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JEE Mathematics 2023 Question with Solution

Determinant question showing D_k as a 3 by 3 determinant with entries 1, 2k, 2k-1 in first row and asking if summation from k equals 1 to n of D_k is 96, then find n.
  • A

    66

  • B

    88

  • C

    44

  • D

    55

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The determinant is

Dk=∣12k2k−1nn2+n+2n2nn2+nn2+n+2∣D_k = \begin{vmatrix} 1 & 2k & 2k - 1 \\ n & n^2 + n + 2 & n^2 \\ n & n^2 + n & n^2 + n + 2 \end{vmatrix}

and

∑k=1nDk=96\sum_{k=1}^{n} D_k = 96

Find: nn

Using linearity of determinant in the first row for the summation,

∣∑k=1n1∑k=1n2k∑k=1n(2k−1)nn2+n+2n2nn2+nn2+n+2∣=96\begin{vmatrix} \sum_{k=1}^{n} 1 & \sum_{k=1}^{n} 2k & \sum_{k=1}^{n} (2k-1) \\ n & n^2 + n + 2 & n^2 \\ n & n^2 + n & n^2 + n + 2 \end{vmatrix} = 96

Now evaluate the sums:

∑k=1n1=n,∑k=1n2k=n(n+1),∑k=1n(2k−1)=n2\sum_{k=1}^{n} 1 = n, \qquad \sum_{k=1}^{n} 2k = n(n+1), \qquad \sum_{k=1}^{n} (2k-1) = n^2

So,

∣nn(n+1)n2nn2+n+2n2nn2+nn2+n+2∣=96\begin{vmatrix} n & n(n+1) & n^2 \\ n & n^2 + n + 2 & n^2 \\ n & n^2 + n & n^2 + n + 2 \end{vmatrix} = 96

Apply row operations:

R2→R2−R1,R3→R3−R1R_2 \rightarrow R_2 - R_1, \qquad R_3 \rightarrow R_3 - R_1

Then the determinant becomes

∣nn2+nn202000n+2∣=96\begin{vmatrix} n & n^2 + n & n^2 \\ 0 & 2 & 0 \\ 0 & 0 & n+2 \end{vmatrix} = 96

Now compute the determinant:

n⋅2⋅(n+2)=96n \cdot 2 \cdot (n+2) = 96

So,

2n(n+2)=962n(n+2) = 96 n(n+2)=48n(n+2) = 48 n2+2n−48=0n^2 + 2n - 48 = 0 (n−6)(n+8)=0(n-6)(n+8) = 0

Hence, since n>0n > 0,

n=6n = 6

Therefore, the correct option is A.

Common mistakes

  • Adding the determinants termwise without using linearity only in the first row. The determinant is not fully linear in all entries together. First combine the summation through the varying row correctly, then evaluate the determinant.

  • Using an incorrect formula for ∑k=1n(2k−1)\sum_{k=1}^{n} (2k-1). This sum is n2n^2, not n(n+1)2\frac{n(n+1)}{2}. Recognize it as the sum of the first nn odd numbers.

  • Making an error in row operations. When applying R2→R2−R1R_2 \rightarrow R_2 - R_1 and R3→R3−R1R_3 \rightarrow R_3 - R_1, the determinant value does not change, but each corresponding entry must be subtracted carefully.

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