Animated Solution for Mathematics - Matrices and Determinants: The sum of distinct values of $\lambda$ for which the system of equations $$\begin{gathered} (\lambda -1)x+(3\lambda +1)y+2\lambda z=0 \\ (\lambda -1)x+(4\lambda -2)y+(\lambda +3)z=0 \\ 2x+(3\lambda +1)y+3(\lambda -1)z=0 \end{gathered}$$ has non-zero solutions, is

Welcome, aspiring mathematician. Today, we are going to unravel a problem that sits at the very heart of linear algebra: the homogeneous system of equations.

When you first look at the system provided, you might feel a sense of dread at the sight of $\lambda$ scattered across every coefficient. But take a breath. The beauty of this problem lies not in brute-force calculation, but in recognizing the underlying structure.

Look closely at the equations. Notice how every single equation equals zero on the right side? This is a homogeneous system.

In the world of linear algebra, a homogeneous system always possesses a 'trivial' solution where $x=y=z=0$. However, the problem asks for non-zero, or non-trivial, solutions. This is a massive hint.

For a homogeneous system to have non-trivial solutions, the system must be linearly dependent. Mathematically, this is equivalent to saying the determinant of the coefficient matrix must be zero: $\Delta =0$. This is our gateway to the solution.

Now, we construct our determinant $\Delta$:

If you try to expand this directly, you are inviting chaos. Instead, let us be surgical. Look at the first column. The first and second rows both start with $\lambda -1$. This is a gift!

By applying the row operation $R_2\to R_2-R_1$, we can turn that second row into something much simpler. The first element becomes zero, the second becomes $(\lambda -3)$, and the third becomes $(3-\lambda )$. Suddenly, the determinant looks like this:

Notice the pattern in the second row? We have $(\lambda -3)$ and $-(\lambda -3)$. We can factor out $(\lambda -3)$ from the entire second row, leaving us with a much cleaner determinant to expand.

With the factor $(\lambda -3)$ safely outside, we expand along the second row. The zero term vanishes, and we are left with two $2\times 2$ determinants.

After careful cross-multiplication and simplification, the entire expression collapses into:

Factoring further, we get $6\lambda (\lambda -3{)}^2=0$. The roots are $\lambda =0$ and $\lambda =3$.

The question asks for the sum of distinct values. Since $3$ is a repeated root, we only count it once. Thus, the sum is $0+3=3$.

This problem teaches us that in JEE Advanced, the path of least resistance is almost always the path of deepest understanding. By using row operations, we avoided a mountain of algebra and arrived at the answer with elegance.

Never rush into expansion; always look for the structure first.