Animated Solution for Mathematics - Matrices and Determinants: For the system of linear equations $x+y+z=6$, $\alpha x+\beta y+7z=3$, $x+2y+3z=14$, which of the following is NOT true ?

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to dissect a system of linear equations that might look intimidating at first glance, but beneath the surface, it hides a beautiful geometric structure.

When you see a system like this, your first instinct might be to reach for heavy machinery like Cramer's Rule or matrix inversion. But wait! Let's pause and observe the system first:

1. $x+y+z=6$ 2. $\alpha x+\beta y+7z=3$ 3. $x+2y+3z=14$

Notice something special? Equations (1) and (3) are completely free of the parameters $\alpha$ and $\beta$. They are the "anchors" of our system. They define a fixed relationship between $x,y,$ and $z$ regardless of what $\alpha$ and $\beta$ might be. This is our golden ticket.

The Strategic Reduction

Instead of diving into determinants, let's use the power of substitution. We have two equations with three variables. Let's subtract equation (1) from equation (3):

Just like that, we have expressed $y$ in terms of $z$. Now, let's take this and plug it back into equation (1) to find $x$:

We have successfully reduced our three-variable system to a single variable, $z$. This is the core of the problem. Now, we take these expressions for $x$ and $y$ and substitute them into the parameter-heavy equation (2):

The Moment of Truth

Let's expand this carefully. Don't rush; precision is the hallmark of an elite student:

Now, group the terms containing $z$ on one side and the constants on the other:

This is the equation $Az=B$ we were looking for, where $A=\alpha -2\beta +7$ and $B=2\alpha -8\beta +3$. This single equation dictates the fate of our entire system.

If $A eq 0$, we have a unique solution. If $A=0$ and $B=0$, we have infinitely many solutions. If $A=0$ and $B eq 0$, we have no solution at all.

Analyzing the Options

Option 4 claims that for every point $(\alpha, \beta) eq (7, 7)$ on the line $\alpha -2\beta +7=0$, the system has infinitely many solutions. Let's test this.

The line $\alpha -2\beta +7=0$ is exactly the condition $A=0$. For infinite solutions, we also need $B=0$. Let's substitute $\alpha =2\beta -7$ (from $A=0$) into $B$:

For $B$ to be zero, we need $\beta =-\frac{11}{4}$. This means infinite solutions occur only at the point where $\beta =-\frac{11}{4}$ and $\alpha =-\frac{25}{2}$.

It is a single, unique point, not an entire line! Therefore, the claim in Option 4 is false. You have just navigated the trap and emerged victorious. Keep this mindset—always look for the simplest path, and never let the parameters intimidate you!