3. Answer the questions below for the matrix
A =

1/2 1/2
1/2 1/2

(a) For which b does the equation Ax = b have a solution? Draw your solution.
(b) Which x satisfy the equation Ax = 0? Draw your solution
(c) Compute the determinant of A.
(d) Is A invertible? Relate your answer to the questions above

Let's go step by step for each part of the question for matrix \( A = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \):

---

### (a) **For which \( b \) does the equation \( A \mathbf{x} = \mathbf{b} \) have a solution?**

We are solving the equation \( A \mathbf{x} = \mathbf{b} \). This matrix equation is equivalent to the following system of equations:

\[
\frac{1}{2}x_1 + \frac{1}{2}x_2 = b_1
\]
\[
\frac{1}{2}x_1 + \frac{1}{2}x_2 = b_2
\]

This system implies that both components \( b_1 \) and \( b_2 \) of vector \( \mathbf{b} \) must be equal, because both equations are identical. Therefore, the equation \( A \mathbf{x} = \mathbf{b} \) has a solution **if and only if \( b_1 = b_2 \)**.

So, the vector \( \mathbf{b} \) must be of the form \( \mathbf{b} = \begin{pmatrix} c \\ c \end{pmatrix} \), where \( c \) is any real number.

#### Graphical Representation:
In two-dimensional space, the solution set for \( b_1 = b_2 \) corresponds to vectors that lie on the line \( y = x \), so \( \mathbf{b} \) must be a point on this line for the equation to have a solution.

---

### (b) **Which \( \mathbf{x} \) satisfy the equation \( A \mathbf{x} = 0 \)?**

To find the solutions to \( A \mathbf{x} = 0 \), we solve the following homogeneous system:

\[
\frac{1}{2}x_1 + \frac{1}{2}x_2 = 0
\]
\[
\frac{1}{2}x_1 + \frac{1}{2}x_2 = 0
\]

Both equations are identical, so we only need to solve:

\[
x_1 + x_2 = 0
\]

Thus, the solutions are of the form \( x_1 = -x_2 \). Therefore, the general solution is:

\[
\mathbf{x} = \begin{pmatrix} t \\ -t \end{pmatrix}
\]

where \( t \) is any real number.

#### Graphical Representation:
In two-dimensional space, the solution set corresponds to all points lying on the line \( x_1 + x_2 = 0 \), which is the line through the origin with a slope of -1.

---

### (c) **Compute the determinant of \( A \)**

The determinant of a 2x2 matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by:

\[
\text{det}(A) = ad - bc
\]

For the matrix \( A = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \), we compute:

\[
\text{det}(A) = \left( \frac{1}{2} \times \frac{1}{2} \right) - \left( \frac{1}{2} \times \frac{1}{2} \right) = \frac{1}{4} - \frac{1}{4} = 0
\]

Thus, the determinant of \( A \) is **0**.

---

### (d) **Is \( A \) invertible? Relate your answer to the questions above.**

A matrix is invertible if and only if its determinant is non-zero. Since the determinant of \( A \) is 0, **\( A \) is not invertible**.

#### Relationship to Previous Questions:
- In part (a), we found that the matrix equation \( A \mathbf{x} = \mathbf{b} \) has a solution only if \( b_1 = b_2 \). This means that the image of \( A \) (the set of possible \( \mathbf{b} \)'s) is not the entire \( \mathbb{R}^2 \) but rather a subspace of it (the line \( y = x \)).
- In part (b), the non-trivial solutions to \( A \mathbf{x} = 0 \) show that the matrix has a non-trivial null space, indicating that \( A \) is not invertible.

---

Would you like a detailed graph, or any other clarifications on this? Here are 5 related questions you could explore:

1. How do you find the null space of a matrix?
2. What are the conditions for a matrix to be invertible?
3. How can the rank of a matrix be related to its invertibility?
4. Can you geometrically interpret the determinant of a matrix?
5. How do eigenvalues relate to the determinant and invertibility of a matrix?

**Tip:** The determinant is a powerful tool for determining many properties of a matrix, including area scaling and invertibility.

3. Answer the questions below for the matrix A = [[1/2, 1/2], [1/2, 1/2]] (a) For which b does the equation Ax = b have a solution? Draw your solution. (b) Which x satisfy the equation Ax = 0? Draw your solution. (c) Compute the determinant of A. (d) Is A invertible? Relate your answer to the questions above.

Explore the matrix equation Ax = b for A = [[1/2, 1/2], [1/2, 1/2]] to determine when a solution exists, find the general solution to Ax = 0, and compute the determinant of A. Learn why this matrix is not invertible, with detailed explanations and graphical representations. Suitable for college-level linear algebra students.

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