Animated Solution for Mathematics - Matrices and Determinants: If $f_r(x),g_r(x),h_r(x),r=1,2,3$ are polynomials in $x$ such that $f_r(a)=g_r(a)=h_r(a),r=1,2,3$ and $F(x)=\left|\begin{array}{ccc}{f}_1(x)& f_2(x)& f_3(x)\\ g_1(x)& g_2(x)& g_3(x)\\ h_1(x)& h_2(x)& h_3(x)\end{array}\right|$ then $F^{\prime }(x)$ at $x=a$ is $\dots \dots \dots$

A Journey into $F^{\prime }(a)$

Imagine you are standing before a $3\times 3$ matrix, a structure that looks intimidatingly complex. You are given a function $F(x)$ defined as the determinant of this matrix, where every single entry is a polynomial in $x$.

Your task is to find the derivative of this determinant at a specific point, $x=a$. At first glance, you might be tempted to expand the determinant, differentiate the resulting polynomial, and then plug in $x=a$.

But wait—stop for a moment. In the world of JEE Advanced, there is almost always a more elegant path. Let us explore the beauty of the differentiation rule for determinants.

Understanding $F(x)$

We have the function defined as:

The problem provides us with a crucial piece of information: at $x=a$, the values of these polynomials are identical, meaning $f_r(a)=g_r(a)=h_r(a)$ for $r=1,2,3$. This is not just a random condition; it is the key that unlocks the entire problem.

When we differentiate a determinant, we do not differentiate every single element. Instead, we use a beautiful property: the derivative of a determinant is the sum of determinants where we differentiate one row at a time, keeping the other rows fixed.

So, $F^{\prime }(x)={\Delta }_1+{\Delta }_2+{\Delta }_3$, where ${\Delta }_1$ has only the first row differentiated, ${\Delta }_2$ has only the second, and ${\Delta }_3$ has only the third.

Now, let us evaluate this sum at $x=a$. Look at ${\Delta }_1(a)$:

In this determinant, the second row is $g_r(a)$ and the third row is $h_r(a)$. But wait! Our condition tells us that $g_r(a)=h_r(a)$.

A fundamental property of determinants states that if any two rows are identical, the determinant is zero. Therefore, ${\Delta }_1(a)=0$.

The same logic applies to ${\Delta }_2(a)$ and ${\Delta }_3(a)$. In ${\Delta }_2(a)$, the first and third rows are identical, so it vanishes. In ${\Delta }_3(a)$, the first and second rows are identical, so it also vanishes.

We are left with $F^{\prime }(a)=0+0+0=0$.

It is truly satisfying, isn't it? What seemed like a daunting task of differentiation turned into a simple observation of determinant properties.

This is the essence of JEE Advanced mathematics—finding the hidden symmetry, the elegant shortcut, and the underlying logic that makes the complex simple. Keep this perspective, and you will find that even the most terrifying problems have a beautiful, simple heart.