Animated Solution for Mathematics - Matrices and Determinants: Let $d\in R$, and $A=\left[\begin{array}{ccc}-2& 4+d& (\sin \theta )-2\\ 1& (\sin \theta )+2& d\\ 5& (2\sin \theta )-d& (-\sin \theta )+2+2d\end{array}\right]$, $\theta \in [0,2\pi ]$. If the minimum value of $\det (A)$ is 8, then a value of d is :

Welcome, future engineer. Today, we are facing a classic JEE Advanced challenge. You see a $3\times 3$ matrix, $A$, filled with trigonometric functions and a variable $d$.

At first glance, it looks like a nightmare of algebra. If you were to expand this determinant directly, you would be wading through a swamp of terms, likely losing your way in a forest of signs and variables.

But here is the secret: in JEE Advanced, the most complex-looking problems often have the most elegant, surgical solutions. We do not use brute force; we use strategy.

Our objective is to find $d$ such that the minimum value of $\det (A)$ is $8$. Before we touch the determinant, we must simplify the matrix. We look at the rows:

Instead of expanding, we perform a row operation: $R_1\to R_1+R_3-2R_2$. We choose this specific operation to eliminate the variables and simplify the structure.

When we calculate the new first row, we see magic happen: - For the first element: $-2+5-2(1)=1$. - For the second element: $(4+d)+(2\sin \theta -d)-2(\sin \theta +2)=0$. - For the third element: $(\sin \theta -2)+(-\sin \theta +2+2d)-2(d)=0$.

Suddenly, our first row is $[1,0,0]$. The complexity has vanished.

Now, our determinant looks like this:

Expanding along the first row is now trivial. We are left with a $2\times 2$ determinant:

Let us expand this carefully. The first part gives us $2\sin \theta +2d\sin \theta -{\sin }^2\theta +4+4d-2\sin \theta$. Notice how $2\sin \theta$ and $-2\sin \theta$ cancel out.

The second part is $-(2d\sin \theta -d^2)$. When we combine everything, the $2d\sin \theta$ terms also cancel out. We are left with a beautiful, clean expression:

Minimization

We are given that the minimum value of this determinant is $8$. We have $\det (A)=(d+2{)}^2-{\sin }^2\theta$.

To minimize this expression, we must subtract the largest possible value of ${\sin }^2\theta$. Since $\sin \theta \in [-1,1]$, the maximum value of ${\sin }^2\theta$ is $1$.

Thus, the minimum value of the determinant is:

Setting this equal to $8$, we get $(d+2{)}^2=9$. Taking the square root, we find $d+2=\pm 3$.

This gives us two possibilities: $d=1$ or $d=-5$. Looking at our options, $-5$ is the one that fits. You have successfully navigated the trap, simplified the chaos, and arrived at the truth.