Animated Solution for Mathematics - Matrices and Determinants: Let $a,b,c$ be real numbers with $a^2+b^2+c^2=1$. Show that the equation $\left|\begin{array}{ccc}ax-by-c& bx+ay& cx+a\\ bx+ay& -ax+by-c& cy+b\\ cx+a& cy+b& -ax-by+c\end{array}\right|=0$ represents a straight line.

Imagine you are sitting in the exam hall. You turn the page, and there it is: a $3\times 3$ determinant filled with variables $a,b,c,x,$ and $y$. It looks like a chaotic mess of terms.

Your first instinct might be to panic, to start expanding it blindly, hoping that the terms will somehow cancel out. But take a deep breath. In JEE Advanced, complexity is rarely the goal; it is a test of your ability to see the underlying structure.

We are given the condition $a^2+b^2+c^2=1$. This is not just a constraint; it is a hint. It screams of vector normalization.

We need to simplify the determinant, and the best way to do that is through row operations. We apply the operation $R_1\to a{R}_1+b{R}_2+c{R}_3$.

Why this specific operation? Because the coefficients $a,b,$ and $c$ are perfectly aligned with the structure of the elements in the determinant.

But remember, we are changing the determinant's value by multiplying the first row by $a$. To keep the equation valid, we must divide the entire determinant by $a$. We are essentially performing a surgical strike on the matrix, simplifying it while keeping the balance of the equation intact.

Now, let's look at the first element, $R_{11}$. When we compute $a(ax-by-c)+b(bx+ay)+c(cx+a)$, something beautiful happens. Let's expand it:

Look closely. The $-aby$ and $+aby$ cancel out. The $-ac$ and $+ac$ vanish. We are left with $x(a^2+b^2+c^2)$.

And because our given condition is $a^2+b^2+c^2=1$, this entire expression collapses into just $x$.

This is the moment where the problem stops being a chore and starts being art. By repeating this logic for the other elements in the first row, we find that the row transforms into $[x,y,1]$. The monster has been tamed.

With our new, clean first row, the rest of the determinant becomes vulnerable. We can now use $R_1$ to eliminate the terms in the second and third rows.

We apply $R_2\to R_2-b{R}_1$ and $R_3\to R_3-c{R}_1$.

This is the 'cleanup' phase. We are systematically removing the clutter. After these operations, the determinant looks significantly lighter. We are no longer fighting the algebra; we are guiding it.

Now, we expand the determinant along the first row. Yes, it is still algebra, but it is controlled, predictable algebra.

As we expand, we get a long polynomial. But look at every single term. There is an $a$ in every one of them! We factor out the $a$, and it cancels perfectly with the $\frac{1}{a}$ we placed outside the determinant earlier.

What remains is the final, elegant factorization:

A Straight Line

We are left with a product of two factors equal to zero. We know that for any real numbers $x$ and $y$, the term $x^2+y^2+1$ is always greater than or equal to $1$.

It can never be zero. Therefore, the only way for the product to be zero is if the second factor is zero:

And there it is. The equation of a straight line. We started with a terrifying matrix and ended with the most fundamental shape in geometry.

This is the beauty of mathematics—no matter how complex the problem appears, there is always a path to simplicity if you look for the underlying structure. You didn't just solve a problem; you decoded it.