Animated Solution for Mathematics - Matrices and Determinants: Let $\lambda$ and $\alpha$ be real. Find the set of all values of $\lambda$ for which the system of linear equations $\lambda x+(\sin \alpha )y+(\cos \alpha )z=0,x+(\cos \alpha )y+(\sin \alpha )z=0,-x+(\sin \alpha )y-(\cos \alpha )z=0$ has a non-trivial solution. For $\lambda =1$, find all values of $\alpha$.

Welcome, student. Today, we are not just solving a system of linear equations; we are exploring the delicate balance between algebra and trigonometry. When you look at the system provided, the first thing that should catch your eye is that the right-hand side is zero for every equation.

This is a homogeneous system, $AX=0$. In the world of linear algebra, this is a special state. It implies that the origin $(0,0,0)$ is always a solution.

But we are hunting for something more—we want the 'non-trivial' solutions, the ones that exist beyond the origin.

Think of the determinant of the coefficient matrix as a gatekeeper. If the determinant is non-zero, the matrix is invertible, and the only path to the solution is the trivial one ($x=0,y=0,z=0$).

But we want to break that gate. We want the system to be singular, meaning the equations are linearly dependent. Therefore, our first and most critical step is to set the determinant of the coefficient matrix to zero:

I know, looking at a $3\times 3$ determinant filled with trigonometric functions can be intimidating. But take a deep breath. We are going to expand this along the first row. Do not panic at the sight of the variables; focus on the structure.

When we expand along the first row, we get:

Look closely at the first term. We have $-\lambda ({\cos }^2\alpha +{\sin }^2\alpha )$. The identity ${\sin }^2\alpha +{\cos }^2\alpha =1$ is your best friend here. It collapses that entire term into just $-\lambda$.

Now, look at the remaining terms. As we distribute the $\sin \alpha$ and $\cos \alpha$, we see a pattern emerging:

Notice the symmetry? We have two instances of $\sin \alpha \cos \alpha$, which combine to give $2\sin \alpha \cos \alpha$. And we have ${\cos }^2\alpha -{\sin }^2\alpha$. These are not random terms; they are the building blocks of double-angle identities.

By substituting the identities $2\sin \alpha \cos \alpha =\sin 2\alpha$ and ${\cos }^2\alpha -{\sin }^2\alpha =\cos 2\alpha$, our complex determinant equation transforms into something remarkably elegant:

This is the heart of the problem. We have reduced a system of three equations into a single, powerful relationship between $\lambda$ and $\alpha$.

To find the range of $\lambda$, we recall that any expression of the form $a\sin \theta +b\cos \theta$ oscillates between $-\sqrt{a^2+b^2}$ and $\sqrt{a^2+b^2}$. With $a=1$ and $b=1$, the range is clearly $[-\sqrt{2},\sqrt{2}]$.

Finally, the problem asks us to find $\alpha$ when $\lambda =1$. We set our equation to:

To solve this, we divide by $\sqrt{2}$ to invoke the compound angle formula:

This becomes $\sin (2\alpha +\frac{\pi }{4})=\frac{1}{\sqrt{2}}$. Solving this for $2\alpha +\frac{\pi }{4}$ gives us the general solutions. By splitting this into cases for even and odd integers $n$, we find that $\alpha =n\pi$ or $\alpha =n\pi +\frac{\pi }{4}$.

We started with a system of equations that seemed to be a chaotic mess of variables and trigonometric functions. Through the systematic application of linear algebra—using the determinant as our tool—and the graceful use of trigonometric identities, we peeled back the layers to reveal a simple, harmonic relationship.

This is the essence of JEE Advanced mathematics: finding the underlying order in what initially appears to be complexity. Keep this mindset, and you will conquer any problem.