Animated Solution for Mathematics - Matrices and Determinants: Let $m$ and $M$ be respectively the minimum and maximum values of $\left|\begin{array}{ccc}{\cos }^{2}x& 1+{\sin }^{2}x& \sin 2x\\ 1+{\cos }^{2}x& {\sin }^{2}x& \sin 2x\\ {\cos }^{2}x& {\sin }^{2}x& 1+\sin 2x\end{array}\right|$. Then the ordered pair $(m,M)$ is equal to :

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of trigonometric functions. We are presented with a $3\times 3$ determinant, and our mission is to find its minimum and maximum values.

Let us define this determinant as a function $f(x)$:

I know what you are thinking: "Do I really have to expand this entire thing?" The answer is a resounding "No." In the world of competitive exams, the brute-force approach is rarely the intended one. We are looking for elegance.

Before we dive into the expansion, let us look for the hidden structure. Notice the first and third rows? They share ${\cos }^{2}x$ in the first column. This is a gift!

By applying the row operation $R_1\to R_1-R_3$, we can turn that first element into a zero. Let us perform this operation:

Look at that! The first row is now $0,1,-1$. That is much cleaner. But we can do better.

Look at the second and third rows. They both have ${\sin }^{2}x$ in the second column. Let us apply $R_2\to R_2-R_3$ to create another zero:

Now that we have created zeros, the expansion becomes a walk in the park. We expand along the first row. The first term is zero, which is fantastic.

The remaining terms are:

Let us evaluate these $2\times 2$ determinants. For the first one, we get $(1)(1+\sin 2x)-(-1)({\cos }^{2}x)=1+\sin 2x+{\cos }^{2}x$. For the second, we get $(1)({\sin }^{2}x)-(0)({\cos }^{2}x)={\sin }^{2}x$.

Putting it all together, we have:

Now, watch the magic happen. When we distribute the negative sign, we get:

We can factor out the negative sign from the squared terms: $-({\cos }^{2}x+{\sin }^{2}x)$. And as every student of trigonometry knows, ${\sin }^{2}x+{\cos }^{2}x=1$.

Our expression simplifies beautifully to:

We have reduced the "beast" to a simple, oscillating function. We know that the range of $\sin 2x$ is $[-1,1]$.

To find the minimum value $m$, we subtract the maximum value of $\sin 2x$ from $-2$:

To find the maximum value $M$, we subtract the minimum value of $\sin 2x$ from $-2$:

Thus, the ordered pair $(m,M)$ is $(-3,-1)$. You see? With a little bit of patience and the right strategy, even the most intimidating problems reveal their simple, elegant cores. Keep practicing, and keep falling in love with the process!