Animated Solution for Mathematics - Matrices and Determinants: The value of $\left|\begin{array}{ccc}(a+1)(a+2)& a+2& 1\\ (a+2)(a+3)& a+3& 1\\ (a+3)(a+4)& a+4& 1\end{array}\right|$ is

Imagine you are sitting in the examination hall. You open your paper, and there it is: a determinant filled with variables:

Your instinct might scream, "Expand it!" But stop. Take a breath. In JEE Advanced, the most complex-looking problems often hide the most elegant, simple solutions. The key is not to calculate; the key is to observe.

Look at the third column. It is not filled with complex polynomials or terrifying variables; it is filled with ones. This is a massive, flashing neon sign from the examiner.

Whenever you see a column or row of ones, you have been handed a gift. Because the determinant is a linear operator, we can manipulate rows and columns to create zeros without changing the value of the determinant.

Zeros are the holy grail of determinants. They turn a $3\times 3$ expansion into a simple $2\times 2$ calculation. Our goal is to turn that third column into a vector that looks like $[1,0,0{]}^T$.

Row Operations

We are going to perform two surgical strikes. We will use row operations to eliminate the ones in the second and third rows. We define our operations: $R_2\to R_2-R_1$ and $R_3\to R_3-R_2$.

Let us execute the first one: $R_2-R_1$. Look at the first column: $(a+2)(a+3)-(a+1)(a+2)$. Do not expand this blindly! Factor out the common term $(a+2)$:

Now look at the second column: $(a+3)-(a+2)=1$. And the third column: $1-1=0$. We have our first zero!

Now, let us repeat this for the third row: $R_3-R_2$. The first column becomes $(a+3)(a+4)-(a+2)(a+3)$. Again, factor out $(a+3)$:

The second column is $(a+4)-(a+3)=1$. The third column is $1-1=0$.

Look at what we have created. Our determinant now stands as:

The complexity has evaporated. We have two zeros in the third column. Now, we expand along this column:

This is just a simple $2\times 2$ determinant. We cross-multiply:

This simplifies to $(2a+4)-(2a+6)$.

Watch closely. The $2a$ terms cancel out perfectly. $2a-2a=0$. We are left with $4-6=-2$.

The variable $a$ is gone. The complexity is gone. We are left with a simple, elegant constant: $-2$.

This is the essence of JEE Advanced mathematics. It is not about brute force; it is about finding the path of least resistance. You didn't just solve a determinant; you mastered the structure of the problem. Remember this feeling the next time you see a wall of variables: look for the ones, create the zeros, and let the algebra do the work for you.