Let's analyze each statement to determine if it's true or false.

Statement Analysis

1. Statement (a): "Não existe um sistema de equações lineares, com 4 incógnitas, possível e indeterminado com grau de determinação 4."

   • Analysis: For a system with 4 unknowns to be indeterminate (meaning there are infinitely many solutions), it must have fewer than 4 independent equations, leading to a system that has free variables. If there is a "degree of indeterminacy" of 4, this would imply the system has no independent equations, making it inconsistent or trivial. Therefore, this statement is true, as a system cannot be "possible and indeterminate" with a full degree of indeterminacy for all variables.

2. Statement (b): "Se $n \geq 2$, $A \in \mathcal{M}_{n \times (n-1)}(\mathbb{R})$, $b \in \mathcal{M}_{n \times 1}(\mathbb{R})$ e a matriz ampliada do sistema de equações lineares $AX = b$ é invertível, então o sistema é impossível."

   • Analysis: In this case, $A$ is an $n \times (n-1)$ matrix, which implies that it has fewer columns than rows, and therefore it cannot be invertible due to its rectangular shape (only square matrices can be invertible). However, the augmented matrix might be invertible if it has full row rank, which could make the system inconsistent if the last column vector $b$ does not lie in the column space of $A$. Thus, this statement is true, as such a system will likely be inconsistent.

3. Statement (c): "Se $S$ é um sistema de equações lineares sobre $\mathbb{R}$ em que o número de equações é superior ao número de incógnitas, então $S$ é impossível."

   • Analysis: When there are more equations than unknowns, the system is "overdetermined." However, it may still be consistent if the additional equations are linearly dependent or redundant, providing solutions that satisfy all equations. Thus, this statement is false as overdetermined systems can still have solutions if there is redundancy.

4. Statement (d): "Se $A \in \mathcal{M}_{n \times n}(\mathbb{R})$ e o sistema $Ax = 0_{n \times 1}$ é possível, então $A$ é invertível."

   • Analysis: The system $Ax = 0$ (the homogeneous system) is always possible (it has at least the trivial solution $x = 0$). However, $A$ being invertible would imply that the only solution is the trivial one. If $A$ has a non-trivial solution to $Ax = 0$, then $A$ is not invertible. Therefore, this statement is false because the existence of a solution does not imply invertibility, as there could be non-trivial solutions.

Summary of Results

• (a) True
• (b) True
• (c) False
• (d) False

Let me know if you want further details on any specific part or have questions on any step.

Additional Questions to Consider

1. What conditions make a system of linear equations inconsistent?
2. How does the rank of a matrix relate to the existence of solutions for linear systems?
3. Can a system with fewer equations than unknowns be determined (have a unique solution)?
4. What role does the determinant play in determining the invertibility of a matrix?
5. How can we use augmented matrices to analyze the consistency of a system?

Tip

For a matrix $A$ to be invertible, it must be square and of full rank. Non-square matrices cannot be invertible by definition.