Animated Solution for Mathematics - Matrices and Determinants: If $D=\left|\begin{array}{ccc}1& 1& 1\\ 1& 1+x& 1\\ 1& 1& 1+y\end{array}\right|$ for $x\ne 0,y\ne 0$, then $D$ is

Welcome, future engineer. Today, we are going to peel back the layers of a classic JEE determinant problem. At first glance, the determinant

might look like a standard, albeit slightly intimidating, algebraic expression. You might be tempted to jump straight into the expansion, multiplying terms and hoping the algebra works out.

But I want to teach you a different way—a way that values elegance, speed, and, most importantly, the structural beauty of linear algebra.

When you see a determinant, your first instinct might be to expand it along the first row. While that is a valid mathematical path, it is often a trap in competitive exams.

You will end up with a long string of terms like $1((1+x)(1+y)-1)-1((1+y)-1)+1(1-(1+x))$. While this will eventually lead you to the correct answer, it is prone to sign errors and consumes precious time.

In the JEE, we don't just want the answer; we want the most efficient path to it.

Look at the matrix again. Do you see the first row? It is all ones. This is a massive hint.

Whenever you see a row or column of ones, your brain should immediately scream, "Create zeros!" We have a powerful tool in our arsenal: the property that adding or subtracting rows does not change the value of the determinant.

Let us apply this. We want to turn the first column into a column with a one at the top and zeros below it. We can achieve this by performing two simple operations: $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$.

When we subtract the first row from the second, the second row becomes: $(1-1),((1+x)-1),(1-1)$, which simplifies beautifully to $0,x,0$.

When we subtract the first row from the third, the third row becomes: $(1-1),(1-1),((1+y)-1)$, which simplifies to $0,0,y$.

Look at what we have created:

This is an upper triangular matrix. The elements below the main diagonal are all zero. This is the "Aha!" moment.

The determinant of a triangular matrix is simply the product of its diagonal elements. Expanding along the first column, we get:

We have arrived at $D=xy$. Because the problem explicitly states that $x eq 0$ and $y eq 0$, we know that $xy$ is a non-zero product.

Therefore, $D$ is clearly divisible by both $x$ and $y$.

This is the power of mathematical intuition. By choosing to simplify the matrix rather than attacking it with brute force, we turned a potentially messy calculation into a simple, elegant result. Keep this mindset as you continue your JEE preparation. Look for the structure, look for the zeros, and always look for the most elegant path. You have got this!