Animated Solution for Mathematics - Matrices and Determinants: Let the system of linear equations $4x+\lambda y+2z=0,2x-y+z=0,\mu x+2y+3z=0,\lambda ,\mu \in \mathbb{R}$ has a non-trivial solution. Then which of the following is true?

Imagine you are standing in a three-dimensional space, looking at three planes. Each equation in our system, $4x+\lambda y+2z=0$, $2x-y+z=0$, and $\mu x+2y+3z=0$, represents a plane passing through the origin.

Because they all pass through $(0,0,0)$, the origin is a guaranteed meeting point. This is what we call the trivial solution.

The JEE Advanced examiner is asking for a non-trivial solution. This means these planes must share more than just a single point; they must intersect along a line or coincide entirely. This is the geometric heart of our problem.

The Gatekeeper of Solutions

To find when these planes share a common line, we must look at the coefficient matrix. For a system of $n$ linear equations in $n$ variables, the existence of a non-trivial solution is tied to the determinant of the coefficient matrix, denoted as $\Delta$.

If $\Delta eq 0$, the system is "well-behaved" and only has the trivial solution. If $\Delta =0$, the system becomes "degenerate," allowing for infinite non-trivial solutions. Our mission is to force the determinant to vanish.

We construct our matrix using the coefficients of $x,y,$ and $z$:

We perform the expansion along the first row using the cofactor expansion method.

First, we take the element $4$ and multiply it by the determinant of the remaining $2\times 2$ matrix:

Next, we move to the second element, $\lambda$. Applying the sign convention, the middle term is negative:

Finally, we take the third element, $2$:

When we combine these pieces, we obtain the following expression:

Grouping the terms, we arrive at $\lambda \mu -6\lambda +2\mu -12=0$. We can factor this by grouping the first two terms and the last two terms:

By pulling out the common factor $(\mu -6)$, we reach the final factored form:

This equation is the key to the entire problem. It tells us that for the system to have a non-trivial solution, either $\lambda =-2$ or $\mu =6$.

Looking at the constraints, we see that $\mu =6$ is the condition that allows $\lambda$ to be any real number. We have successfully navigated the algebraic landscape and uncovered the hidden constraint.

Remember, in JEE Advanced, the math is not just about calculation; it is about finding the path of least resistance through the logic. You have mastered the system—well done!