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Let's break down the exercise step by step.

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### **Problem Setup:**
- The linear application \( f \) from \( \mathbb{R}^3 \) to \( \mathbb{R}^3 \) is defined by \( f(x, y, z) = (z, 0, x) \), for all \( (x, y, z) \in \mathbb{R}^3 \).

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### **Q42: Finding the Matrix \( A \)**
- The goal is to find the matrix representation of the linear transformation \( f \) with respect to the canonical basis of \( \mathbb{R}^3 \).

Given \( f(x, y, z) = (z, 0, x) \), let's apply this transformation to the canonical basis vectors:
1. \( f(1, 0, 0) = (0, 0, 1) \)
2. \( f(0, 1, 0) = (0, 0, 0) \)
3. \( f(0, 0, 1) = (1, 0, 0) \)

So, the matrix representation \( A \) is:
\[
A = \begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
1 & 0 & 0
\end{pmatrix}
\]

Thus, the correct option is **(b)**.

---

### **Q43: Determining if the Application \( f \) is Injective, Surjective, or an Automorphism**

We analyze the matrix \( A \):
\[
A = \begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
1 & 0 & 0
\end{pmatrix}
\]

- **Injectivity**: \( f \) is not injective because the second row is entirely zeros, which means the rank of the matrix is less than 3, specifically 2. The null space is non-trivial.
  
- **Surjectivity**: The rank of the matrix \( A \) is 2, meaning it maps to a 2-dimensional subspace of \( \mathbb{R}^3 \), so \( f \) is not surjective either.

- **Automorphism**: An automorphism requires the transformation to be both injective and surjective (i.e., bijective), which is not the case here.

Thus, the correct option is **(b)** (non-injective).

---

### **Q44: Characteristic Polynomial of Matrix \( A \)**

To find the characteristic polynomial \( P_A(X) \) of matrix \( A \), we compute the determinant of \( A - X I \), where \( I \) is the identity matrix.

\[
A - X I = \begin{pmatrix}
-X & 0 & 1 \\
0 & -X & 0 \\
1 & 0 & -X
\end{pmatrix}
\]

The determinant is:
\[
\det(A - X I) = \det \begin{pmatrix}
-X & 0 & 1 \\
0 & -X & 0 \\
1 & 0 & -X
\end{pmatrix}
= (-X)(-X)(-X) + (1)(-X)(1) = -X^3 + X
\]

Thus, the characteristic polynomial is:
\[
P_A(X) = X(X^2 - 1) = X(X - 1)(X + 1)
\]

The correct option is **(a)**.

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### Summary:
- **Q42**: The matrix is **(b)**.
- **Q43**: The application is **(b)** (non-injective).
- **Q44**: The characteristic polynomial is **(a)**.

Would you like a more detailed explanation on any step? Here are some related questions you might find useful:

1. How can we determine injectivity and surjectivity from a matrix?
2. What is the null space of a matrix, and how is it related to injectivity?
3. How do you compute the characteristic polynomial of a general matrix?
4. Why does an automorphism require both injectivity and surjectivity?
5. Can a matrix that is not injective still be invertible?

**Tip:** To find the rank of a matrix quickly, focus on identifying the number of linearly independent rows or columns.

Soit l'application linéaire f de R^3 dans R^3 définie par f(x, y, z) = (z, 0, x), pour tout (x, y, z) ∈ R^3.

Solve linear algebra problems involving matrix representations of transformations, injectivity, surjectivity, and the characteristic polynomial in R^3. This exercise is ideal for university students studying Linear Algebra.

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