To show that the set $B = \{ \mathbf{u}, \mathbf{v}, \mathbf{w} \}$ where:

$\mathbf{u} = (1, -1, 2), \quad \mathbf{v} = (1, 1, 1), \quad \mathbf{w} = (-1, 0, 1)$

is a basis of $\mathbb{R}^3$, we need to demonstrate two things:

1. Linear Independence: The vectors $\mathbf{u}, \mathbf{v}, \mathbf{w}$ are linearly independent.
2. Span: The vectors $\mathbf{u}, \mathbf{v}, \mathbf{w}$ span $\mathbb{R}^3$.

Step 1: Check Linear Independence

To check if the vectors are linearly independent, we set up the equation:

$c_1 \mathbf{u} + c_2 \mathbf{v} + c_3 \mathbf{w} = \mathbf{0}$

where $\mathbf{0} = (0, 0, 0)$ is the zero vector in $\mathbb{R}^3$. Substituting the vectors:

$c_1 (1, -1, 2) + c_2 (1, 1, 1) + c_3 (-1, 0, 1) = (0, 0, 0)$

This leads to the following system of linear equations:

c_1 + c_2 - c_3 = 0 \\ -c_1 + c_2 = 0 \\ 2c_1 + c_2 + c_3 = 0 \end{cases}$$ We can write this system in matrix form as: $$\begin{bmatrix} 1 & 1 & -1 \\ -1 & 1 & 0 \\ 2 & 1 & 1 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ To check for linear independence, we need to determine if the only solution to this system is $$c_1 = c_2 = c_3 = 0$$. #### Row Reduction to Echelon Form Let's perform Gaussian elimination to reduce the matrix to echelon form: $$\begin{bmatrix} 1 & 1 & -1 \\ -1 & 1 & 0 \\ 2 & 1 & 1 \end{bmatrix}$$ 1. **Add the first row to the second row**: $$\begin{bmatrix} 1 & 1 & -1 \\ 0 & 2 & -1 \\ 2 & 1 & 1 \end{bmatrix}$$ 2. **Subtract 2 times the first row from the third row**: $$\begin{bmatrix} 1 & 1 & -1 \\ 0 & 2 & -1 \\ 0 & -1 & 3 \end{bmatrix}$$ 3. **Add $$\frac{1}{2}$$ times the second row to the third row**: $$\begin{bmatrix} 1 & 1 & -1 \\ 0 & 2 & -1 \\ 0 & 0 & \frac{5}{2} \end{bmatrix}$$ Since the matrix is now in echelon form and there is a pivot (leading entry) in every column, the only solution to the system is $$c_1 = 0, c_2 = 0, c_3 = 0$$. Thus, the vectors are linearly independent. ### Step 2: Check Span To check if $$B = \{\mathbf{u}, \mathbf{v}, \mathbf{w}\}$$ spans $$\mathbb{R}^3$$, we need to verify that any vector $$\mathbf{x} = (x_1, x_2, x_3) \in \mathbb{R}^3$$ can be written as a linear combination of $$\mathbf{u}, \mathbf{v}, \mathbf{w}$$. Since there are 3 linearly independent vectors in $$\mathbb{R}^3$$, they automatically span $$\mathbb{R}^3$$. ### Conclusion Since the set $$B = \{\mathbf{u}, \mathbf{v}, \mathbf{w}\}$$ consists of three linearly independent vectors in $$\mathbb{R}^3$$ and there are three dimensions in $$\mathbb{R}^3$$, the set $$B$$ is a basis for $$\mathbb{R}^3$$. Would you like more details or have any questions? ### Related Questions: 1. How do you determine if a set of vectors spans a vector space? 2. What is the significance of a basis in a vector space? 3. How can we find the coordinate vector of a vector relative to a given basis? 4. What is the geometric interpretation of linearly independent vectors? 5. Can a set of more than three vectors be a basis for $$\mathbb{R}^3$$? ### Tip: To quickly check if a set of vectors is a basis, you can form a matrix with the vectors as columns and check if the matrix is invertible (i.e., has a non-zero determinant).