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 staring at a massive, intimidating determinant on your JEE Advanced paper. It is filled with $x$'s, $a$'s, $b$'s, and $c$'s. Your first instinct might be to panic or, worse, to start expanding it blindly.

But stop. Take a deep breath. In the world of competitive mathematics, complexity is often just a mask for elegance.

We are given the function:

We are asked to find $f(50)$. If you try to plug $50$ into every single $x$, you will be trapped in an algebraic nightmare. The secret lies in the condition $a-2b+c=1$. This is not just a random equation; it is your roadmap.

Decoding the Condition

Look closely at the first column of our determinant: $x+a$, $x+b$, and $x+c$. Now, look at the condition $a-2b+c=1$.

The coefficients $1,-2,1$ are staring right at you. This is the key to the lock. We need to manipulate the rows so that the first column transforms into something manageable.

We invoke the most powerful tool in our determinant arsenal: row operations. Specifically, we will perform the operation $R_1\to R_1-2R_2+R_3$. This operation perfectly mirrors the structure of our given condition.

Executing the Operation

Let us apply this operation to the first column. We take the element in $R_1$, subtract twice the element in $R_2$, and add the element in $R_3$.

The math looks like this: $(x+a)-2(x+b)+(x+c)$. If we group the terms, we get $(x-2x+x)+(a-2b+c)$.

The $x$ terms vanish into thin air, leaving us with $0+(a-2b+c)$. Since we know $a-2b+c=1$, the entire first element of our new first row becomes $1$.

But wait, the magic doesn't stop there. We must apply this same operation to the other columns.

For the second column, we have $(x+2)-2(x+3)+(x+4)$. Again, the $x$ terms cancel: $x-2x+x=0$. The constants $2-6+4$ also equal $0$.

For the third column, we have $(x+1)-2(x+2)+(x+3)$. The $x$ terms cancel, and the constants $1-4+3$ also equal $0$. Our first row is now a beautiful, simple $[1,0,0]$.

The Beauty of Invariance

Now, our determinant looks like this:

Expanding along the first row is now trivial. We get:

Calculating this $2\times 2$ determinant gives us $(x+3)(x+3)-(x+2)(x+4)$. Expanding these, we get $(x^2+6x+9)-(x^2+6x+8)$.

The $x^2$ and $6x$ terms cancel out perfectly, leaving us with $9-8=1$. The function $f(x)$ is not a function of $x$ at all; it is a constant!

No matter what value you plug in for $x$, the result is always $1$. Therefore, $f(50)=1$. This is the elegance of mathematics—the ability to see through the noise and find the simple truth hidden underneath.