Animated Solution for Mathematics - Matrices and Determinants: Let $\alpha \in (0,\infty )$ and $A=\left[\begin{array}{ccc}1& 2& \alpha \\ 1& 0& 1\\ 0& 1& 2\end{array}\right]$. If $\det (\text{adj}(2A-A^T)\cdot \text{adj}(A-2A^T))=2^8$, then $(\det (A){)}^2$ is equal to:

Imagine you are standing before a complex matrix equation. It looks intimidating, doesn't it? A $3\times 3$ matrix $A$ filled with variables, and an equation involving adjoints and transposes that seems to stretch across the page.

But here is the secret: in the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask together.

We are given:

The condition provided is:

The first instinct might be to start calculating the adjoints. Stop! That is the trap. Instead, let us use the properties of determinants.

We know that $\det (PQ)=\det (P)\cdot \det (Q)$ and $\det (\text{adj}(M))=(\det (M){)}^{n-1}$. Since our matrix is $3\times 3$, $n=3$, so $\det (\text{adj}(M))=(\det (M){)}^2$.

This transforms our equation into something much more manageable:

Now, look at the two terms inside the determinants: $2A-A^T$ and $A-2A^T$. If we let $M=2A-A^T$, then $A-2A^T=-(2A-A^T)=-M$.

Using the property $\det (kM)=k^n\det (M)$, we find that:

When we plug this back into our equation, the negative sign disappears because of the squaring:

This simplifies beautifully to $\det (M)=\pm 4$.

Now we must find the actual matrix $M=2A-A^T$. Computing $2A$ and $A^T$ is straightforward, and subtracting them gives us:

Expanding this determinant along the second row (a clever choice because of the zeros!) gives us:

Setting this equal to $\pm 4$, we get two potential values for $\alpha$. Since $\alpha >0$, we reject $-5/3$ and accept $\alpha =1$.

Finally, we calculate $\det (A)$ with $\alpha =1$:

The question asks for $(\det (A){)}^2$, so $(-4{)}^2=16$. We have navigated the complexity and arrived at the truth.