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The document contains a mathematical exam in Greek focused on Linear Algebra. Let's solve problem 3, which involves finding the dimension and a basis for the row space and column space of a given matrix \( A \).

Given matrix \( A \):

\[
A = \begin{pmatrix}
1 & 2 & 0 & 2 & 1 \\
-1 & -2 & 1 & 1 & 0 \\
1 & 2 & -3 & -7 & -2
\end{pmatrix}
\]

### Row Space

To find the row space, we need to row reduce the matrix \( A \) to its row echelon form.

First, we start with the given matrix:
\[
\begin{pmatrix}
1 & 2 & 0 & 2 & 1 \\
-1 & -2 & 1 & 1 & 0 \\
1 & 2 & -3 & -7 & -2
\end{pmatrix}
\]

Perform row operations to simplify it:

1. Add the first row to the second row:
\[
R2 = R2 + R1
\]

\[
\begin{pmatrix}
1 & 2 & 0 & 2 & 1 \\
0 & 0 & 1 & 3 & 1 \\
1 & 2 & -3 & -7 & -2
\end{pmatrix}
\]

2. Subtract the first row from the third row:
\[
R3 = R3 - R1
\]

\[
\begin{pmatrix}
1 & 2 & 0 & 2 & 1 \\
0 & 0 & 1 & 3 & 1 \\
0 & 0 & -3 & -9 & -3
\end{pmatrix}
\]

3. Add 3 times the second row to the third row:
\[
R3 = R3 + 3R2
\]

\[
\begin{pmatrix}
1 & 2 & 0 & 2 & 1 \\
0 & 0 & 1 & 3 & 1 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
\]

The row echelon form shows that the first and second rows are linearly independent, and the third row is zero.

### Basis and Dimension

- **Row space** basis: The non-zero rows of the row echelon form matrix form a basis for the row space.

\[
\left\{
\begin{pmatrix}
1 & 2 & 0 & 2 & 1
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 1 & 3 & 1
\end{pmatrix}
\right\}
\]

- **Dimension of the row space**: The number of non-zero rows in the row echelon form, which is 2.

### Column Space

The columns of the original matrix corresponding to the leading ones in the row echelon form form a basis for the column space.

\[
\left\{
\begin{pmatrix}
1 \\
-1 \\
1
\end{pmatrix},
\begin{pmatrix}
2 \\
-2 \\
2
\end{pmatrix},
\begin{pmatrix}
0 \\
1 \\
-3
\end{pmatrix}
\right\}
\]

- **Dimension of the column space**: The number of leading ones in the row echelon form, which is 3.

**Summary:**
- Dimension of the row space: 2
- Basis for the row space: \(\{(1, 2, 0, 2, 1), (0, 0, 1, 3, 1)\}\)
- Dimension of the column space: 3
- Basis for the column space: \(\{(1, -1, 1), (2, -2, 2), (0, 1, -3)\}\)

Would you like more details or have any questions?

Here are 5 numbered questions for further exploration:
1. How do you determine the rank of a matrix?
2. What is the difference between row space and column space?
3. How do you find the null space of a matrix?
4. What is the significance of the row echelon form?
5. Can you explain how to find the eigenvalues of a matrix?

**Tip:** Always double-check your row operations during matrix reduction to avoid errors in finding the basis and dimension.

Learn how to find the dimension and basis for the row space and column space of a matrix using linear algebra principles. This problem involves detailed row operations and basis determination, focusing on matrix properties and dimensionality. Suitable for advanced college-level students and enthusiasts.

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