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JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Determinants

Let

D_{k} = 1 n n 2 k n^{2} + n + 2 n^{2} + n 2 k - 1 n^{2} n^{2} + n + 2

. If

\sum_{k = 1}^{n} D_{k} = 96

, then

n

is equal to ______.

Answer: 6

Solution

Using the property that the sum of determinants equals the determinant with summed elements:

k = 1 \sum n D_{k} = n n n n^{2} + n n^{2} + n + 2 n^{2} + n n^{2} n^{2} n^{2} + n + 2 = 96

Applying row operations

R_{2} \to R_{2} - R_{1}

and

R_{3} \to R_{3} - R_{1}

:

n 00 n^{2} + n 20 n^{2} 0 n + 2 = 96

Expanding along the first column:

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

2 n (n + 2) = 96

n (n + 2) = 48

n^{2} + 2 n - 48 = 0

(n + 8) (n - 6) = 0

Since

n

must be positive,

n = 6

.

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Integer answer

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