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

Imagine you are standing before a system of linear equations. At first glance, they might look like a jumble of variables and coefficients, but there is a hidden symmetry here. We are dealing with a homogeneous system:

Because every equation equals zero, we know immediately that the trivial solution $(0,0,0)$ is always a possibility. But the problem whispers a secret: there exists a non-zero solution.

In the world of linear algebra, a homogeneous system possesses a non-trivial solution if and only if the determinant of its coefficient matrix is exactly zero. This is a fundamental geometric truth about the linear independence of the vectors formed by these equations.

Now, let us construct our determinant $\Delta$ from the coefficients of $x,y,$ and $z$:

Many students, in their haste, would dive straight into expanding this $3\times 3$ determinant. While that would eventually lead to the answer, it is a path filled with potential algebraic pitfalls.

Instead, let us use the art of row and column operations to simplify our lives. Look at the second and third columns. Notice that the elements in the second column are $2a,3b,4c$ and in the third are $a,b,c$.

If we perform the operation $C_2\to C_2-2C_3$, we can transform the second column into something much cleaner. The new second column becomes $2a-2a=0$, $3b-2b=b$, and $4c-2c=2c$. Our determinant now looks like this:

With our simplified determinant, the expansion becomes a breeze. Expanding along the first row, we get:

The first term simplifies to $-bc$. The second term is zero. The third term is $a(2c-b)$, which is $2ac-ab$.

Putting it all together, we have $-bc+2ac-ab=0$. Rearranging this, we find the elegant relation:

To reveal the nature of the progression, we divide the entire equation by the product $abc$:

This simplifies beautifully to:

This is the classic condition for Harmonic Progression. Since the reciprocals $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in Arithmetic Progression, the original terms $a,b,c$ must be in Harmonic Progression.