Animated Solution for Mathematics - Matrices and Determinants: If $\alpha ,\beta \ne 0$, and $f(n)={\alpha }^n+{\beta }^n$ and $\left|\begin{array}{ccc}3& 1+f(1)& 1+f(2)\\ 1+f(1)& 1+f(2)& 1+f(3)\\ 1+f(2)& 1+f(3)& 1+f(4)\end{array}\right|=K(1-\alpha {)}^2(1-\beta {)}^2(\alpha -\beta {)}^2$, then $K$ is equal to:

Hello, future engineers! Today, we are going to peel back the layers of a problem that, at first glance, looks like a terrifying $3\times 3$ wall of algebra.

You see a determinant, you see powers of $\alpha$ and $\beta$, and your instinct might be to start expanding it row by row. But stop! In the world of JEE Advanced, brute force is rarely the intended path. Let's look at this with the eyes of a detective.

The Pattern Hunt

We are given $f(n)={\alpha }^n+{\beta }^n$. The determinant is filled with terms like $1+f(1)$, $1+f(2)$, and so on.

Let's look at that first element, $3$. If we write $3$ as $1+1+1$, and recall that $1=1^0$, ${\alpha }^0=1$, and ${\beta }^0=1$, we can rewrite $3$ as $1^0+{\alpha }^0+{\beta }^0$, which is $1+f(0)$.

Now, look at the general term: $1+f(n)=1^n+{\alpha }^n+{\beta }^n$. Suddenly, the entire determinant is composed of sums of powers!

This is the 'Aha!' moment. Whenever you see a determinant where each element is a sum of products, your mind should immediately jump to matrix multiplication. We are looking at a hidden product of two matrices.

The Matrix Decomposition

Let our determinant be $\Delta$. Because of the structure we just uncovered, we can express $\Delta$ as the product of two matrices, $A$ and $A^T$.

Specifically, $\Delta =\det (A)\cdot \det (A^T)$. Since the determinant of a matrix is equal to the determinant of its transpose, we have $\Delta =(\det A{)}^2$.

The matrix $A$ is the famous Vandermonde matrix:

This is a beautiful, symmetric structure. Our problem has now collapsed from a complex determinant expansion into the simple task of calculating the determinant of $A$ and squaring it.

The Vandermonde Magic

To calculate $\det A$, we want to create as many zeros as possible. Let's apply column operations: $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$.

This gives us:

Expanding along the first row, we get a $2\times 2$ determinant. We can factor out $(\alpha -1)$ from the first column and $(\beta -1)$ from the second column.

This leaves us with a very simple $2\times 2$ determinant of:

Calculating this is straightforward: $(\beta +1)-(\alpha +1)=\beta -\alpha$. Thus, $\det A=(\alpha -1)(\beta -1)(\beta -\alpha )$.

The Final Comparison

We are almost there! We know $\Delta =(\det A{)}^2$. Squaring our result, we get:

Notice that $(\beta -\alpha {)}^2=(\alpha -\beta {)}^2$, and $(\alpha -1{)}^2=(1-\alpha {)}^2$. So, $\Delta =(1-\alpha {)}^2(1-\beta {)}^2(\alpha -\beta {)}^2$.

Comparing this to the expression given in the problem, $K(1-\alpha {)}^2(1-\beta {)}^2(\alpha -\beta {)}^2$, it is clear that $K=1$.

See how the complexity vanished? By identifying the structure, we turned a mountain into a molehill. Keep this mindset, and you will conquer any problem the JEE throws at you!