Animated Solution for Mathematics - Matrices and Determinants: If $A=\left[\begin{array}{ccc}{e}^t& e^{-t}\cos t& e^{-t}\sin t\\ e^t& -e^{-t}\cos t-e^{-t}\sin t& -e^{-t}\sin t+e^{-t}\cos t\\ e^t& 2e^{-t}\sin t& -2e^{-t}\cos t\end{array}\right]$ Then A is-

Mastering the Matrix

Welcome, future engineer. When you first look at this matrix $A$, it is designed to intimidate you. You see exponentials, you see trigonometric functions, and you see a $3\times 3$ grid that looks like a nightmare to calculate.

But here is the secret of JEE Advanced: Complexity is often just a mask for elegance.

The Art of Factoring

Never dive headfirst into a determinant calculation. If you start expanding this matrix as it is, you will drown in a sea of $e^t$ and $e^{-t}$ terms. Instead, look for the pattern.

Notice that the first column is dominated by $e^t$. The second and third columns are dominated by $e^{-t}$. We can pull these out!

By factoring $e^t$ from the first column and $e^{-t}$ from both the second and third columns, we transform the matrix into something much friendlier. We are left with an external scalar factor of:

Suddenly, the matrix inside the determinant is purely trigonometric. We have stripped away the noise to reveal the signal.

The Strategic Zeros

Now that we have a cleaner matrix, we need to make it even simpler. Our goal is to create zeros, as they are the best friends of a determinant expansion.

By applying the row operations $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$, we clear out the first column.

Imagine you are standing on the wedge of this problem. You have successfully reduced the first column to a single $1$ at the top, followed by two $0$s. Now, expanding along this column is not a chore—it is a breeze. We only need to calculate the determinant of the remaining $2\times 2$ sub-matrix.

The Grand Finale

This is where the magic happens. We are left with a $2\times 2$ determinant involving terms like $(-2\cos t-\sin t)$ and $(2\sin t-\cos t)$. When you perform the cross-multiplication $ad-bc$, you might feel a moment of panic as the expressions grow.

But stay calm. When you expand the squares, something beautiful occurs. The cross-terms—those pesky $4\sin t\cos t$ terms—will perfectly cancel each other out.

You are left with:

Factoring out the $5$, we are left with the fundamental identity: ${\sin }^{2}t+{\cos }^{2}t=1$.

We arrive at the final result:

Think about this result. The exponential function $e^{-t}$ is strictly positive for all $t\in \mathbb{R}$. It never touches zero.

Therefore, the determinant is never zero. This means the matrix $A$ is invertible for all real values of $t$.

You didn't just solve a problem; you navigated through a storm of variables to find a calm, constant truth. That is the essence of mathematics. Keep this mindset, and no matrix will ever intimidate you again.