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. write each vector in a column

To ensure the vectors \((0, -1, 2, 1)\), \((a, -1, 0, 1)\), and \((1, -1, b, 0)\) span a subspace of dimension 2, they must be linearly dependent. This means that one vector can be expressed as a linear combination of the others.

Let's write the vectors as columns in a matrix:

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

To find conditions on \(a\) and \(b\) such that the dimension of the column space is 2, we can set the determinant of a \(2 \times 2\) submatrix to zero. 

Let's consider the submatrix formed by the first and the second columns:

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

The determinant of a \(2 \times 2\) matrix is:

\[
\text{det} = 0 \cdot (-1) - (-1) \cdot 2 = 2 \neq 0 \text{ (not dependent)}
\]

Next, let's consider the submatrix formed by the first and the third columns:

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

For this determinant to be zero, we can calculate:

\[
\text{det} = 0 \cdot (-1) - (-1) \cdot 2 = 2
\]

This condition doesn't help. Instead, we can set the third vector as a linear combination of the first two:

\[
(1, -1, b, 0) = x(0, -1, 2, 1) + y(a, -1, 0, 1)
\]

From this, we can derive the equations:

1. \(x(0) + y(a) = 1\) (1st component)
2. \(x(-1) + y(-1) = -1\) (2nd component)
3. \(x(2) + y(0) = b\) (3rd component)
4. \(x(1) + y(1) = 0\) (4th component)

From equation 1, \(ya = 1\).  
From equation 4, \(x + y = 0\) implies \(y = -x\).

Substituting \(y\) into the first equation gives:

\[
-xa = 1 \quad \Rightarrow \quad a = -\frac{1}{x}
\]

From equation 2:

\[
-x - y = -1 \quad \Rightarrow \quad -x + x = -1 \Rightarrow \quad x = 1 \implies y = -1
\]

Now substituting \(x = 1\) in \(a\):

\[
a = -1
\]

Finally, substituting \(x\) and \(y\) into equation 3:

\[
2(1) + 0(-1) = b \quad \Rightarrow \quad b = 2
\]

Thus, the values of \(a\) and \(b\) that allow the vectors to generate a subspace of dimension 2 are:

\[
a = -1, \quad b = 2
\]

Would you like further details or clarifications? Here are some related questions:

1. What conditions ensure linear independence among vectors?
2. How do you calculate the rank of a matrix?
3. Can you provide examples of linearly dependent and independent sets of vectors?
4. What is the significance of the dimension of a subspace in linear algebra?
5. How do these vectors behave in different dimensions? 

**Tip:** Always check the determinant of submatrices to assess linear dependence!

Explore how to determine the values of a and b in the vectors (0,−1,2,1),(a,−1,0,1),(1,−1,b,0) to create a subspace of dimension 2. This solution details the conditions for linear dependence and includes a step-by-step explanation. Ideal for advanced high school students.

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