Animated Solution for Mathematics - Matrices and Determinants: Let $a-2b+c=1$, If $f(x)=\left|\begin{array}{ccc}x+a& x+2& x+1\\ x+b& x+3& x+2\\ x+c& x+4& x+3\end{array}\right|$, then:

Imagine you are sitting in the examination hall, the clock is ticking, and you are presented with a $3\times 3$ determinant. At first glance, it looks like a monster. You see $x$ everywhere—in every single entry.

Your instinct might be to dive in, expand along the first row, and start multiplying terms. But stop. Take a breath.

In JEE Advanced, whenever you see a problem that looks like a computational nightmare, there is almost always a hidden geometric or algebraic symmetry waiting to be exploited. This problem is a classic example of that.

We are given the function:

We are also given the constraint $a-2b+c=1$. If you try to expand this directly, you are essentially signing up for a long, tedious algebraic slog.

Look at the columns. Look at the constants: $2,3,4$ in the second column and $1,2,3$ in the third. They are in arithmetic progression. This is the 'spark' we need.

Why do we choose the operation $R_1\to R_1-2R_2+R_3$? Let us break down the logic. We want to eliminate the variable $x$ because it is the source of all our trouble.

If we look at the first column, we have $(x+a)$, $(x+b)$, and $(x+c)$. If we perform the operation $R_1-2R_2+R_3$, the $x$ terms become $x-2x+x$, which is exactly $0$.

This is the beauty of linear algebra. By choosing the coefficients $1,-2,1$, we are essentially creating a 'filter' that removes the variable part of the matrix. Let us apply this to the first column:

We are given that $a-2b+c=1$. Just like that, the first entry of our new first row becomes $1$.

Now, let us apply the same operation to the second and third columns. For the second column, we have:

And for the third column:

Do you see what happened? The entire first row has collapsed into $[1,0,0]$. This is a massive victory.

We have transformed a complex determinant into something incredibly simple. The determinant now looks like this:

Expanding this along the first row is trivial. We simply take the element $1$ and multiply it by the determinant of the remaining $2\times 2$ minor matrix.

We are left with:

Now, we evaluate this $2\times 2$ determinant using the standard cross-multiplication rule: $(ad-bc)$.

Expanding these binomials, we get:

Watch closely as the terms cancel out. $x^2$ minus $x^2$ is $0$. $6x$ minus $6x$ is $0$. We are left with $9-8$, which is $1$.

This is the elegance of the problem. We started with a matrix full of $x$ terms, and through the power of row operations, we proved that the function $f(x)$ is actually a constant function: $f(x)=1$.

It does not matter if $x$ is $50$, $-50$, or a billion; the output will always be $1$. So, when the question asks for $f(50)$ or $f(-50)$, the answer is staring us in the face. It is $1$.