Animated Solution for Mathematics - Matrices and Determinants: If $f(x)=\left|\begin{array}{ccc}1& x& x+1\\ 2x& x(x-1)& (x+1)x\\ 3x(x-1)& x(x-1)(x-2)& (x+1)x(x-1)\end{array}\right|$ then $f(100)$ is equal to

Welcome, JEE warrior. Today, we face a problem that looks like a monster but is, in reality, a paper tiger. You are presented with a $3\times 3$ determinant:

You are asked to find $f(100)$. The trap is set. The number $100$ is a siren song, calling you to plug it in and drown in a sea of arithmetic. Do not fall for it.

In the JEE, when you see a complex expression, your first instinct should not be calculation; it should be investigation.

Let us look at the columns of this determinant, $C_1$, $C_2$, and $C_3$. There is a rhythm here, a hidden symmetry. Let us test a hypothesis: is the third column simply the sum of the first two?

Let us check row by row. For the first row, we have $1+x=x+1$. It works perfectly.

Now, the second row: This is exactly the element in the third column.

Finally, the third row: $3x(x-1)+x(x-1)(x-2)$. Let us factor out the common term $x(x-1)$: This, too, matches the third column perfectly.

We have discovered that for every row, $C_1+C_2=C_3$. This means the third column is a linear combination of the first two. In the language of linear algebra, the columns are linearly dependent.

When you see this, you should feel a surge of excitement. It means the determinant is zero. We can prove this by applying the column operation $C_3\to C_3-(C_1+C_2)$.

This operation transforms the third column into a column of zeros. A determinant with a column of zeros is always zero. Thus, $f(x)=0$ for all $x$.

Whether $x$ is $100$, $1$, or $10^6$, the answer is the same. The complexity vanishes, leaving only the elegant truth.

The correct option is (a). Remember, the JEE is not testing your ability to calculate; it is testing your ability to see.