Q89: Let $A = \begin{bmatrix} 2 & 1 & 4 \\... - Mathematics - ICAR AIEEA PG 2024 - Statistical Sciences | Agrigyan

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).