Animated Solution for Mathematics - Matrices and Determinants: An ordered pair $(\alpha ,\beta )$ for which the system of linear equations $(1+\alpha )x+\beta y+z=2$, $\alpha x+(1+\beta )y+z=3$, $\alpha x+\beta y+2z=2$ has a unique solution is

A Journey into Linear Systems

Welcome, future engineer. Today, we are not just solving a system of equations; we are exploring the very architecture of space.

When you look at a system of linear equations, I want you to stop seeing just variables and coefficients. I want you to see three planes in three-dimensional space. The question asks for a unique solution, which is the mathematical equivalent of asking: "When do these three planes intersect at exactly one point?"

The Gatekeeper of Uniqueness

In the world of linear algebra, the determinant of the coefficient matrix, which we denote as $\Delta$, is the ultimate gatekeeper. If $\Delta eq 0$, the system is well-behaved, and a unique solution exists.

If $\Delta =0$, the system collapses—either into a state of infinite solutions or no solution at all. Our mission is to ensure the system remains well-behaved. We start by constructing our matrix from the given equations:

Extracting the coefficients, we define our determinant $\Delta$ as:

The Surgical Strike (Row Operations)

Now, many students would jump straight into expanding this $3\times 3$ determinant. But you are not "many students." You are a strategist.

Expanding this directly is a recipe for a sign error. Instead, let's use the power of row operations to simplify the matrix. We want to create zeros, as they are our best friends in determinant calculations.

First, let's perform $R_1\to R_1-R_2$. Look at the first column: $(1+\alpha )-\alpha =1$. Look at the second column: $\beta -(1+\beta )=-1$. Look at the third column: $1-1=0$.

Our determinant transforms into:

See that? We have a zero in the first row! Let's keep going. Let's perform $R_2\to R_2-R_3$.

In the second row, $\alpha -\alpha =0$, $(1+\beta )-\beta =1$, and $1-2=-1$. Now our determinant is beautifully sparse:

The Final Verdict

Now, expansion is trivial. Expanding along the first row, we get:

For a unique solution, we require $\Delta eq 0$. Therefore, our condition is $\alpha + \beta + 2 eq 0$, or simply $\alpha + \beta eq -2$.

This is the "Master Key" we were looking for. Any pair $(\alpha ,\beta )$ that sums to $-2$ will cause the system to fail the uniqueness test.

Let's test our options. For $(1,-3)$, the sum is $-2$. For $(-3,1)$, the sum is $-2$. For $(-4,2)$, the sum is $-2$.

All these options lead to $\Delta =0$, which means they are traps! Only the pair $(2,4)$ gives us a sum of $6$, which is clearly not $-2$. Thus, $\Delta = 8 eq 0$.

Mathematics is not about memorizing formulas; it is about recognizing the structure of the problem. By using row operations, we turned a daunting $3\times 3$ determinant into a simple linear expression.

You have successfully navigated the trap and found the unique solution. Keep this mindset—simplify, strategize, and solve—and you will conquer any problem the JEE throws at you.