Animated Solution for Mathematics - Matrices and Determinants: If $\left|\begin{array}{ccc}x-4& 2x& 2x\\ 2x& x-4& 2x\\ 2x& 2x& x-4\end{array}\right|=(A+Bx)(x-A{)}^2$, then the ordered pair $(A,B)$ is equal to :

Imagine you are standing before this $3\times 3$ determinant. At first glance, it looks like a dense wall of variables. But look closer. Do you see the pattern?

The elements on the main diagonal are $x-4$, and the off-diagonal elements are all $2x$. This is not a random arrangement; it is a beautiful, symmetric structure. In the world of JEE Advanced, symmetry is your best friend. It is a silent signal that a clever transformation is waiting to be discovered.

Our goal is to simplify this matrix. We want to create a row or column where all elements are identical, allowing us to factor them out.

If we add the elements of the first column, the second column, and the third column together, what happens? For any row, we get $(x-4)+2x+2x=5x-4$. This is a constant!

This realization leads us to the operation $C_1\to C_1+C_2+C_3$. By replacing the first column with this sum, we transform our determinant into:

We have successfully pulled out the factor $(5x-4)$. Now, we are left with a column of ones. This is the moment where many students rush to expand, but hold your horses! The golden rule of determinants is to use those ones to create zeros.

With a column of ones, the path forward is clear. We use the first row as our pivot. By applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$, we systematically clear the first column:

Look at that! We have created an upper triangular matrix. The elements below the main diagonal are all zero. This is the moment of triumph.

The determinant of an upper triangular matrix is simply the product of its diagonal elements. We do not need to perform a long, tedious expansion. We simply multiply: $1\cdot (-(x+4))\cdot (-(x+4))$, which simplifies beautifully to $(5x-4)(x+4{)}^2$.

The problem asks us to compare our result with the form $(A+Bx)(x-A{)}^2$. We have:

By comparing the squared terms, $(x+4{)}^2$ and $(x-A{)}^2$, we identify that $A=-4$. Now, substitute $A=-4$ into the linear term $(A+Bx)$.

We get $(-4+Bx)$. Comparing this to our linear factor $(5x-4)$, it becomes crystal clear that $B=5$.

And there you have it! The ordered pair $(A,B)$ is $(-4,5)$. You have navigated the complexity, respected the symmetry, and arrived at the solution with elegance. Keep this mindset—look for the pattern, simplify, and conquer.