Math Problem Statement
Solve the following system of equations in the least square sense X1-x2=4, 3x1+2x2=1, -2x1+4x2=3
Solution
To solve the given system of equations in the least squares sense, we first represent it in matrix form as , where:
- is the coefficient matrix of the system,
- is the vector of unknowns ,
- is the vector of constants.
The given system of equations is:
x_1 - x_2 &= 4, \\ 3x_1 + 2x_2 &= 1, \\ -2x_1 + 4x_2 &= 3. \end{aligned}$$ ### Step 1: Write the system in matrix form This can be written as: $$\begin{pmatrix} 1 & -1 \\ 3 & 2 \\ -2 & 4 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ 3 \end{pmatrix}$$ So: $$A = \begin{pmatrix} 1 & -1 \\ 3 & 2 \\ -2 & 4 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 4 \\ 1 \\ 3 \end{pmatrix}.$$ ### Step 2: Least Squares Solution The least squares solution is given by: $$\mathbf{x} = (A^T A)^{-1} A^T \mathbf{b}.$$ #### Step 2.1: Compute $$A^T A$$ First, calculate $$A^T$$, the transpose of $$A$$: $$A^T = \begin{pmatrix} 1 & 3 & -2 \\ -1 & 2 & 4 \end{pmatrix}.$$ Now, calculate $$A^T A$$: $$A^T A = \begin{pmatrix} 1 & 3 & -2 \\ -1 & 2 & 4 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 3 & 2 \\ -2 & 4 \end{pmatrix} = \begin{pmatrix} 14 & 2 \\ 2 & 21 \end{pmatrix}.$$ #### Step 2.2: Compute $$A^T \mathbf{b}$$ Now, calculate $$A^T \mathbf{b}$$: $$A^T \mathbf{b} = \begin{pmatrix} 1 & 3 & -2 \\ -1 & 2 & 4 \end{pmatrix} \begin{pmatrix} 4 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 26 \end{pmatrix}.$$ #### Step 2.3: Solve for $$\mathbf{x}$$ Now, solve for $$\mathbf{x}$$ by calculating: $$\mathbf{x} = (A^T A)^{-1} A^T \mathbf{b}.$$ First, find the inverse of $$A^T A$$: $$(A^T A)^{-1} = \frac{1}{14 \times 21 - 2^2} \begin{pmatrix} 21 & -2 \\ -2 & 14 \end{pmatrix} = \frac{1}{290} \begin{pmatrix} 21 & -2 \\ -2 & 14 \end{pmatrix}.$$ Now, compute $$\mathbf{x}$$: $$\mathbf{x} = \frac{1}{290} \begin{pmatrix} 21 & -2 \\ -2 & 14 \end{pmatrix} \begin{pmatrix} 3 \\ 26 \end{pmatrix} = \frac{1}{290} \begin{pmatrix} 63 - 52 \\ -6 + 364 \end{pmatrix} = \frac{1}{290} \begin{pmatrix} 11 \\ 358 \end{pmatrix}.$$ Thus: $$\mathbf{x} = \begin{pmatrix} \frac{11}{290} \\ \frac{358}{290} \end{pmatrix} = \begin{pmatrix} 0.0379 \\ 1.2345 \end{pmatrix}.$$ ### Final Answer: The least squares solution is approximately: $$x_1 = 0.0379, \quad x_2 = 1.2345.$$ Would you like further clarification or more details? Here are some related questions: 1. How do you interpret a least squares solution geometrically? 2. Can a system of equations have multiple least squares solutions? 3. How does the least squares method handle overdetermined systems? 4. What role does matrix inversion play in the least squares method? 5. How is the residual error computed in least squares problems? **Tip:** The least squares method minimizes the sum of the squared differences between the observed and predicted values, making it ideal for solving inconsistent systems.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Least Squares Method
Matrix Operations
Formulas
Least squares solution: x = (A^T A)^{-1} A^T b
Matrix transpose: A^T
Matrix multiplication
Theorems
Least Squares Theorem
Matrix Inversion Theorem
Suitable Grade Level
Undergraduate - Linear Algebra
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