Animated Solution for Mathematics - Matrices and Determinants: The number of values of $\alpha$ for which the system of equations : $x+y+z=\alpha$, $\alpha x+2\alpha y+3z=-1$, $x+3\alpha y+5z=4$ is inconsistent, is

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a system of linear equations; we are investigating a geometric conflict.

Imagine three planes floating in three-dimensional space. Each equation in our system, $x+y+z=\alpha$, $\alpha x+2\alpha y+3z=-1$, and $x+3\alpha y+5z=4$, represents one of these planes.

Usually, these three planes intersect at a single, beautiful point—a unique solution. However, sometimes the universe of mathematics is not so cooperative, leading to an inconsistent system where no common ground exists for all three.

Cramer's Rule

To navigate this, we turn to the elegant machinery of Cramer's Rule. This diagnostic tool dictates that the existence of a unique solution depends entirely on the main determinant, $D$, formed by the coefficients of our variables.

If $D eq 0$, the planes intersect at a single point. If $D=0$, the system is either inconsistent (no solution) or dependent (infinitely many solutions).

Our mission is to find the value of $\alpha$ that forces the system into the 'inconsistent' state. The condition is clear: $D=0$ AND at least one of the auxiliary determinants ($D_x,D_y,\text{or }{D}_z$) must be non-zero.

The Main Determinant

Let us construct our main determinant, $D$, by extracting the coefficients of $x,y,\text{and }z$:

We expand this along the first row. We take the first element, $1$, and multiply it by the determinant of the remaining $2\times 2$ matrix, then subtract the second element multiplied by its minor, and finally add the third element multiplied by its minor:

Simplifying the expression term by term, we obtain:

Factoring out the $3$, we reveal the elegance of the result: $D=3({\alpha }^2-2\alpha +1)$, which is a perfect square. Thus, $D=3(\alpha -1{)}^2$. Setting $D=0$ gives us our critical value: $\alpha =1$.

The Verification Trap

Many students stop here, but $D=0$ is only a necessary condition. We must verify that the system is truly inconsistent by checking if at least one auxiliary determinant is non-zero.

We replace the first column of $D$ with the constant terms from the right-hand side ($1,-1,4$) to calculate $D_x$:

Expanding this, we get:

Since $7 eq 0$, we have confirmed that for $\alpha =1$, the system is indeed inconsistent. The planes do not meet at a common point; they are in a state of perpetual disagreement.

We have found that $\alpha =1$ is the only value that creates this inconsistency. The question asks for the number of such values, and we have found exactly one.

This journey through determinants and consistency conditions is the heart of linear algebra. It teaches us that even when equations seem to conflict, there is a rigorous, logical structure governing that conflict.

The final answer is 1.