show that B={u,v,w} where u=(1,-1,2), v=(1,1,1) and w=(-1,0,1) is a basis of R^3

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:

\[
\begin{cases}
c_1 + c_2 - c_3 = 0 \\
-c_1 + c_2 = 0 \\
2c_1 + c_2 + c_3 = 0
\end{cases}
\]

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).

Learn how to verify that the set { (1, -1, 2), (1, 1, 1), (-1, 0, 1) } forms a basis of R^3. Detailed step-by-step explanation includes checking for linear independence and span in vector spaces. Suitable for college-level students studying linear algebra.

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