Math Problem Statement

Consider the following vectors: x1 = (-3, 1, 6), x2 = (2, -3, -4), x3 = (1, z, -2). (a) For which value(s) of z does the equation x3 = c1 * x1 + c2 * x2 have a solution? (b) For which value(s) of z are {x1, x2, x3} linearly dependent?

Solution

The given problem provides three vectors:

x1=(−316),x2=(2−3−4),x3=(1z−2)\mathbf{x_1} = \begin{pmatrix} -3 \\ 1 \\ 6 \end{pmatrix}, \quad \mathbf{x_2} = \begin{pmatrix} 2 \\ -3 \\ -4 \end{pmatrix}, \quad \mathbf{x_3} = \begin{pmatrix} 1 \\ z \\ -2 \end{pmatrix}

It asks for:

(a) For which value(s) of zz does the equation x3=c1x1+c2x2\mathbf{x_3} = c_1 \mathbf{x_1} + c_2 \mathbf{x_2} have a solution?

(b) For which value(s) of zz are the vectors {x1,x2,x3}\{\mathbf{x_1}, \mathbf{x_2}, \mathbf{x_3}\} linearly dependent?

Part (a) Solution:

To solve x3=c1x1+c2x2\mathbf{x_3} = c_1 \mathbf{x_1} + c_2 \mathbf{x_2}, we need to write the system of equations:

(1z−2)=c1(−316)+c2(2−3−4)\begin{pmatrix} 1 \\ z \\ -2 \end{pmatrix} = c_1 \begin{pmatrix} -3 \\ 1 \\ 6 \end{pmatrix} + c_2 \begin{pmatrix} 2 \\ -3 \\ -4 \end{pmatrix}

This gives the following system:

-3c_1 + 2c_2 &= 1 \\ c_1 - 3c_2 &= z \\ 6c_1 - 4c_2 &= -2 \end{aligned}$$ We will solve this system to find conditions on $$z$$. ### Part (b) Solution: For the vectors $$\{\mathbf{x_1}, \mathbf{x_2}, \mathbf{x_3}\}$$ to be linearly dependent, there must exist constants $$c_1, c_2, c_3$$, not all zero, such that: $$c_1 \mathbf{x_1} + c_2 \mathbf{x_2} + c_3 \mathbf{x_3} = 0$$ This leads to a homogeneous system of equations. We will analyze the determinant of the matrix formed by $$\mathbf{x_1}, \mathbf{x_2}, \mathbf{x_3}$$ to determine for which values of $$z$$ the vectors are linearly dependent. I'll solve both parts of the problem step by step. Let's begin by solving the system for part (a).For part (a), the system $$\mathbf{x_3} = c_1 \mathbf{x_1} + c_2 \mathbf{x_2}$$ has a solution if we can express $$c_1$$ and $$c_2$$ in terms of $$z$$. The solutions are: $$c_1 = -\frac{2z}{7} - \frac{3}{7}, \quad c_2 = -\frac{3z}{7} - \frac{1}{7}$$ Thus, the equation $$\mathbf{x_3} = c_1 \mathbf{x_1} + c_2 \mathbf{x_2}$$ has a solution for **any value of $$z$$**, as there are no restrictions on $$z$$ in this case. ### Part (b) Solution: To find for which values of $$z$$ the vectors $$\mathbf{x_1}, \mathbf{x_2}, \mathbf{x_3}$$ are linearly dependent, we need to solve the determinant of the matrix formed by these vectors being zero. Let's proceed to calculate that determinant.It seems like I can’t do more advanced data analysis right now. Please try again later. Would you like me to walk through how to compute the determinant manually for part (b) instead, or assist with something else? Here are 5 related questions you might consider: 1. How do you interpret the meaning of linear dependence between vectors geometrically? 2. What happens when the determinant of a matrix of vectors equals zero? 3. Can you generalize conditions for three vectors to be linearly independent in $$\mathbb{R}^3$$? 4. How do systems of linear equations relate to vector spaces? 5. What methods can we use to find if a set of vectors forms a basis for a vector space? **Tip:** Linear dependence means that one vector in the set can be written as a combination of the others!

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Math Problem Analysis

Mathematical Concepts

Linear algebra
Vector equations
Linear dependence

Formulas

x3 = c1 * x1 + c2 * x2
Determinants of 3x3 matrices

Theorems

Linear dependence criterion
System of linear equations

Suitable Grade Level

Undergraduate Linear Algebra

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