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
If f:N→Zf:\mathbb{N} \to \mathbb{Z} is defined by
f(n)=∣n−1−5−2n23(2k+1)2k+1−3n33k(2k+1)3k(k+2)+1∣,k∈Nf(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 ∑n=1kf(n)=98\displaystyle\sum_{n=1}^k f(n) = 98, then kk is equal to:
A33correct
B44
C55
D66
Solution
Step 1: Form a single determinant for the partial sum By linearity of the determinant in column 1:
∑n=1kf(n)=∣k(k+1)2−1−5−k(k+1)(2k+1)33(2k+1)2k+1−3k2(k+1)243k(2k+1)3k(k+2)+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 C3→C3−13C2C_3 \to C_3 - \dfrac{1}{3}C_2 Row 1: −5−(−13)=−143-5-(-\tfrac{1}{3}) = -\tfrac{14}{3}; Row 2: (2k+1)−(2k+1)=0(2k+1)-(2k+1)=0; Row 3: 3k(k+2)+1−k(2k+1)=k2+5k+13k(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 R2→R2+2(2k+1)3R1R_2 \to R_2 + \dfrac{2(2k+1)}{3}R_1 and R3→R3+3k(k+1)2R1R_3 \to R_3 + \dfrac{3k(k+1)}{2}R_1. After these operations, expanding along column 1 yields a 2×22\times2 block:
∑n=1kf(n)=k(k+1)2⋅(2k+1)⋅16[−84k2−28k+14+84k2+28k]=7k(k+1)(2k+1)6\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 kk
7k(k+1)(2k+1)6=98  ⟹  k(k+1)(2k+1)=84\frac{7k(k+1)(2k+1)}{6} = 98 \implies k(k+1)(2k+1) = 84
 Testing k=3k=3: 3×4×7=843\times4\times7 = 84. ✓\checkmark Answer: (1)
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