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For question 7(a), let's solve the problem step by step:

**Given:**
The linear mapping \( F: \mathbb{R}^4 \to \mathbb{R}^3 \) is defined as:

\[
F(x, y, z, t) = (-x + y + 2t, x + 2z - t, x + y + 3z - 3t).
\]

**Tasks:**
1. Find a basis and the dimension of the image of \( F \).
2. Find a basis and the dimension of the kernel of \( F \).

---

### Step 1: Write \( F(x, y, z, t) \) in matrix form

We express \( F \) as a matrix transformation:
\[
F(x, y, z, t) = A \begin{bmatrix} x \\ y \\ z \\ t \end{bmatrix},
\]
where \( A \) is a \( 3 \times 4 \) matrix. By reading the coefficients of \( x, y, z, t \) in the components of \( F \):
\[
A = \begin{bmatrix}
-1 & 1 & 0 & 2 \\
1 & 0 & 2 & -1 \\
1 & 1 & 3 & -3
\end{bmatrix}.
\]

---

### Step 2: Row reduce \( A \) to find the rank (image dimension)

Perform row reduction on \( A \):

1. Start with:
\[
A = \begin{bmatrix}
-1 & 1 & 0 & 2 \\
1 & 0 & 2 & -1 \\
1 & 1 & 3 & -3
\end{bmatrix}.
\]

2. Add Row 1 to Row 2 and Row 3:
\[
\begin{bmatrix}
-1 & 1 & 0 & 2 \\
0 & 1 & 2 & 1 \\
0 & 2 & 3 & -1
\end{bmatrix}.
\]

3. Replace Row 3 with Row 3 - 2 × Row 2:
\[
\begin{bmatrix}
-1 & 1 & 0 & 2 \\
0 & 1 & 2 & 1 \\
0 & 0 & -1 & -3
\end{bmatrix}.
\]

4. Multiply Row 3 by -1:
\[
\begin{bmatrix}
-1 & 1 & 0 & 2 \\
0 & 1 & 2 & 1 \\
0 & 0 & 1 & 3
\end{bmatrix}.
\]

5. Eliminate the 2 in Row 2, Column 3 by subtracting 2 × Row 3 from Row 2:
\[
\begin{bmatrix}
-1 & 1 & 0 & 2 \\
0 & 1 & 0 & -5 \\
0 & 0 & 1 & 3
\end{bmatrix}.
\]

6. Eliminate the 1 in Row 1, Column 2 by subtracting Row 2 from Row 1:
\[
\begin{bmatrix}
-1 & 0 & 0 & 7 \\
0 & 1 & 0 & -5 \\
0 & 0 & 1 & 3
\end{bmatrix}.
\]

Finally, multiply Row 1 by -1:
\[
\begin{bmatrix}
1 & 0 & 0 & -7 \\
0 & 1 & 0 & -5 \\
0 & 0 & 1 & 3
\end{bmatrix}.
\]

---

### Step 3: Interpret the row-reduced form

- The rank of \( A \) (number of pivots) is 3.
- Dimension of the image of \( F \): \( \text{dim(Im } F) = 3 \).
- A basis for the image of \( F \) consists of the rows of the row-reduced matrix:
\[
\text{Basis for Im } F: \{ (1, 0, 0, -7), (0, 1, 0, -5), (0, 0, 1, 3) \}.
\]

---

### Step 4: Find the kernel of \( F \)

The kernel is the solution space to \( A \mathbf{x} = 0 \). Since the rank of \( A \) is 3, the nullity of \( A \) is:
\[
\text{nullity of } A = 4 - 3 = 1.
\]

Thus, the kernel is 1-dimensional. From the row-reduced form, we write:
\[
\begin{bmatrix}
1 & 0 & 0 & -7 \\
0 & 1 & 0 & -5 \\
0 & 0 & 1 & 3
\end{bmatrix}
\begin{bmatrix} x \\ y \\ z \\ t \end{bmatrix} = 0.
\]

This gives:
\[
x - 7t = 0 \quad \Rightarrow \quad x = 7t,
\]
\[
y - 5t = 0 \quad \Rightarrow \quad y = 5t,
\]
\[
z + 3t = 0 \quad \Rightarrow \quad z = -3t.
\]

Let \( t = t \). The kernel is spanned by:
\[
\text{Basis for Ker } F: \{ (7, 5, -3, 1) \}.
\]

---

### Final Answer:
1. **Image:**
   - Dimension: 3.
   - Basis: \( \{ (1, 0, 0, -7), (0, 1, 0, -5), (0, 0, 1, 3) \} \).

2. **Kernel:**
   - Dimension: 1.
   - Basis: \( \{ (7, 5, -3, 1) \} \). 

Would you like a step-by-step diagram or further clarification?

---

**Expanding Questions:**
1. How is the dimension of the kernel and image related to the rank-nullity theorem?
2. What is the geometric interpretation of the image and kernel of a linear map?
3. Can you construct a similar example with a 4×4 matrix and find its kernel and image?
4. How do eigenvalues relate to the kernel of a matrix?
5. What role does row reduction play in determining the properties of a matrix?

**Tip:** Always verify row reduction steps carefully to avoid calculation errors!

Let F: R^4 → R^3 be the linear mapping defined by F(x, y, z, t) = (-x + y + 2t, x + 2z - t, x + y + 3z - 3t). Find a basis and the dimension of the image of F. Find a basis and the dimension of the kernel of the map F.

Learn how to solve a linear algebra problem involving a linear mapping from R^4 to R^3. Find the basis and dimension of the image and kernel using rank-nullity theorem and matrix row reduction. Perfect for undergraduate students studying linear algebra.

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