Matrices & DeterminantsmediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Matrices & Determinants: Defined Vmatrix Vmatrix Equal (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question

f:\mathbb{N} \to \mathbb{Z}

is defined by

f(n) = \begin{vmatrix} n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}, \quad k \in \mathbb{N}

and

\displaystyle\sum_{n=1}^k f(n) = 98

, then

k

is equal to:

3

correct

4

5

6

Solution

Step 1: Form a single determinant for the partial sum By linearity of the determinant in column 1:

\sum_{n=1}^k f(n) = \begin{vmatrix} \dfrac{k(k+1)}{2} & -1 & -5 \\[6pt] -\dfrac{k(k+1)(2k+1)}{3} & 3(2k+1) & 2k+1 \\[6pt] -\dfrac{3k^2(k+1)^2}{4} & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}

Step 2: Apply the column operation

C_3 \to C_3 - \dfrac{1}{3}C_2

Row 1:

-5-(-\tfrac{1}{3}) = -\tfrac{14}{3}

; Row 2:

(2k+1)-(2k+1)=0

; Row 3:

3k(k+2)+1-k(2k+1)=k^2+5k+1

. Step 3: Apply row operations to eliminate column 1 entries in rows 2 and 3 Use

R_2 \to R_2 + \dfrac{2(2k+1)}{3}R_1

and

R_3 \to R_3 + \dfrac{3k(k+1)}{2}R_1

. After these operations, expanding along column 1 yields a

2\times2

block:

\sum_{n=1}^k f(n) = \frac{k(k+1)}{2}\cdot(2k+1)\cdot\frac{1}{6}\bigl[-84k^2-28k+14+84k^2+28k\bigr] = \frac{7k(k+1)(2k+1)}{6}

Step 4: Solve for

k

\frac{7k(k+1)(2k+1)}{6} = 98 \implies k(k+1)(2k+1) = 84

Testing

k=3

3\times4\times7 = 84

\checkmark

Answer: (1)

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