Animated Solution for Mathematics - Matrices and Determinants: The solution set of the equation $\left|\begin{array}{ccc}1& 4& 20\\ 1& -2& 5\\ 1& 2x& 5x^2\end{array}\right|=0$ is .........

Imagine you are standing in front of this matrix, a $3\times 3$ grid of numbers. The standard, brute-force approach is to expand it, to grind through the arithmetic of minors and cofactors.

But in the world of JEE Advanced, we are not just calculators; we are detectives. We look for the hidden structure, the elegant shortcut that turns a tedious calculation into a moment of clarity.

Look at the determinant again:

Notice the third row, $R_3=[1,2x,5x^2]$. It is the only place where our variable $x$ lives. This is a massive clue.

If we were to expand this determinant along the third row, we would be multiplying terms by $x$ and $x^2$. This tells us immediately that the resulting equation is a quadratic in $x$.

A quadratic equation, by the Fundamental Theorem of Algebra, can have at most two roots. We are not just solving an equation; we are hunting for two specific values.

Here is the secret weapon: the property of determinants that states if any two rows are identical, the determinant is zero.

This is not just a rule; it is a geometric reality. If two rows are identical, the vectors they represent are linearly dependent, and the volume of the parallelepiped they define collapses to zero.

We don't need to expand; we just need to make the rows match.

Let us force the third row to be a carbon copy of the first row, $R_1=[1,4,20]$. We set the corresponding elements equal:

1. $2x=4\Rightarrow x=2$

2. $5x^2=20\Rightarrow x^2=4\Rightarrow x=\pm 2$

The common value that satisfies both is $x=2$. If we plug $x=2$ back into $R_3$, it becomes $[1,4,20]$, which is identical to $R_1$. The determinant vanishes. We have our first root!

Now, let us try to make the third row identical to the second row, $R_2=[1,-2,5]$:

1. $2x=-2\Rightarrow x=-1$

2. $5x^2=5\Rightarrow x^2=1\Rightarrow x=\pm 1$

The common value here is $x=-1$. If we plug $x=-1$ into $R_3$, it becomes $[1,-2,5]$, which is identical to $R_2$. The determinant vanishes again. We have our second root!

We have found our two roots, $x=2$ and $x=-1$, without ever performing a full expansion. We used the structure of the matrix to our advantage.

The final solution set is $\{-1,2\}$.

This is the beauty of mathematics—when you stop fighting the problem and start understanding its soul, the solution reveals itself with grace.