Analysis of matrix inversion properties:

• Incorrect: The inverse $A^{-1}$ of an invertible matrix is always invertible because $\det(A^{-1}) = 1/\det(A) \ne 0$.
• Correct: Inverting the inverse matrix returns the original matrix: $(A^{-1})^{-1} = A$.
• Incorrect: Inversion follows the reversal law: $(AB)^{-1} = B^{-1} A^{-1}$.
• Incorrect: An involution returns $A$, not $A^{-1}$.

Row reduction of matrix A:
Given matrix:

Apply elementary row operations:

• $R_2 \to R_2 - 2R_1$:
  $$R_2 = [2-2, 6-4, -3-0, -3 - (-2)] = [0, 2, -3, -1]$$
• $R_3 \to R_3 - 3R_1$:
  $$R_3 = [3-3, 10-6, -6-0, -5 - (-3)] = [0, 4, -6, -2]$$
• Now observe $R_3 = 2 R_2$. Apply $R_3 \to R_3 - 2R_2$:
  $$R_3 = [0-0, 4-4, -6 - (-6), -2 - (-2)] = [0, 0, 0, 0]$$
  The row echelon form is:
  $$\begin{bmatrix} 1 & 2 & 0 & -1 \\ 0 & 2 & -3 & -1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
  There are exactly 2 non-zero linearly independent rows. Thus, $\text{rank}(A) = 2$.
• Incorrect: Rank 0 applies only to the null matrix.
• Correct: Rank is exactly 2.
• Incorrect: Rank cannot exceed the row count (3).
• Incorrect: Row 3 is a linear combination of Row 1 and Row 2, so rank cannot be 3.

Analysis of subspace theorems:

• Incorrect: The linear sum $W_1 + W_2$ is always a subspace of $V$.
• Incorrect: The union $W_1 \cup W_2$ is a subspace if and only if one is contained in the other ($W_1 \subseteq W_2$ or $W_2 \subseteq W_1$).
• Correct: A fundamental theorem in linear algebra states that the intersection $W_1 \cap W_2$ of any two subspaces of a vector space $V$ is always a subspace of $V$ (closed under vector addition and scalar multiplication, and containing the zero vector).
• Incorrect: Subspaces are not necessarily orthogonal.

Row reduction of matrix B to RREF:
Given upper triangular matrix:

Determinant is $\det(B) = 5 \times 2 \times 7 = 70 \ne 0$. Because $\det(B) \ne 0$, matrix $B$ is non-singular of full rank 3.
By Gauss-Jordan elimination:

• Divide each row by its pivot: leading entries become 1.
• Eliminate entries above pivots: all off-diagonal entries become 0.
  Every non-singular square matrix has a unique Reduced Row Echelon Form equal to the identity matrix $I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
• Correct: Identity matrix $I_3$.
• Incorrect: Permuted identity matrix.
• Incorrect: Leading entry in row 1 must be 1, not 5.
• Incorrect: Off-diagonal elements above pivots must be zero in reduced echelon form.