Math Problem Statement
Let \(\mathcal{B}\) be the basis of \(\mathbb{R}^2\) consisting of \(\mathbf{b}_1 = \begin{bmatrix} 1 \\ 3 \end{bmatrix}\) and \(\mathbf{b}_2 = \begin{bmatrix} -1 \\ 2 \end{bmatrix}\). For the vector \(\mathbf{x} = \begin{bmatrix} -1 \\ -1 \end{bmatrix}\), find its coordinates in the basis \(\mathcal{B}\).
Solution
To find the coordinates of the vector in the basis , consisting of and , we solve the equation:
This can be rewritten as a system of linear equations:
1 & -1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \end{bmatrix}$$ Let me solve this system for $$c_1$$ and $$c_2$$.import numpy as np # Coefficient matrix and vector A = np.array([[1, -1], [3, 2]]) b = np.array([-1, -1]) # Solve for c1 and c2 c = np.linalg.solve(A, b) carray([-0.6, 0.4])array([-0.6, 0.4])successThe coordinates of the vector $$\mathbf{x}$$ in the basis $$\mathcal{B}$$ are: $$\mathbf{x}_\mathcal{B} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} -0.6 \\ 0.4 \end{bmatrix}$$ This means: - $$c_1 = -0.6$$ - $$c_2 = 0.4$$ Would you like a step-by-step explanation or clarification? Here are five related questions to deepen understanding: 1. What does it mean for a vector to have coordinates in a different basis? 2. How do you compute the inverse of the basis matrix manually? 3. Can this approach be extended to three-dimensional space? 4. What happens if the basis vectors are not linearly independent? 5. How does this process relate to transforming between coordinate systems? **Tip**: Always verify that the given basis vectors are linearly independent before solving for coordinates.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Change of Basis
Coordinate Systems
Formulas
\(\mathbf{x} = c_1 \mathbf{b}_1 + c_2 \mathbf{b}_2\)
Solving a system of linear equations \(A \mathbf{c} = \mathbf{b}\)
Theorems
Linear Independence of Basis Vectors
Invertibility of Square Matrices
Suitable Grade Level
Undergraduate Linear Algebra (or Advanced High School)