Given the vectors (0,−1,2,1),(a,−1,0,1),(1,−1,b,0) calculate a and b so they generate a subspace of dimension 2. the answer i got is a = 0 and b = 2 is it correct anf if it is can u explain in detail how i got to this answer and why this is the answer

To determine the values of \(a\) and \(b\) for which the vectors \((0, -1, 2, 1)\), \((a, -1, 0, 1)\), and \((1, -1, b, 0)\) generate a subspace of dimension 2, we need to analyze the linear independence of these vectors.

### Steps to Analyze Linear Independence

1. **Form a Matrix**: We arrange the vectors as rows in a matrix:
   \[
   A = \begin{bmatrix}
   0 & -1 & 2 & 1 \\
   a & -1 & 0 & 1 \\
   1 & -1 & b & 0
   \end{bmatrix}
   \]

2. **Row Reduction**: We will perform row operations to bring this matrix to row echelon form (REF) or reduced row echelon form (RREF) and find conditions for linear dependence.

3. **Determine Conditions for Linear Dependence**: For the dimension of the span to be 2, we need exactly one of the vectors to be a linear combination of the others. This will happen if the rank of the matrix \(A\) is 2.

### Performing Row Operations

Start with the matrix:
\[
A = \begin{bmatrix}
0 & -1 & 2 & 1 \\
a & -1 & 0 & 1 \\
1 & -1 & b & 0
\end{bmatrix}
\]

We can swap row 1 and row 3 (to have a leading 1 at the top):
\[
\begin{bmatrix}
1 & -1 & b & 0 \\
a & -1 & 0 & 1 \\
0 & -1 & 2 & 1
\end{bmatrix}
\]

Next, eliminate the first element in the second row:
\[
R_2 \leftarrow R_2 - a R_1
\]
\[
\begin{bmatrix}
1 & -1 & b & 0 \\
0 & a-1 & -ab & 1 \\
0 & -1 & 2 & 1
\end{bmatrix}
\]

Now eliminate the first element in the third row (which is already 0):
Next, we can simplify \(R_3\):
\[
R_3 \leftarrow R_3 + (a-1)R_2
\]

This leads us to:
\[
\begin{bmatrix}
1 & -1 & b & 0 \\
0 & a-1 & -ab & 1 \\
0 & 0 & 2 - (a-1)(-ab) & 1 - (a-1)(1)
\end{bmatrix}
\]

### Set Conditions for Rank 2

To ensure the rank is 2, we want the last row to become 0. The conditions for that row to equal zero gives us:
1. \(2 - (a-1)(-ab) = 0\)
2. \(1 - (a-1)(1) = 0\)

From the second equation:
\[
1 - (a-1) = 0 \implies a - 1 = 1 \implies a = 2
\]

Now substituting \(a = 2\) into the first equation:
\[
2 - (2-1)(-2b) = 0 \implies 2 + 2b = 0 \implies 2b = -2 \implies b = -1
\]

### Conclusion

The values of \(a\) and \(b\) should be \(a = 2\) and \(b = -1\) for the vectors to span a subspace of dimension 2. 

If you found \(a = 0\) and \(b = 2\), it seems to be incorrect. The correct analysis leads to \(a = 2\) and \(b = -1\) to meet the condition.

If you want further details or have any questions, let me know! 

### Further Questions:
1. What does it mean for vectors to be linearly independent?
2. How does changing the values of \(a\) and \(b\) affect the dimension of the span?
3. Can you give an example of three vectors that are linearly independent in \(\mathbb{R}^4\)?
4. What is the significance of the rank of a matrix in linear algebra?
5. How do row operations affect the solutions to a system of equations? 

**Tip:** Always check the linear combinations and determinants when analyzing vector independence.

Explore how to determine the values of a and b for the vectors (0,−1,2,1), (a,−1,0,1), and (1,−1,b,0) to generate a subspace of dimension 2. This problem involves linear independence, row reduction techniques, and conditions for linear dependence, making it ideal for Grades 11-12 students.

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