Let's solve the given problem using matrices.

Problem Statement

Find the coordinates $x_1, x_2, x_3$ of the vector $\mathbf{v} = (m, 0, 1)$ in the basis $\{ \mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3 \}$, where: $\mathbf{u}_1 = (1, 0, 0), \quad \mathbf{u}_2 = (1, 1, 0), \quad \mathbf{u}_3 = (0, -1, 1)$

This means we need to express $\mathbf{v}$ as a linear combination of $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$: $\mathbf{v} = x_1 \mathbf{u}_1 + x_2 \mathbf{u}_2 + x_3 \mathbf{u}_3$

Step-by-Step Solution

1. Write the vector equation in terms of the basis vectors:

$(m, 0, 1) = x_1 (1, 0, 0) + x_2 (1, 1, 0) + x_3 (0, -1, 1)$

2. Expand the right-hand side:

$(m, 0, 1) = (x_1, 0, 0) + (x_2, x_2, 0) + (0, -x_3, x_3)$

3. Combine like terms:

$(m, 0, 1) = (x_1 + x_2, x_2 - x_3, x_3)$

4. Set up the system of equations by comparing components:

x_1 + x_2 = m \\ x_2 - x_3 = 0 \\ x_3 = 1 \end{cases}$$ 5. **Solve the system of equations:** From the third equation, we get: $$x_3 = 1$$ From the second equation, substitute $$x_3 = 1$$: $$x_2 - 1 = 0 \implies x_2 = 1$$ From the first equation, substitute $$x_2 = 1$$: $$x_1 + 1 = m \implies x_1 = m - 1$$ ### Solution The coordinates of $$\mathbf{v}$$ in the given basis are: $$(x_1, x_2, x_3) = (m - 1, 1, 1)$$ Would you like any more details or have any questions about this solution? ### Relative Questions 1. How do you determine the coordinates of a vector in a different basis? 2. What are the properties of a basis in vector space? 3. How do you solve a system of linear equations using matrices? 4. What is the importance of the linear independence of basis vectors? 5. How do you change coordinates from one basis to another? 6. What is the rank of a matrix, and how is it related to solving systems of linear equations? 7. How can you verify if a given set of vectors forms a basis? 8. How do you find the inverse of a matrix? ### Tip When solving systems of linear equations, it’s often useful to use matrix methods such as row reduction or the inverse matrix method for more efficient and systematic solutions.