Animated Solution for Mathematics - Matrices and Determinants: Let the number 2,b,c be in an A.P. and $A=\left|\begin{array}{ccc}1& 1& 1\\ 2& b& c\\ 4& b^2& c^2\end{array}\right|$. If $\det (A)\in [2,16]$, then c lies in the interval :

Imagine you are standing before a complex matrix. At first glance, it looks like a daunting wall of numbers:

Many students would immediately reach for the standard expansion method, calculating minors and cofactors, risking a sign error in the process.

But as an elite aspirant, you must learn to see the soul of the matrix. This is a classic Vandermonde determinant, a structure that appears frequently in advanced mathematics. By recognizing this, we transform a tedious calculation into a moment of pure mathematical clarity.

We are told that $2,b,c$ form an Arithmetic Progression. This is our key.

An A.P. is defined by its constant rhythm—the common difference $d$. If we start at $2$, the next term $b$ is simply $2+d$, and the term after that, $c$, is $2+2d$.

By defining everything in terms of $d$, we collapse the complexity of the problem into a single variable. This is the essence of problem-solving: reducing the unknown until it has nowhere to hide.

For any matrix of the form

the determinant is elegantly given by $(x_2-x_1)(x_3-x_2)(x_3-x_1)$. In our case, $x_1=2$, $x_2=b$, and $x_3=c$.

Substituting these values, we get:

Now, watch the magic happen. We know that $(b-2)=d$ and $(c-b)=d$. Furthermore, the distance from $2$ to $c$ is $(c-2)=2d$.

Substituting these into our determinant expression, we find:

The entire matrix, with all its rows and columns, simplifies down to a simple cubic expression. The complexity has vanished, leaving behind a clear path forward.

Solving the Inequality

We are given the condition that $\det (A)\in [2,16]$. This translates to the inequality:

Dividing by $2$, we get $1\le d^3\le 8$. Taking the cube root of all parts, we find that the common difference $d$ must lie between $1$ and $2$, inclusive:

But we are not done yet. The question asks for the interval of $c$, not $d$. We recall our earlier relation: $c=2+2d$.

If $d=1$, then $c=2+2(1)=4$. If $d=2$, then $c=2+2(2)=6$.

Thus, as $d$ ranges from $1$ to $2$, $c$ ranges from $4$ to $6$. The final interval is $c\in [4,6]$.

Take a moment to appreciate what you have just done. You didn't just solve a matrix problem; you identified a pattern, used the properties of sequences to simplify the algebra, and navigated an inequality with precision.

This is the mindset of a topper. When you face the JEE Advanced paper, remember this: look for the structure, simplify the variables, and trust the logic. You have the tools; now go forth and conquer.