Sure! Let's use row reduction to determine the values of \(a\) and \(b\) so that the vectors \((0, -1, 2, 1)\), \((a, -1, 0, 1)\), and \((1, -1, b, 0)\) generate a subspace of dimension 2.

We can write the vectors as rows in a matrix:

\[
\begin{bmatrix}
0 & -1 & 2 & 1 \\
a & -1 & 0 & 1 \\
1 & -1 & b & 0
\end{bmatrix}
\]

Next, we will perform row reduction to achieve row echelon form.

### Step 1: Swap Rows
Since the first row starts with 0, we can swap the first and third rows:

\[
\begin{bmatrix}
1 & -1 & b & 0 \\
a & -1 & 0 & 1 \\
0 & -1 & 2 & 1
\end{bmatrix}
\]

### Step 2: Eliminate First Column Below Pivot
Now, we'll eliminate the first column entries below the pivot (1):

- For Row 2: \(R_2 - a \cdot R_1\)
- For Row 3: \(R_3\) (remains the same as it has a zero in the first column)

Calculating \(R_2 - aR_1\):

\[
\begin{aligned}
R_2: & \quad (a - a \cdot 1, -1 + a \cdot 1, 0 - a \cdot b, 1 - a \cdot 0) \\
& = (0, -1 + a, -ab, 1)
\end{aligned}
\]

The updated matrix becomes:

\[
\begin{bmatrix}
1 & -1 & b & 0 \\
0 & -1 + a & -ab & 1 \\
0 & -1 & 2 & 1
\end{bmatrix}
\]

### Step 3: Eliminate Second Column
Next, we eliminate the entry in the second column of Row 3. We do this by adding \(R_2\) to \(R_3\):

\[
R_3 + \left(\frac{1 - a}{-1 + a}\right) R_2
\]

Calculating \(R_3 + R_2\):

\[
\begin{aligned}
R_3: & \quad (0 + 0, -1 + (-1 + a), 2 - ab, 1 + 1) \\
& = (0, a - 2, 2 - ab, 2)
\end{aligned}
\]

Now the matrix looks like:

\[
\begin{bmatrix}
1 & -1 & b & 0 \\
0 & -1 + a & -ab & 1 \\
0 & a - 2 & 2 - ab & 2
\end{bmatrix}
\]

### Step 4: Set the Determinant Condition
For the subspace to be of dimension 2, we need the last row to be a linear combination of the first two rows, which means:

1. The rows must be linearly dependent.
2. The determinant of the matrix should be zero.

This requires solving the condition from the last row:

To express the third row in terms of the first two, we can set up equations based on the second row being a pivot. The last row should be a scalar multiple of the second row:

\[
k(0, -1 + a, -ab, 1) = (0, a - 2, 2 - ab, 2)
\]

From here, we can deduce:

1. \(k(0) = 0\)
2. \(k(-1 + a) = a - 2\)
3. \(k(-ab) = 2 - ab\)
4. \(k = 2\)

### Solve the System
From \(k(-1 + a) = a - 2\):

\[
2(-1 + a) = a - 2 \quad \Rightarrow \quad -2 + 2a = a - 2
\]

This simplifies to:

\[
2a - a = 0 \quad \Rightarrow \quad a = 0
\]

Now substituting \(a = 0\) into the third row condition:

\[
2(-0) = 2 - 0 \quad \Rightarrow \quad 2 = 2 \quad \text{(True)}
\]

Finally, substituting \(a = 0\) into our original rows and checking:

We can re-evaluate \(b\) using a similar approach. Choosing conditions from \(k(-ab) = 2 - ab\):

With \(a = 0\):

\[
-0b = 2 \quad \Rightarrow \quad b \text{ can be any value.}
\]

### Conclusion
Thus, the condition for \(a\) and \(b\) to ensure that the dimension of the generated subspace is 2 is:

\[
a = 0 \quad \text{and } b \text{ can be any value.}
\]

If you have more questions or need further clarification, let me know! Here are some related questions:

1. What is the significance of row echelon form in linear algebra?
2. How can we determine if a set of vectors is linearly independent?
3. Can you explain the concept of the rank of a matrix?
4. What role do pivot positions play in matrix row reduction?
5. How does the dimension of a subspace relate to the number of vectors? 

**Tip:** Always check if your rows can be expressed as combinations of one another when determining linear dependence!

Can you do it through row reducing, please?

Learn to use row reduction techniques to determine the values of parameters in a vector set, ensuring the generated subspace has a specific dimension. This tutorial provides a step-by-step guide, covering essential linear algebra concepts such as row echelon form and linear dependence.

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