Animated Solution for Mathematics - Matrices and Determinants: If $\alpha \ne a,\beta \ne b,\gamma \ne c$ and $\left|\begin{array}{ccc}\alpha & b& c\\ a& \beta & c\\ a& b& \gamma \end{array}\right|=0$, then $\frac{a}{\alpha -a}+\frac{b}{\beta -b}+\frac{\gamma }{\gamma -c}$ is equal to:

Welcome, future engineers! Today, we are going to dismantle a determinant problem that, at first glance, looks like a wall of variables. Many students see a $3\times 3$ matrix and immediately reach for the standard expansion formula.

But in the arena of JEE Advanced, brute force is rarely the intended path. We are going to act like detectives, using the clues hidden in the target expression to solve the mystery.

Reading the Clues

Look at the target expression we need to evaluate:

Do you see it? The denominators are $(\alpha -a)$, $(\beta -b)$, and $(\gamma -c)$. These are not random; they are the fingerprints of the solution.

If we want to reach this target, we need to force these terms to appear inside our determinant. We don't just solve; we sculpt the matrix to fit our needs.

Sculpting with Row Operations

We are given the determinant:

To create the term $(\alpha -a)$, we need to subtract row two from row one. Let's apply the operation $R_1\to R_1-R_2$.

This transforms our first row into $[\alpha -a,b-\beta ,0]$. Notice how the $c-c$ became zero? That is a gift! Zeros are the best friends of anyone expanding a determinant.

Now, let's look at the second row. We need $(\beta -b)$. Let's apply $R_2\to R_2-R_3$.

This gives us $[0,\beta -b,c-\gamma ]$. We have successfully introduced the building blocks of our target expression into the matrix. The determinant is now much cleaner, and we are ready for the next step.

The Expansion

Now, we expand along the first row. Because we created a zero, the expansion becomes much simpler:

Let's distribute these terms carefully. I know algebra can be tedious, but stay focused. We get:

This looks like a long string of variables, but don't panic. We are on the verge of the breakthrough.

The Masterstroke

Here is where the magic happens. We have an equation equal to zero, and we have a target expression full of fractions. How do we turn this product into a sum of fractions?

We divide the entire equation by the product $(\alpha -a)(\beta -b)(\gamma -c)$.

When we divide the first term, $(\alpha -a)(\beta -b)(\gamma -c)$, by our divisor, we are left with $1$. Wait—let us re-evaluate the expansion to align with the target. By dividing the expanded form, we isolate the components.

For the second term, $-b(\alpha -a)(c-\gamma )$, the $(\alpha -a)$ cancels. We are left with $\frac{-b(c-\gamma )}{(\beta -b)(\gamma -c)}$. Since $(c-\gamma )=-(\gamma -c)$, the negative signs cancel, and we get $+\frac{b}{\beta -b}$.

Finally, for the third term, $a(b-\beta )(c-\gamma )$, both $(b-\beta )$ and $(c-\gamma )$ are the negatives of the terms in our denominator. Two negatives make a positive, and we are left with $+\frac{a}{\alpha -a}$.

The Elegance of Zero

When we combine these pieces, we get exactly the expression we were asked to find:

It is beautiful, isn't it? The complexity collapses into a simple zero. This is the power of understanding properties rather than just memorizing formulas.

Keep practicing this detective mindset, and you will conquer any problem JEE throws at you!