Animated Solution for Mathematics - Matrices and Determinants: If the system of linear equations $2x+2ay+az=0,2x+3by+bz=0,2x+4cy+cz=0$, where $a,b,c\in \mathbb{R}$ are non-zero and distinct; has non-zero solution, then

Welcome, future engineer! Today, we are going to peel back the layers of a beautiful problem in linear algebra.

At first glance, you see a system of three linear equations: $2x+2ay+az=0$, $2x+3by+bz=0$, and $2x+4cy+cz=0$.

It looks like a standard system, but notice the right-hand side—it is all zeros. This is a homogeneous system.

In the world of JEE, this is a massive hint. A homogeneous system always admits the trivial solution, $x=0,y=0,z=0$.

But the problem demands a non-zero solution. This is the key that unlocks the door.

It tells us that the equations are not independent; they are locked in a geometric dance where they essentially describe the same space. Mathematically, this means the determinant of the coefficient matrix must be zero:

Now, I know the temptation is to jump straight into expanding this determinant. But hold on! If you expand this directly, you are inviting a storm of algebra that will likely lead to a sign error.

We are smarter than that. Let us use the power of row operations to create some breathing room.

By performing $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$, we transform our determinant into something much cleaner:

Look at that! The first column is now dominated by zeros. This makes the expansion along the first column a trivial task. We are left with:

Now, we enter the phase of careful, patient expansion. Do not rush this. Multiply the terms inside the brackets:

Notice the beauty of mathematics here? The $2a^2$ terms cancel out perfectly. It is as if the universe is rewarding your patience.

After combining the remaining terms, we are left with:

Rearranging this, we get the elegant relation:

We are almost there. We have the relation $ab+bc=2ac$, but it doesn't quite look like the standard form yet. This is where we use the final trick in our toolkit.

We divide the entire equation by the product $abc$:

The terms cancel out with satisfying precision, leaving us with:

This, my friend, is the classic condition for an Arithmetic Progression. If $\frac{1}{a}+\frac{1}{c}=\frac{2}{b}$, then by definition, $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.

You have successfully navigated the trap, simplified the complexity, and arrived at the truth. Keep this mindset—look for the structure, simplify before you calculate, and trust the process. You are doing great!