Math Problem Statement

solve this

The rows of a matrix M produce three vectors r1, r2, r3 (when taken as columns). Find two sets of y1, y2, y3 such that the linear combination y1r1 + y2r2 + y3r3 = 0.

Solution

The problem asks to find two sets of y1,y2,y3y_1, y_2, y_3 such that the linear combination of the vectors r1,r2,r3r_1, r_2, r_3 results in the zero vector:

y1r1+y2r2+y3r3=0y_1 \mathbf{r_1} + y_2 \mathbf{r_2} + y_3 \mathbf{r_3} = 0

Where the vectors r1,r2,r3\mathbf{r_1}, \mathbf{r_2}, \mathbf{r_3} are given as:

r1=(147),r2=(258),r3=(369)\mathbf{r_1} = \begin{pmatrix} 1 \\ 4 \\ 7 \end{pmatrix}, \quad \mathbf{r_2} = \begin{pmatrix} 2 \\ 5 \\ 8 \end{pmatrix}, \quad \mathbf{r_3} = \begin{pmatrix} 3 \\ 6 \\ 9 \end{pmatrix}

The equation to solve is:

y1(147)+y2(258)+y3(369)=(000)y_1 \begin{pmatrix} 1 \\ 4 \\ 7 \end{pmatrix} + y_2 \begin{pmatrix} 2 \\ 5 \\ 8 \end{pmatrix} + y_3 \begin{pmatrix} 3 \\ 6 \\ 9 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}

This gives us the following system of linear equations:

1. & \quad y_1 + 2y_2 + 3y_3 = 0 \\ 2. & \quad 4y_1 + 5y_2 + 6y_3 = 0 \\ 3. & \quad 7y_1 + 8y_2 + 9y_3 = 0 \\ \end{aligned}$$ To solve this system, we can set up the augmented matrix and perform row reduction (Gaussian elimination) or recognize that the vectors are linearly dependent (since the third vector is a linear combination of the first two). ### Step 1: Form the matrix equation $$\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$ ### Step 2: Apply Gaussian elimination Notice that row 3 can be obtained by adding row 1 and row 2 (times appropriate coefficients), confirming that the vectors are indeed linearly dependent. ### Step 3: Find the solutions This system has an infinite number of solutions. To find two sets, let's express $$y_1, y_2, y_3$$ in terms of a free variable $$y_3 = t$$. From the first equation: $$y_1 = -2y_2 - 3t$$ From the second equation: $$4(-2y_2 - 3t) + 5y_2 + 6t = 0$$ Simplifying: $$-8y_2 - 12t + 5y_2 + 6t = 0 \implies -3y_2 - 6t = 0 \implies y_2 = -2t$$ Substituting back: $$y_1 = -2(-2t) - 3t = 4t - 3t = t$$ Thus, the general solution is: $$y_1 = t, \quad y_2 = -2t, \quad y_3 = t$$ ### Step 4: Two specific solutions 1. Set $$t = 1$$: $$y_1 = 1, \quad y_2 = -2, \quad y_3 = 1$$ 2. Set $$t = 2$$: $$y_1 = 2, \quad y_2 = -4, \quad y_3 = 2$$ ### Final Answer Two sets of solutions for $$y_1, y_2, y_3$$ are: 1. $$(1, -2, 1)$$ 2. $$(2, -4, 2)$$ Would you like more details, or do you have any questions? --- Here are 5 questions related to this topic: 1. How do you determine if a set of vectors is linearly dependent? 2. What is the significance of the free variable in a system of linear equations? 3. How can Gaussian elimination be used to solve a system of equations? 4. What are the possible geometric interpretations of the solution to this problem? 5. What would happen if the vectors were linearly independent? **Tip:** Always check for linear dependence by computing the determinant or by performing row operations to avoid unnecessary calculations.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Vector Spaces
Linear Dependence

Formulas

y1r1 + y2r2 + y3r3 = 0

Theorems

Linear Dependence Theorem

Suitable Grade Level

Undergraduate Level

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