} = -1$ and $y = \frac{D_y}{D} = \frac{-16}{-4} = 4$. |
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| minimax | Cramer's Rule is a method for solving a system of linear equations with as many equations as unknowns, provided that the system has a unique solution. It uses determinants to solve the system. Given the system of equations: 1) $2x + 3y = 10$ 2) $4x + 4y = 12$ First, we need to find the determinant of the coefficient matrix, which we'll call D: $D = \begin{vmatrix} 2 & 3 \\ 4 & 4 \\ \end{vmatrix} = (2)(4) - (3)(4) = 8 - 12 = -4$ Next, we find the determinant of the matrix with the first column replaced by the constants from the right-hand side of the equations, which we'll call D_x: $D_x = \begin{vmatrix} 10 & 3 \\ 12 & 4 \\ \end{vmatrix} = (10)(4) - (3)(12) = 40 - 36 = 4$ Then, we find the determinant of the matrix with the second column replaced by the constants, which we'll call D_y: $D_y = \begin{vmatrix} 2 & 10 \\ 4 & 12 \\ \end{vmatrix} = (2)(12) - (10)(4) = 24 - 40 = -16$ Now, we can find the values of x and y using Cramer's Rule: $x = \frac{D_x}{D} = \frac{4}{-4} = -1$ $y = \frac{D_y}{D} = \frac{-16}{-4} = 4$ Therefore, the solution to the system of equations is $x = -1$ and $y = 4$. |
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| kimi | To solve the system of linear equations using Cramer's rule, we first need to write down the coefficient matrix, the constant matrix, and then find the determinants. Given the system of equations: \[ 2x + 3y = 10 \] \[ 4x + 4y = 12 \] The coefficient matrix (A) is: \[ A = \begin{bmatrix} 2 & 3 \\ 4 & 4 \end{bmatrix} \] The constant matrix (B) is: \[ B = \begin{bmatrix} 10 \\ 12 \end{bmatrix} \] First, we calculate the determinant of A (denoted as |A|): \[ |A| = (2 \cdot 4) - (3 \cdot 4) = 8 - 12 = -4 \] Now, we create a new matrix by replacing the first column of A with the constant matrix B. Let's call this new matrix A_x: \[ A_x = \begin{bmatrix} |
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| wenxinyiyan | We have the two linear equations: $2x + 3y = 10$ $4x + 4y = 12$ We can solve these equations for $x$ and $y$ using Cramer's rule. First, we need to find the determinant of the coefficient matrix, which we'll call $D$. $D = \begin{vmatrix} 2 & 3 \\ 4 & 4 \end{vmatrix} = 2 \times 4 - 3 \times 4 = 8 - 12 = -4$ Since $D$ is non-zero, we can proceed with Cramer's rule. Next, we'll find $D_x$ and $D_y$, which are the determinants of the matrices obtained by replacing the columns of the coefficient matrix with the constant terms of the equations. $D_x = \begin{vmatrix} 10 & 3 \\ 12 & 4 \end{vmatrix} = 10 \times 4 - 3 \times 12 = 40 - 36 = 4$ $D_y = \begin{vmatrix} 2 & 10 \\ 4 & 12 \end{vmatrix} = 2 \times 12 - 10 \times 4 = 24 - 40 = -16$ Now, we can find $x$ and $y$ using Cramer's rule: $x = \frac{D_x}{D} = \frac{4}{-4} = -1$ $y = \frac{D_y}{D} = \frac{-16}{-4} = 4$ So, the solution to the system of equations is $x = -1$ and $y = 4$. |
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| zhipu | To solve the system of linear equations using Cramer's Rule, we need to follow these steps: 1. Write the system of equations in matrix form \(Ax = b\), where \(A\) is the coefficient matrix, \(x\) is the column matrix of variables, and \(b\) is the column matrix of constants. 2. Calculate the determinant of the coefficient matrix \(A\), denoted as \(D\). 3. Calculate the determinants of the matrices formed by replacing the columns of \(A\) with the column matrix \(b\), denoted as \(D_x\) and \(D_y\). 4. Use Cramer's Rule to find the values of \(x\) and \(y\). Given the system of equations: \[ 2x + 3y = 10 \] \[ 4x + 4y = 12 \] Let's write this in matrix form: \[ A = \begin{pmatrix} 2 & 3 \\ 4 & 4 \end{pmatrix}, \quad x = \begin{pmatrix} x \\ y \end{pmatrix}, \quad b = \begin{pmatrix} 10 \\ 12 \end{pmatrix} \] ### Step 1: Calculate the determinant \(D\) of matrix \(A\) \[ D = \det(A) = \begin{vmatrix} 2 & 3 \\ 4 & 4 \end{vmatrix} = (2 \cdot 4) - (3 \cdot 4) = 8 - 12 = -4 \] ### Step 2: Calculate \(D_x\) (replace the first column of \(A\) with \(b\)) \[ D_x = \begin{vmatrix} 10 & 3 \\ 12 & 4 \end{vmatrix} = (10 \cdot 4) - (3 \cdot 12) = 40 - 36 = 4 \] ### Step 3: Calculate \(D_y\) (replace the second column of \(A\) with \(b\)) \[ D_y = \begin{vmatrix} 2 & 10 \\ 4 & 12 \end{vmatrix} = (2 \cdot 12) - (10 \cdot 4) = 24 - 40 = -16 \] ### Step 4: Use Cramer's Rule to find \(x\) and \(y\) \[ x = \frac{D_x}{D} = \frac{4}{-4} = -1 \] \[ y = \frac{D_y}{D} = \frac{-16}{-4} = 4 \] ### Solution The solution to the system of equations is: \[ x = -1 \] \[ y = 4 \] Therefore, the values of \(x\) and \(y\) that satisfy the given system of equations are \(x = -1\) and \(y = 4\). |
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| gpto1 | To solve the system of equations using **Cramer's Rule**, we'll follow these steps: **Given equations:** \[ \begin{cases} 2x + 3y = 10 \quad \text{(1)} \\ 4x + 4y = 12 \quad \text{(2)} \end{cases} \] **Cramer's Rule Overview:** For a system of two linear equations: \[ \begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases} \] The solutions are: \[ x = \frac{D_x}{D}, \quad y = \frac{D_y}{D} \] where: - \( D \) is the determinant of the coefficients: \[ D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = a_1 b_2 - a_2 b_1 \] - \( D_x \) is the determinant when the constants replace the \( x \)-coefficients: \[ D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = c_1 b_2 - c_2 b_1 \] - \( D_y \) is the determinant when the constants replace the \( y \)-coefficients: \[ D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = a_1 c_2 - a_2 c_1 \] --- ### **Step 1: Compute the Determinant \( D \)** Extract coefficients from the equations: - From Equation (1): \( a_1 = 2 \), \( b_1 = 3 \) - From Equation (2): \( a_2 = 4 \), \( b_2 = 4 \) Calculate \( D \): \[ D = \begin{vmatrix} 2 & 3 \\ 4 & 4 \end{vmatrix} = (2)(4) - (4)(3) = 8 - 12 = \mathbf{-4} \] --- ### **Step 2: Compute the Determinant \( D_x \)** Replace \( x \)-coefficients with constants: - Constants: \( c_1 = 10 \), \( c_2 = 12 \) Calculate \( D_x \): \[ D_x = \begin{vmatrix} 10 & 3 \\ 12 & 4 \end{vmatrix} = (10)(4) - (12)(3) = 40 - 36 = \mathbf{4} \] --- ### **Step 3: Compute the Determinant \( D_y \)** Replace \( y \)-coefficients with constants: Calculate \( D_y \): \[ D_y = \begin{vmatrix} 2 & 10 \\ 4 & 12 \end{vmatrix} = (2)(12) - (4)(10) = 24 - 40 = \mathbf{-16} \] --- ### **Step 4: Solve for \( x \) and \( y \)** Compute \( x \): \[ x = \frac{D_x}{D} = \frac{4}{-4} = \mathbf{-1} \] Compute \( y \): \[ y = \frac{D_y}{D} = \frac{-16}{-4} = \mathbf{4} \] --- ### **Solution** The solution to the system is: \[ \boxed{\begin{cases} x = -1 \\ y = 4 \end{cases}} \] **Verification:** Plug \( x = -1 \) and \( y = 4 \) back into the original equations to verify. **Equation (1):** \[ 2(-1) + 3(4) = -2 + 12 = 10 \quad \checkmark \] **Equation (2):** \[ 4(-1) + 4(4) = -4 + 16 = 12 \quad \checkmark \] Both equations are satisfied, confirming the solution is correct. --- **Answer:** By using Cramer's Rule, the solution to the system is \( x = -1 \) and \( y = 4 \). |
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| deepseekv3 | To solve the system of linear equations using **Cramer's Rule**, follow these steps: ### Given Equations: \[ \begin{cases} 2x + 3y = 10 \\ 4x + 4y = 12 \end{cases} \] ### Step 1: Write the system in matrix form The system of equations can be represented as: \[ AX = B \] where: \[ A = \begin{pmatrix} 2 & 3 \\ 4 & 4 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} 10 \\ 12 \end{pmatrix} \] ### Step 2: Compute the determinant of \(A\) The determinant of \(A\) is calculated as: \[ \det(A) = (2)(4) - (3)(4) = 8 - 12 = -4 \] ### Step 3: Compute the determinant of \(A_x\) Replace the first column of \(A\) with \(B\) to form \(A_x\): \[ A_x = \begin{pmatrix} 10 & 3 \\ 12 & 4 \end{pmatrix} \] The determinant of \(A_x\) is: \[ \det(A_x) = (10)(4) - (3)(12) = 40 - 36 = 4 \] ### Step 4: Compute the determinant of \(A_y\) Replace the second column of \(A\) with \(B\) to form \(A_y\): \[ A_y = \begin{pmatrix} 2 & 10 \\ 4 & 12 \end{pmatrix} \] The determinant of \(A_y\) is: \[ \det(A_y) = (2)(12) - (10)(4) = 24 - 40 = -16 \] ### Step 5: Solve for \(x\) and \(y\) using Cramer's Rule Using Cramer's Rule: \[ x = \frac{\det(A_x)}{\det(A)} = \frac{4}{-4} = -1 \] \[ y = \frac{\det(A_y)}{\det(A)} = \frac{-16}{-4} = 4 \] ### Final Answer: \[ x = -1, \quad y = 4 \] |
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