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### Problem 8

Given the matrix \( A \) and the vector \( \alpha \):

\[ A = \begin{pmatrix} 1 & 2 & -2 \\ 2 & 1 & 2 \\ 3 & 0 & 4 \end{pmatrix}, \quad \alpha = \begin{pmatrix} a \\ 1 \\ 1 \end{pmatrix} \]

We are to find \( a \) such that \( A\alpha \) and \( \alpha \) are linearly dependent.

First, compute \( A\alpha \):

\[ A\alpha = \begin{pmatrix} 1 & 2 & -2 \\ 2 & 1 & 2 \\ 3 & 0 & 4 \end{pmatrix} \begin{pmatrix} a \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1(a) + 2(1) + (-2)(1) \\ 2(a) + 1(1) + 2(1) \\ 3(a) + 0(1) + 4(1) \end{pmatrix} = \begin{pmatrix} a \\ 2a + 3 \\ 3a + 4 \end{pmatrix} \]

Since \( A\alpha \) and \( \alpha \) are linearly dependent, there exists a scalar \( \lambda \) such that:

\[ \begin{pmatrix} a \\ 2a + 3 \\ 3a + 4 \end{pmatrix} = \lambda \begin{pmatrix} a \\ 1 \\ 1 \end{pmatrix} \]

This implies three equations:

1. \( a = \lambda a \)
2. \( 2a + 3 = \lambda \)
3. \( 3a + 4 = \lambda \)

From the first equation, if \( a \neq 0 \):

\[ \lambda = 1 \]

Substitute \( \lambda = 1 \) into the other two equations:

2. \( 2a + 3 = 1 \)  =>  \( 2a = -2 \)  =>  \( a = -1 \)
3. \( 3a + 4 = 1 \)  =>  \( 3a = -3 \)  =>  \( a = -1 \)

Thus, \( a = -1 \).

### Problem 9

Given the vectors \( \alpha_1, \alpha_2, \alpha_3 \):

\[ \alpha_1 = \begin{pmatrix} 1 \\ 2 \\ -1 \\ 0 \end{pmatrix}, \quad \alpha_2 = \begin{pmatrix} 1 \\ 1 \\ 0 \\ 2 \end{pmatrix}, \quad \alpha_3 = \begin{pmatrix} 2 \\ 1 \\ 1 \\ a \end{pmatrix} \]

We need to determine \( a \) such that the dimension of the space spanned by \( \alpha_1, \alpha_2, \alpha_3 \) is 2.

The vectors must be linearly dependent, which means the determinant of the matrix formed by these vectors must be zero. Form the matrix:

\[ M = \begin{pmatrix} 1 & 1 & 2 \\ 2 & 1 & 1 \\ -1 & 0 & 1 \\ 0 & 2 & a \end{pmatrix} \]

For \( \alpha_1, \alpha_2, \alpha_3 \) to be linearly dependent, the rank of \( M \) must be less than 3. Thus, we need to find \( a \) such that the determinant of the matrix formed by any 3 rows of \( M \) is zero.

Choose the 3x3 submatrix formed by the first three rows and first three columns:

\[ \begin{vmatrix} 1 & 1 & 2 \\ 2 & 1 & 1 \\ -1 & 0 & 1 \end{vmatrix} = 1 \cdot \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 1 \\ -1 & 1 \end{vmatrix} + 2 \cdot \begin{vmatrix} 2 & 1 \\ -1 & 0 \end{vmatrix} \]

Calculate the determinants:

\[ = 1 \cdot (1) - 1 \cdot (2 \cdot 1 - 1 \cdot -1) + 2 \cdot (2 \cdot 0 - (-1) \cdot 1) \]
\[ = 1 - 1 \cdot (2 + 1) + 2 \cdot (1) \]
\[ = 1 - 3 + 2 = 0 \]

Thus, this determinant is already zero, indicating that the vectors \( \alpha_1, \alpha_2, \alpha_3 \) are linearly dependent for any value of \( a \).

Therefore, the vectors span a space of dimension 2 for any \( a \). So, \( a \) can be any real number.

Would you like more details or have any questions? Here are five more questions you might consider asking:

1. How do you find the inverse of a matrix?
2. Can you explain the concept of eigenvalues and eigenvectors?
3. What is the process for solving a system of linear equations using matrix methods?
4. How can you determine if a set of vectors is linearly independent?
5. What are the applications of linear algebra in real-world problems?

**Tip:** When dealing with matrix operations, always check the dimensions of your matrices and vectors to ensure that the operations are valid.

Explore how to determine scalar values and vector dependencies in matrices. This problem involves finding scalar 'a' such that matrices A*alpha and alpha are linearly dependent. Understand concepts like matrix multiplication and determinant to solve this problem effectively.

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