Animated Solution for Mathematics - Matrices and Determinants: Consider the following system of equations $\alpha x+2y+z=1$, $2\alpha x+3y+z=1$, $3x+\alpha y+2z=\beta$ For some $\alpha ,\beta \in \mathbb{R}$. Then which of the following is NOT correct.

Imagine you are standing in a three-dimensional space, looking at three planes defined by our equations. Each equation represents a flat surface. A unique solution exists only if these three planes intersect at a single, precise point.

But what happens when the planes are parallel, or when they intersect along a line, or when they are all the same plane? This is the physical reality behind the algebraic condition $D=0$.

When the determinant $D$ of the coefficient matrix is zero, the planes do not intersect at a single point. They are either parallel (no solution) or they intersect in a way that creates a line or plane of solutions (infinitely many solutions).

We begin by setting up the determinant of the coefficient matrix:

This determinant is the gatekeeper; it tells us whether the system is well-behaved. Expanding along the first row, we calculate:

As we simplify this, we must be meticulous. Distributing the terms, we get:

Combining the like terms, we arrive at the elegant quadratic:

Factoring this, we find $D=(\alpha -3)(\alpha +1)$. This tells us that the system's behavior changes dramatically at $\alpha =3$ and $\alpha =-1$. These are our critical values.

Let us focus on the case $\alpha =-1$. At this value, $D=0$, meaning the system is either inconsistent or dependent.

To distinguish between these, we look at the numerator determinant $D_x$. We replace the first column of our matrix with the constants from the right-hand side: $1,1,$ and $\beta$:

Expanding this, we get:

Simplifying this, we find $D_x=7-4+2\beta -1-3\beta$, which reduces to $D_x=2-\beta$. This is a profound result.

If $\beta eq 2$, then $D_x eq 0$, and the system has no solution. However, if $\beta =2$, then $D_x=0$. In this specific scenario, checking $D_y$ and $D_z$ also yields zero, confirming that the system has infinitely many solutions.

We are now equipped to evaluate the options. Statement (2) claims that for $\alpha =-1$, there is no solution for all $\beta \in \mathbb{R}$.

We have just proven that this is false because when $\beta =2$, the system actually has infinitely many solutions. Therefore, statement (2) is the incorrect one.

This problem teaches us that parameters are not just variables; they are the knobs that tune the geometry of the system. By understanding the determinant, we have mastered the system's behavior.