Math Problem Statement

Express the vector \(\vec{w} = \begin{bmatrix} -4 \\ 14 \end{bmatrix}\) as a linear combination of \(\vec{v_1}, \vec{v_2}, \vec{v_3}\), where \(\vec{v_1} = \begin{bmatrix} 1 \\ -2 \end{bmatrix}\), \(\vec{v_2} = \begin{bmatrix} -5 \\ 10 \end{bmatrix}\), and \(\vec{v_3} = \begin{bmatrix} 2 \\ -5 \end{bmatrix}\).

Solution

The problem involves expressing the vector w⃗\vec{w} as a linear combination of the vectors v1⃗,v2⃗,v3⃗\vec{v_1}, \vec{v_2}, \vec{v_3} with unknown coefficients a,b,ca, b, c, i.e.:

w⃗=av1⃗+bv2⃗+cv3⃗.\vec{w} = a\vec{v_1} + b\vec{v_2} + c\vec{v_3}.

Here are the given vectors:

\vec{v_1} = \begin{bmatrix} 1 \\ -2 \end{bmatrix}, \ \vec{v_2} = \begin{bmatrix} -5 \\ 10 \end{bmatrix}, \ \vec{v_3} = \begin{bmatrix} 2 \\ -5 \end{bmatrix}.$$ We now solve for $$a, b, c$$ by setting up the following system of equations: $$\begin{bmatrix} -4 \\ 14 \end{bmatrix} = a \begin{bmatrix} 1 \\ -2 \end{bmatrix} + b \begin{bmatrix} -5 \\ 10 \end{bmatrix} + c \begin{bmatrix} 2 \\ -5 \end{bmatrix}.$$ This expands to two equations: 1. **For the first component (top row):** $$-4 = a(1) + b(-5) + c(2) \quad \Rightarrow \quad a - 5b + 2c = -4.$$ 2. **For the second component (bottom row):** $$14 = a(-2) + b(10) + c(-5) \quad \Rightarrow \quad -2a + 10b - 5c = 14.$$ Thus, the system of equations becomes: $$\begin{aligned} a - 5b + 2c &= -4, \\ -2a + 10b - 5c &= 14. \end{aligned}$$ We can solve this system of equations for $$a, b, c$$. Let me calculate these values. ### Solving the system: 1. Solve for $$a$$ in terms of $$b$$ and $$c$$ using the first equation: $$a = -4 + 5b - 2c.$$ 2. Substitute $$a$$ into the second equation: $$-2(-4 + 5b - 2c) + 10b - 5c = 14.$$ Simplify: $$8 - 10b + 4c + 10b - 5c = 14.$$ Combine like terms: $$8 - c = 14.$$ Solve for $$c$$: $$c = -6.$$ 3. Substitute $$c = -6$$ back into the expression for $$a$$: $$a = -4 + 5b - 2(-6).$$ Simplify: $$a = -4 + 5b + 12 \quad \Rightarrow \quad a = 8 + 5b.$$ 4. Now substitute $$a = 8 + 5b$$ and $$c = -6$$ into the first equation to solve for $$b$$: $$(8 + 5b) - 5b + 2(-6) = -4.$$ Simplify: $$8 + 5b - 5b - 12 = -4.$$ Combine terms: $$-4 = -4.$$ This equation holds true for any value of $$b$$. Thus, $$b$$ is a **free variable**. ### Final Solution: The solution can be expressed as: $$a = 8 + 5b, \quad b = b \ (\text{free variable}), \quad c = -6.$$ In other words, there are infinitely many solutions because the vectors $$\vec{v_1}, \vec{v_2}, \vec{v_3}$$ are **linearly dependent**. For any value of $$b$$, the corresponding values of $$a$$ and $$c$$ will satisfy the equation. --- Would you like me to explain further or verify the results? Here are 5 related questions to explore: 1. How can we determine if vectors are linearly dependent or independent? 2. How do you solve systems of equations using matrices (e.g., Gaussian elimination)? 3. What is the significance of free variables in a system of linear equations? 4. How do you find the rank of a matrix formed by vectors? 5. Can a vector always be expressed as a linear combination of other vectors? **Tip:** To check if vectors are linearly independent, calculate the determinant of the matrix formed by arranging them as columns. If the determinant is zero, the vectors are dependent.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Linear Combinations
Systems of Linear Equations
Vector Spaces

Formulas

\(\vec{w} = a\vec{v_1} + b\vec{v_2} + c\vec{v_3}\)
Linear system: \(Ax = b\)

Theorems

Linear Dependence and Independence

Suitable Grade Level

Undergraduate or Advanced High School (Grades 11-12)

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