Animated Solution for Mathematics - Matrices and Determinants: For all values of $A,B,C$ and $P,Q,R$ show that $\left|\begin{array}{ccc}\cos (A-P)& \cos (A-Q)& \cos (A-R)\\ \cos (B-P)& \cos (B-Q)& \cos (B-R)\\ \cos (C-P)& \cos (C-Q)& \cos (C-R)\end{array}\right|=0$.

Imagine you are standing before a massive, imposing fortress of a problem. You see a $3\times 3$ determinant, and every single entry is a trigonometric expression involving differences of angles.

It feels like a nightmare of algebra, doesn't it? You might be tempted to start expanding it using the standard determinant formula, but stop!

In JEE Advanced, when you see a structure that looks like a sum of products, you are not looking at a calculation problem; you are looking at a structural problem. The problem is not asking you to calculate; it is asking you to see.

The secret lies in the identity $\cos (A-P)=\cos A\cos P+\sin A\sin P$. This is your key to the fortress.

We are going to decompose this. We are going to take this matrix and split it into two. Why? Because matrix multiplication is the hidden language of these determinants.

By adding a zero term, we force the matrix into a $3\times 3$ form. We write each element as:

This is the 'Aha!' moment. We are essentially reverse-engineering the process of matrix multiplication.

Let us construct our two matrices, $M_1$ and $M_2$. For $M_1$, we take the first parts of our products: the $\cos A$ and $\sin A$ terms.

The first column becomes $(\cos A,\cos B,\cos C{)}^T$, the second column becomes $(\sin A,\sin B,\sin C{)}^T$, and the third column is filled with zeros because we only had two terms in our expansion. So:

Now for $M_2$. To make the row-by-column multiplication work, the rows of $M_2$ must match the columns of $M_1$.

The first row is $(\cos P,\cos Q,\cos R)$, the second row is $(\sin P,\sin Q,\sin R)$, and the third row is $(0,0,0)$. Thus:

Now, look at the beauty of what we have created. The fundamental property of determinants tells us that if any single row or column is completely zero, the value of the entire determinant is zero.

Look at $M_1$. Its third column is entirely zero. Therefore, $\det (M_1)=0$.

Now look at $M_2$. Its third row is entirely zero. Therefore, $\det (M_2)=0$.

Since our original determinant $\Delta =\det (M_1)\times \det (M_2)$, we have $\Delta =0\times 0=0$.

The massive, intimidating problem has collapsed into a beautiful, simple zero. This is the elegance of mathematics. It is not about brute force; it is about seeing the hidden structure.