Math Problem Statement
solve step-by-step breakdown
Express the vector v = (8, -6, -1)^T in R^3 in terms of the vectors v1 = (2, 3, 1)^T, v2 = (5, -3, 0)^T, v3 = (4, 3, 2)^T.
Solution
To express the vector as a linear combination of the vectors:
\mathbf{v}_2 = \begin{pmatrix} 5 \\ -3 \\ 0 \end{pmatrix}, \, \mathbf{v}_3 = \begin{pmatrix} 4 \\ 3 \\ 2 \end{pmatrix},$$ we write: $$\mathbf{v} = c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3.$$ This means solving the equation: $$\begin{pmatrix} 8 \\ -6 \\ -1 \end{pmatrix} = c_1 \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} + c_2 \begin{pmatrix} 5 \\ -3 \\ 0 \end{pmatrix} + c_3 \begin{pmatrix} 4 \\ 3 \\ 2 \end{pmatrix}.$$ This expands to the system of equations: $$2c_1 + 5c_2 + 4c_3 = 8,$$ $$3c_1 - 3c_2 + 3c_3 = -6,$$ $$c_1 + 0c_2 + 2c_3 = -1.$$ ### Step 1: Write the system in matrix form The system can be written as: $$\begin{bmatrix} 2 & 5 & 4 \\ 3 & -3 & 3 \\ 1 & 0 & 2 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 8 \\ -6 \\ -1 \end{bmatrix}.$$ ### Step 2: Solve using row reduction or substitution To solve for $$c_1, c_2,$$ and $$c_3$$, we will reduce the augmented matrix: $$\begin{bmatrix} 2 & 5 & 4 & 8 \\ 3 & -3 & 3 & -6 \\ 1 & 0 & 2 & -1 \end{bmatrix}.$$ #### Row Reduction 1. Divide Row 1 by 2 to simplify: $$\begin{bmatrix} 1 & 2.5 & 2 & 4 \\ 3 & -3 & 3 & -6 \\ 1 & 0 & 2 & -1 \end{bmatrix}.$$ 2. Subtract 3 × Row 1 from Row 2 and subtract Row 1 from Row 3: $$\begin{bmatrix} 1 & 2.5 & 2 & 4 \\ 0 & -10.5 & -3 & -18 \\ 0 & -2.5 & 0 & -5 \end{bmatrix}.$$ 3. Divide Row 2 by -10.5 to simplify: $$\begin{bmatrix} 1 & 2.5 & 2 & 4 \\ 0 & 1 & 0.286 & 1.714 \\ 0 & -2.5 & 0 & -5 \end{bmatrix}.$$ 4. Add 2.5 × Row 2 to Row 3: $$\begin{bmatrix} 1 & 2.5 & 2 & 4 \\ 0 & 1 & 0.286 & 1.714 \\ 0 & 0 & 0.714 & -0.714 \end{bmatrix}.$$ 5. Divide Row 3 by 0.714: $$\begin{bmatrix} 1 & 2.5 & 2 & 4 \\ 0 & 1 & 0.286 & 1.714 \\ 0 & 0 & 1 & -1 \end{bmatrix}.$$ 6. Back-substitute to simplify Rows 1 and 2: - Subtract 2 × Row 3 from Row 1. - Subtract 0.286 × Row 3 from Row 2. Final matrix: $$\begin{bmatrix} 1 & 0 & 0 & 3 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -1 \end{bmatrix}.$$ ### Step 3: Solution $$c_1 = 3, \, c_2 = 2, \, c_3 = -1.$$ Thus, the vector $$\mathbf{v}$$ can be expressed as: $$\mathbf{v} = 3\mathbf{v}_1 + 2\mathbf{v}_2 - \mathbf{v}_3.$$ Do you have any questions or would like more details? #### Related Questions: 1. How can row reduction be applied more efficiently in solving linear equations? 2. What is the geometric interpretation of expressing a vector as a combination of others? 3. Can this method be extended to higher dimensions or other vector spaces? 4. How does linear independence affect the ability to solve such equations? 5. What are alternative methods to row reduction for solving these equations? **Tip:** Always check if the given vectors form a basis of the space to ensure a unique solution exists.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Vector Spaces
Linear Combinations
Formulas
v = c1*v1 + c2*v2 + c3*v3
Matrix equation: [A]*[C] = [b]
Theorems
Systems of Linear Equations
Row Reduction
Suitable Grade Level
College-Level or Advanced High School
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