गणित एवं संख्यात्मक विश्लेषण

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.

Analysis of matrix transpose algebraic properties:

• Incorrect: The transpose of a transpose returns the original matrix: $(A^T)^T = A$, not $A^T$.
• Incorrect: The reversal law applies to products: $(AB)^T = B^T A^T$.
• Incorrect: For a scalar $r$, transpose does not apply to scalars: $(rA)^T = r A^T$.
• Correct: For any two conformable matrices $A$ and $B$ of identical dimensions, the transpose of their sum is strictly equal to the sum of their individual transposes: $(A + B)^T = A^T + B^T$.