Linear independence theorems in vector spaces:
A set consisting of a single non-zero vector $S = \{v\}$ where $v \ne 0$ is always linearly independent, because $c \cdot v = 0$ with $v \ne 0$ strictly implies $c = 0$.
Why others are false:

• is false: By convention, the empty set $\emptyset$ is linearly independent.
• is false: A set containing the zero vector $\{0\}$ or dependent vectors is not independent.
• is false: Many non-empty sets (like basis sets) are linearly independent.

A matrix satisfying $A^2 = A$ is called an idempotent matrix.

Let $\lambda$ be an eigenvalue of $A$ with corresponding non-zero eigenvector $v$:

$$Av = \lambda v$$

Multiplying by $A$:

$$A^2 v = A(Av) = A(\lambda v) = \lambda(Av) = \lambda(\lambda v) = \lambda^2 v$$

Since $A^2 = A$:

$$Av = \lambda^2 v \implies \lambda v = \lambda^2 v$$ $$(\lambda^2 - \lambda) v = 0$$

Because $v \ne 0$:

$$\lambda^2 - \lambda = 0 \implies \lambda(\lambda - 1) = 0 \implies \lambda = 0 \text{ or } \lambda = 1$$

The rank of a $3 \times 3$ matrix is 3 if its determinant is non-zero.

Calculate $\det(A)$ expanding along the first row:

$$\det(A) = 2 \begin{vmatrix} 2 & 5 \\ -1 & 1 \end{vmatrix} - 1 \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} + 4 \begin{vmatrix} 3 & 2 \\ 0 & -1 \end{vmatrix}$$ $$= 2(2(1) - 5(-1)) - 1(3(1) - 5(0)) + 4(3(-1) - 2(0))$$ $$= 2(2 + 5) - 1(3 - 0) + 4(-3 - 0)$$ $$= 2(7) - 3 + 4(-3) = 14 - 3 - 12 = 14 - 15 = -1$$

Since $\det(A) = -1 \ne 0$, the matrix $A$ is non-singular. Consequently, its rank is 3 (Option 4).

Let the given equations be:

• $x_1 - 3x_2 + 4x_3 = 3$
• $2x_1 - 5x_2 + 7x_3 = 6$
• $3x_1 - 8x_2 + 11x_3 = 11$
  
  Notice the linear relationship among the left-hand sides:
  $$\text{LHS}_1 + \text{LHS}_2 = (x_1 + 2x_1) + (-3x_2 - 5x_2) + (4x_3 + 7x_3) = 3x_1 - 8x_2 + 11x_3 = \text{LHS}_3$$
  
  Now compare the corresponding right-hand sides:
  $$\text{RHS}_1 + \text{RHS}_2 = 3 + 6 = 9$$
  However, the right-hand side of Equation 3 is:
  $$\text{RHS}_3 = 11$$
  
  This gives the contradiction $9 = 11$, meaning $\text{rank}(A) = 2 \ne \text{rank}(A|B) = 3$. Therefore, the system of equations is inconsistent and no solution exists. Option 4 is correct.