Suppose that Ax = b is a system of equations. For simplicity in this problem, we’ll assume that A is 3 × 3, but the statement below holds for all m × n matrices. Show that if x = (x, y, z) T is a solution and x ′ = (x ′ , y′ , z′ ) T is another distinct solution, then the system has infinitely many solutions. That is, it’s impossible to have exactly two solutions (which, by extension, means it’s also impossible to have more than two yet still finitely many solutions). Hint: Think of these two solutions as points in R 3 . How do you parameterize the line containing these two points? What if you plug in any point on this line into the matrix equation?

Let's analyze the problem step by step. Given that \( A \) is a \( 3 \times 3 \) matrix and \( \mathbf{b} \) is a vector in \( \mathbb{R}^3 \), we are working with the system of linear equations \( A\mathbf{x} = \mathbf{b} \).

### Step 1: Define the two solutions

Suppose \( \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \) and \( \mathbf{x'} = \begin{pmatrix} x' \\ y' \\ z' \end{pmatrix} \) are two distinct solutions to the equation \( A\mathbf{x} = \mathbf{b} \). This means that:
\[
A\mathbf{x} = \mathbf{b}
\quad \text{and} \quad
A\mathbf{x'} = \mathbf{b}
\]

### Step 2: Parameterize the line containing these two points

The solutions \( \mathbf{x} \) and \( \mathbf{x'} \) can be thought of as points in \( \mathbb{R}^3 \). The line passing through these two points can be parameterized as:
\[
\mathbf{x_t} = (1 - t)\mathbf{x} + t\mathbf{x'}
\]
where \( t \) is a real number that parameterizes the line.

### Step 3: Substitute the parameterized solution into the original equation

Let's see if any point \( \mathbf{x_t} \) on this line is also a solution to \( A\mathbf{x} = \mathbf{b} \). Substituting \( \mathbf{x_t} \) into the equation gives:
\[
A\mathbf{x_t} = A\left((1 - t)\mathbf{x} + t\mathbf{x'}\right)
\]

By the linearity of matrix multiplication:
\[
A\mathbf{x_t} = (1 - t)A\mathbf{x} + tA\mathbf{x'}
\]

Since \( A\mathbf{x} = \mathbf{b} \) and \( A\mathbf{x'} = \mathbf{b} \), we can substitute these into the equation:
\[
A\mathbf{x_t} = (1 - t)\mathbf{b} + t\mathbf{b} = \mathbf{b}
\]

### Step 4: Conclusion

We have shown that for any value of \( t \), the point \( \mathbf{x_t} \) lies on the line connecting \( \mathbf{x} \) and \( \mathbf{x'} \) and satisfies the equation \( A\mathbf{x_t} = \mathbf{b} \). 

This implies that there are infinitely many solutions along this line. Therefore, if there are two distinct solutions \( \mathbf{x} \) and \( \mathbf{x'} \), then the system \( A\mathbf{x} = \mathbf{b} \) must have infinitely many solutions.

**Conclusion:** It is impossible to have exactly two solutions (or finitely many solutions greater than two) for a system of linear equations. If there are at least two distinct solutions, then the system has infinitely many solutions.

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Would you like more details or have any questions?

Here are 5 related questions:
1. How do the number of solutions of a linear system relate to the rank of the coefficient matrix?
2. What is the geometric interpretation of the solution set of a system of linear equations in \( \mathbb{R}^3 \)?
3. How does the concept of linear independence relate to the number of solutions in a system?
4. What are the conditions under which a linear system has a unique solution?
5. How can the determinant of a matrix help in determining the number of solutions of a system?

**Tip:** When dealing with systems of linear equations, remember that the existence of two distinct solutions usually indicates an infinite number of solutions due to the linear nature of the system.

Explore the concept of solutions in systems of linear equations, focusing on how the existence of two distinct solutions leads to infinitely many solutions. Learn about matrix multiplication, line parameterization, and the implications for linear algebra.

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