Animated Solution for Mathematics - Matrices and Determinants: If $\alpha +\beta +\gamma =2\pi$, then the system of equations $x+(\cos \gamma )y+(\cos \beta )z=0$, $(\cos \gamma )x+y+(\cos \alpha )z=0$, $(\cos \beta )x+(\cos \alpha )y+z=0$ has :

Solving the Homogeneous Mystery

Welcome, future engineers! Today, we are going to dive into a problem that sits at the beautiful intersection of linear algebra and trigonometry. It is a classic JEE Advanced challenge that tests not just your calculation speed, but your ability to see the hidden structure within a system of equations.

We are presented with a system of three linear equations:

Notice something crucial? The right-hand side of every equation is zero. This is a homogeneous system of linear equations, which we can write in matrix form as $AX=0$.

In the world of linear algebra, homogeneous systems are special. They always possess the trivial solution $(x,y,z)=(0,0,0)$. The real question is: do they possess non-trivial solutions?

The answer lies entirely in the determinant of the coefficient matrix, $\Delta$. If $\Delta eq 0$, the system has a unique solution (the trivial one). If $\Delta =0$, the system has infinitely many solutions. Our mission is to find $\Delta$ and see if it vanishes.

Let us construct our matrix $A$ and find its determinant $\Delta$:

This matrix is symmetric, which is a hint that we should expect some elegant cancellations. Expanding along the first row, we get:

Expanding this, we obtain:

Grouping the terms, we arrive at the expression:

Now, we must use our constraint: $\alpha +\beta +\gamma =2\pi$. This is the key that unlocks the problem. We can rewrite this as $\gamma =2\pi -(\alpha +\beta )$.

Taking the cosine of both sides, we use the property $\cos (2\pi -\theta )=\cos \theta$ to find that $\cos \gamma =\cos (\alpha +\beta )$.

Substituting this into our expression for $\Delta$, we replace $\cos \gamma$ with $\cos (\alpha +\beta )$. We also use the identity $1-{\cos }^2\alpha ={\sin }^2\alpha$. The expression becomes:

This is where the magic happens. We use the identity $2\cos \alpha \cos \beta =\cos (\alpha +\beta )+\cos (\alpha -\beta )$. Substituting this into our equation:

Expanding the term in the brackets:

The $-{\cos }^2(\alpha +\beta )$ and $+{\cos }^2(\alpha +\beta )$ terms cancel out perfectly! We are left with:

Using the identity $\cos (A-B)\cos (A+B)={\cos }^{2}A-{\sin }^{2}B$, we get:

Rearranging the terms, we see the final beauty:

We have successfully shown that $\Delta =0$. In the realm of homogeneous linear equations, a determinant of zero is the signal that the system is linearly dependent, meaning there are infinitely many non-trivial solutions.

You have just navigated a complex path of algebraic and trigonometric manipulation to reach a simple, elegant truth. Keep practicing this kind of structural thinking—it is exactly what separates the good from the elite in JEE Advanced.