Animated Solution for Mathematics - Matrices and Determinants: The set of all values of $\lambda$ for which the system of linear equations $x-2y-2z=\lambda x$, $x+2y+z=\lambda y$, $-x-y=\lambda z$ has a non-trivial solution.

Welcome, fellow traveler on the path to JEE mastery! Today, we are diving into the heart of linear algebra. We are presented with a system of equations that, at first glance, looks like a tangled mess of variables and parameters.

But remember, in mathematics, complexity is often just a mask for underlying symmetry. Let us peel back that mask together.

The Homogeneous Transformation

We start with the system:

Our first instinct might be to panic at the sight of $\lambda$ scattered across the right-hand side. But take a breath. We need to bring order to this chaos.

By subtracting the $\lambda$ terms from both sides, we group the variables into a beautiful, homogeneous structure:

Now, we have the classic form $AX=0$. This is the language of linear transformations.

We are asking: for what values of $\lambda$ does this transformation collapse the space such that a non-zero vector $X$ is mapped to the origin? This happens if and only if the matrix $A$ is not invertible, which is equivalent to saying its determinant is zero: $\det (A)=0$.

The Determinant's Power

Let us construct our matrix $A$ and set its determinant to zero:

This determinant represents the volume of the parallelepiped formed by the column vectors of $A$. If this volume is zero, the vectors are linearly dependent, and the system has non-trivial solutions.

Let us expand this along the first row:

The Art of Cancellation

Now, let us simplify this step-by-step. It looks intimidating, but watch how the terms dance together.

First term: $(1-\lambda )[-{\lambda }^2+2\lambda +1]$. Inside the bracket: $(2-\lambda )(-\lambda )=-2\lambda +{\lambda }^2$. So, ${\lambda }^2-2\lambda +1$, which is $(\lambda -1{)}^2$. Thus, the first term is $(1-\lambda )(\lambda -1{)}^2$.

Second term: $2(-\lambda +1)=2(1-\lambda )$.

Third term: $-2(-1+2-\lambda )=-2(1-\lambda )$.

Look at that! The second and third terms are $2(1-\lambda )$ and $-2(1-\lambda )$. They vanish into thin air!

We are left with:

Since $(1-\lambda )=-(\lambda -1)$, this becomes:

The Singleton Set

Solving $-(\lambda -1{)}^3=0$ gives us exactly one solution: $\lambda =1$. The set of all values of $\lambda$ is simply $\{1\}$.

Because this set contains only one element, it is, by definition, a singleton set.

Isn't it beautiful? We started with a system that seemed to have many moving parts, and through the rigorous application of linear algebra, we found that only one specific value of $\lambda$ allows for a non-trivial solution. Keep this mindset—look for the structure, trust the process, and the math will always reveal its secrets.