Determinant Dk Sum From k=1 to n Equals 96, Find n

JEE Main 2023 — 10 April, Shift 2 (moderate confidence). Previous Year Question.

Problem

Let $D_k$ be the determinant of the matrix:

$$ D_k = \begin{vmatrix} 1 & 2k & 2k-1 \\ n & n^2+n+2 & n^2 \\ n & n^2+n & n^2+n+2 \end{vmatrix} $$

If $\displaystyle\sum_{k=1}^{n} D_k = 96$, find the value of $n$.

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Key insight. Only the first row of $D_k$ actually depends on $k$ — the other two rows are fixed. Since a determinant is linear in any single row, summing $D_k$ over $k$ is the same as replacing that first row with the sum of its entries over $k$, and computing one determinant instead of $n$ of them.

Approach

This is the kind of determinant-sum problem where the individual entries change with $k$ only in the first row, so summing $D_k$ over $k$ before expanding — using $\sum_{k=1}^n 2k = n(n+1)$ and $\sum_{k=1}^n (2k-1) = n^2$ — collapses the sum into a single, much simpler determinant. From there, a couple of row operations and one expansion leave a clean quadratic in $n$.

Solution

Step 1 — Collapse the sum into a single determinant

Because rows 2 and 3 of $D_k$ don’t involve $k$ at all, and a determinant is linear in each row separately, summing $D_k$ from $k=1$ to $n$ is the same as summing only the first row’s entries over $k$ and keeping rows 2 and 3 as they are:

$$\sum_{k=1}^n 1 = n, \qquad \sum_{k=1}^n 2k = n(n+1), \qquad \sum_{k=1}^n (2k-1) = n^2$$

So the whole sum equals one determinant, with the summed row on top:

$$\sum_{k=1}^{n} D_k = \begin{vmatrix} n & n^2+n & n^2 \\ n & n^2+n+2 & n^2 \\ n & n^2+n & n^2+n+2 \end{vmatrix}$$

Step 2 — Row-reduce to expose the zeros

Row operations $R_2 \to R_2 – R_1$ and $R_3 \to R_3 – R_1$ don’t change the value of a determinant, and here they clear out most of the first column:

$$\begin{vmatrix} n & n^2+n & n^2 \\ 0 & 2 & 0 \\ 0 & 0 & n+2 \end{vmatrix}$$

(Row 2’s middle entry drops to $(n^2+n+2)-(n^2+n)=2$; row 3’s last entry drops to $(n^2+n+2)-n^2=n+2$ — every other difference is zero.)

Step 3 — Expand and solve for n

With two rows now almost entirely zero, expanding along the first column leaves only the $n$ in the top-left corner times the $2\times 2$ minor below it:

$$n \cdot \begin{vmatrix} 2 & 0 \\ 0 & n+2 \end{vmatrix} = n \cdot 2(n+2) = 2n(n+2)$$

Setting this equal to 96:

$$2n(n+2) = 96 \implies n(n+2) = 48 \implies n^2+2n-48=0$$

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

Since $n$ indexes the size of the matrix and the range of the sum, $n$ must be a positive integer, ruling out $n=-8$.

Answer

Common mistakes

• Expanding $D_k$ term-by-term for a general $k$ before summing, rather than summing the entries first — this makes the algebra far messier than it needs to be, since it means expanding $n$ separate $3\times 3$ determinants instead of one.
• Forgetting to discard the negative root. The quadratic $n^2+2n-48=0$ has two roots, $6$ and $-8$, but $n$ here is both a matrix dimension and an upper summation limit — it has to be a positive integer, so $-8$ isn’t a valid answer even though it solves the equation.

Practise next

• Let $D_k=\begin{vmatrix} 2k & 1 \\ n^2 & n \end{vmatrix}$. If $\displaystyle\sum_{k=1}^{n} D_k = 144$, find $n$. The point is the same one this problem turns on: a determinant is linear in the row holding $k$, so the sum can be taken inside the determinant before it is expanded.