Animated Solution for Mathematics - Matrices and Determinants: Let $D_k=\left|\begin{array}{ccc}1& 2k& 2k-1\\ n& n^2+n+2& n^2\\ n& n^2+n& n^2+n+2\end{array}\right|$. If ${\sum }_{k=1}^{n-1}{D}_k=96$, then $n$ is equal to ________.

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dismantle a problem that, at first glance, looks like a terrifying wall of variables.

We have a determinant $D_k$ where the index $k$ dances through the first row, and we are asked to find the sum of these determinants. Many students see this and immediately reach for the expansion formula—a path that leads to a swamp of algebraic exhaustion.

But we are not going to do that. We are going to look for the hidden symmetry.

Imagine you are standing before the matrix:

The first thing to notice is that the second and third rows are completely independent of $k$. They are static, like mountains in a landscape, while the first row is the only thing changing as $k$ varies.

Because the determinant is a multilinear function of its rows, we can push the summation operator ${\sum }_{k=1}^n$ right inside the determinant. It is like finding a secret door that lets us bypass the chaos.

Now, we must evaluate the sums of the elements in the first row. We have three distinct tasks:

1. The first element: ${\sum }_{k=1}^{n}1$. This is simply adding $1$ to itself $n$ times, which gives us $n$.

2. The second element: ${\sum }_{k=1}^{n}2k$. We pull the constant $2$ out, and we are left with the sum of the first $n$ natural numbers, $2\times \frac{n(n+1)}{2}$, which simplifies beautifully to $n^2+n$.

3. The third element: ${\sum }_{k=1}^n(2k-1)$. This is the classic sum of the first $n$ odd numbers. We can write this as $\sum 2k-\sum 1$, which is $(n^2+n)-n=n^2$.

Look at that! The first row has transformed into $[n,n^2+n,n^2]$. The complexity has vanished, replaced by a clean, structured row.

Now our determinant looks like this:

Do you see the pattern? The rows are almost identical. In the world of determinants, whenever you see rows that are nearly the same, you should immediately think of row operations.

We want to create zeros to make the expansion trivial. Let us perform $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$.

Subtracting the first row from the second gives us $[0,2,0]$. Subtracting the first row from the third gives us $[0,0,n+2]$. Our matrix is now an upper triangular matrix:

For an upper triangular matrix, the determinant is simply the product of the diagonal elements. This is the moment of truth.

We multiply $n\times 2\times (n+2)$ to get $2n^2+4n$. We are told this sum equals $96$. So, we set up our final equation:

Dividing by $2$, we get $n^2+2n-48=0$. Factoring this quadratic, we find $(n+8)(n-6)=0$.

Since $n$ must be a positive integer, we arrive at our destination: $n=6$.