Animated Solution for Mathematics - Matrices and Determinants: If $y(x)=\left|\begin{array}{ccc}\sin x& \cos x& \sin x+\cos x+1\\ 27& 28& 27\\ 1& 1& 1\end{array}\right|,x\in \mathbb{R}$, then $\frac{d^{2}y}{d{x}^2}+y$ is equal to

Mastering Determinants in JEE Advanced

Welcome, future engineers. Today, we are going to dissect a problem that, at first glance, looks like a chaotic mess of trigonometry and large integers. You see a $3\times 3$ determinant, and your instinct might be to panic, grab your pen, and start expanding along the first row.

Stop. Take a breath. In the arena of JEE Advanced, the most complex-looking problems often have the most elegant, almost poetic, solutions. Our goal today is not just to solve for $\frac{d^{2}y}{d{x}^2}+y$, but to understand the 'why' behind the simplification.

The Detective Work

Let us look at our function:

When you stare at this, do not see numbers. See patterns. Look at the first column ($C_1$) and the third column ($C_3$).

In $C_1$, we have $\sin x$, $27$, and $1$. In $C_3$, we have $\sin x+\cos x+1$, $27$, and $1$. Do you see it? The elements of $C_1$ are literally embedded inside $C_3$.

This is the 'Spark' moment. In mathematics, whenever you see a column that is a linear combination of other columns, you are looking at a simplification waiting to happen. We don't need to expand this yet; we need to perform a surgical strike using determinant properties.

The Surgical Strike

We want to create zeros. Zeros are our best friends in linear algebra. If we can turn the third column into a column of zeros and a single non-zero term, the expansion becomes trivial.

We apply the column operation: $C_3\to C_3-C_1$. Let's see what happens to each row:

1. Row 1: $(\sin x+\cos x+1)-(\sin x)=\cos x+1$ 2. Row 2: $27-27=0$ 3. Row 3: $1-1=0$

Look at the beauty of that! Our determinant has transformed into:

We have successfully created two zeros in the third column. Now, expanding this is no longer a nightmare; it is a simple arithmetic exercise.

The Collapse

Expanding along the third column ($C_3$), we get:

Now, we evaluate the $2\times 2$ determinant:

So, our function, which started as a terrifying $3\times 3$ matrix, has collapsed into:

Take a moment to appreciate this. We have stripped away the complexity and reduced a matrix to a simple trigonometric function. This is the essence of JEE problem-solving: reducing the unknown to the known.

The Calculus Finale

Now that we have $y(x)=-\cos x-1$, the calculus part is straightforward, but we must be precise. We need to find $\frac{d^{2}y}{d{x}^2}+y$.

First, let's find the first derivative, $\frac{dy}{dx}$:

Since the derivative of $-\cos x$ is $\sin x$ and the derivative of a constant is $0$, we get:

Next, we find the second derivative, $\frac{d^{2}y}{d{x}^2}$:

Finally, we combine these into our target expression:

Watch the terms cancel out. The $\cos x$ and the $-\cos x$ vanish into thin air, leaving us with the final result:

We started with a complex determinant, performed a strategic column operation, simplified the expression, and applied basic calculus to arrive at a clean, constant answer. This problem was not about brute force; it was about pattern recognition and the confidence to manipulate the structure of the problem before diving into the calculations. Keep this mindset, and you will find that even the most intimidating JEE problems are just puzzles waiting to be solved.