Q158 / 269 · NDA 2025 (Phase 1) · Mathematics

Consider the following statements in respect of the determinant $\Delta = \begin{vmatrix} k(k+2) & 2k+1 & 1 \\ 2k+1 & k+2 & 1 \\ 3 & 3 & 1 \end{vmatrix}$:
I. $\Delta$ is positive if $k > 0$.
II. $\Delta$ is negative if $k < 0$.
III. $\Delta$ is zero if $k = 0$.
How many of the statements given above are correct?

  • A
    None
  • Correct answer
    B
    One
  • C
    Two
  • D
    All three

Solution

Given determinant is $\Delta = \begin{vmatrix} k(k+2) & 2k+1 & 1 \\ 2k+1 & k+2 & 1 \\ 3 & 3 & 1 \end{vmatrix}$

Applying row operations $R_1 \rightarrow R_1 - R_2$ and $R_2 \rightarrow R_2 - R_3$:

$\Delta = \begin{vmatrix} k^2-1 & k-1 & 0 \\ 2k-2 & k-1 & 0 \\ 3 & 3 & 1 \end{vmatrix}$

Expanding along the third column:

$\Delta = 1 \cdot \begin{vmatrix} k^2-1 & k-1 \\ 2k-2 & k-1 \end{vmatrix}$

$\Delta = (k^2-1)(k-1) - (2k-2)(k-1)$

$\Delta = (k-1) [ k^2 - 1 - 2k + 2 ]$

$\Delta = (k-1) (k^2 - 2k + 1)$

$\Delta = (k-1)^3$

Evaluating the given statements:

I. For $k = \dfrac{1}{2} > 0$, $\Delta = \left(\dfrac{1}{2} - 1\right)^3 = -\dfrac{1}{8} < 0$. Thus, statement I is incorrect.

II. For $k < 0$, $k - 1 < -1$, which implies $(k-1)^3 < 0$. Thus, statement II is correct.

III. For $k = 0$, $\Delta = (0-1)^3 = -1 \neq 0$. Thus, statement III is incorrect.

Only one statement is correct.

Answer: One

Hence, the correct answer is option: 'B'.

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