Animated Solution for Mathematics - Matrices and Determinants: The system of equations $\alpha x+y+z=\alpha -1$, $x+\alpha y+z=\alpha -1$, $x+y+\alpha z=\alpha -1$ has infinite solutions, if $\alpha$ is

A Detective Story

Imagine you are standing in a 3D coordinate space. You have three planes, each defined by a linear equation.

Your goal is to find the value of $\alpha$ that makes these three planes dance together in perfect harmony, creating not just one point of intersection, but an infinite number of them. This is the essence of the problem we are tackling today.

It is not just about crunching numbers; it is about understanding the structural integrity of a system.

The Determinant as a Gatekeeper

We start with our system of equations:

1) $\alpha x+y+z=\alpha -1$

2) $x+\alpha y+z=\alpha -1$

3) $x+y+\alpha z=\alpha -1$

Before we dive into the algebra, let us think like architects. A system of three linear equations in three variables represents three planes.

If the determinant of the coefficient matrix, $\Delta$, is non-zero, these planes intersect at a single, unique point. But we are looking for infinite solutions.

This means the planes must be dependent—they must either be coincident or intersect along a common line. The gatekeeper for this condition is the determinant $\Delta$. We must set $\Delta =0$ to even begin our search.

The Algebra of the Determinant

Let us construct our coefficient matrix and calculate its determinant:

Expanding this determinant is where many students stumble, but let us do it with elegance. Expanding along the first row, we get:

Now, look closely. Do you see the common factor? It is $(\alpha -1)$. Let us factor it out to simplify our lives:

Factoring the quadratic part further, we get:

Setting $\Delta =0$, we find our critical values: $\alpha =1$ and $\alpha =-2$. These are the only two values where the system could possibly have infinite solutions.

The Reality Check

Now, we must be detectives. We have two suspects: $\alpha =1$ and $\alpha =-2$. We must test them to see which one actually produces infinite solutions.

If we test $\alpha =1$, the system becomes:

All three equations are identical! Geometrically, these are coincident planes. This is the definition of infinite solutions, as every point on this plane is a solution.

Now, let us test $\alpha =-2$:

If we add these three equations together, the left side becomes $(-2+1+1)x+(1-2+1)y+(1+1-2)z=0$, and the right side becomes $-3-3-3=-9$.

We get $0=-9$, which is impossible. This system is inconsistent; it has no solution.

The JEE Mindset

Mathematically, $\alpha =1$ is the correct condition for infinite solutions.

However, in the high-stakes environment of competitive exams, you might sometimes find that the provided options or the official key point toward $\alpha =-2$. This is a classic trap.

Always remember: the determinant condition $\Delta =0$ is necessary, but the consistency check is the final, decisive step. Never skip it. You have the tools, you have the logic—now go forth and conquer these systems with confidence!