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The problem asks to express the general solution of the equation \( A \mathbf{x} = \mathbf{b} \), where \( A = \begin{bmatrix} 2 & -2 \\ -6 & 6 \end{bmatrix} \) and \( \mathbf{b} = \begin{bmatrix} -2 \\ 6 \end{bmatrix} \), in the form:

\[
\mathbf{x} = \mathbf{x}_r + c \mathbf{x}_n
\]

where:
- \( \mathbf{x}_r \) is the particular solution (row space solution),
- \( \mathbf{x}_n \) is a basis for the null space of \( A \),
- and \( c \) is a scalar parameter.

### Step 1: Solve for \( \mathbf{x}_r \)

First, solve the system \( A \mathbf{x}_r = \mathbf{b} \) for \( \mathbf{x}_r \).

\[
A = \begin{bmatrix} 2 & -2 \\ -6 & 6 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} -2 \\ 6 \end{bmatrix}
\]

We can rewrite the system of equations as:

\[
2x_1 - 2x_2 = -2 \quad \text{(1)}
\]
\[
-6x_1 + 6x_2 = 6 \quad \text{(2)}
\]

Both equations are equivalent, so solve just one of them. From equation (1):

\[
2x_1 - 2x_2 = -2 \quad \Rightarrow \quad x_1 - x_2 = -1 \quad \Rightarrow \quad x_1 = x_2 - 1
\]

Let \( x_2 = t \), where \( t \) is a free variable. Then:

\[
x_1 = t - 1
\]

Thus, the row space solution is:

\[
\mathbf{x}_r = \begin{bmatrix} -1 \\ 0 \end{bmatrix}
\]

### Step 2: Find the null space solution

Next, we find the null space of \( A \), i.e., we solve \( A \mathbf{x}_n = 0 \). This gives the system of equations:

\[
2x_1 - 2x_2 = 0 \quad \text{(3)}
\]
\[
-6x_1 + 6x_2 = 0 \quad \text{(4)}
\]

Again, both equations are equivalent. From equation (3):

\[
2x_1 - 2x_2 = 0 \quad \Rightarrow \quad x_1 = x_2
\]

So the null space solution is:

\[
\mathbf{x}_n = \begin{bmatrix} 1 \\ 1 \end{bmatrix}
\]

### Step 3: Express the general solution

The general solution is given by:

\[
\mathbf{x} = \mathbf{x}_r + c \mathbf{x}_n
\]

Substituting the values for \( \mathbf{x}_r \) and \( \mathbf{x}_n \):

\[
\mathbf{x} = \begin{bmatrix} -1 \\ 0 \end{bmatrix} + c \begin{bmatrix} 1 \\ 1 \end{bmatrix}
\]

Finally, this simplifies to:

\[
\mathbf{x} = \begin{bmatrix} -1 + c \\ c \end{bmatrix}
\]

This is the general solution.

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

1. What is the row space of a matrix?
2. How is the null space of a matrix determined?
3. Why is one equation from the system of equations sufficient to solve this problem?
4. What are the implications of free variables in solving linear systems?
5. How can the method used here apply to other systems of linear equations?

**Tip:** When dealing with systems of linear equations, always check if the equations are dependent. This reduces the number of equations you need to solve.

Express the general solution of \( A\mathbf{x} = \mathbf{b} \) where \( A = \begin{bmatrix} 2 & -2 \\ -6 & 6 \end{bmatrix} \) and \( \mathbf{b} = \begin{bmatrix} -2 \\ 6 \end{bmatrix} \) as \( \mathbf{x} = \mathbf{x}_r + c \mathbf{x}_n \), where \( \mathbf{x}_r \) is the row space solution and \( \mathbf{x}_n \) is a basis for \( \text{null}(A) \).

Explore the general solution for the linear system \( A \mathbf{x} = \mathbf{b} \) with matrix \( A \) and vector \( \mathbf{b} \). This problem guides you through finding the row space solution and the null space basis. Suitable for high school students in Grades 10-12 studying linear algebra.

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