Animated Solution for Mathematics - Matrices and Determinants: Let $A$ be a square matrix such that $A{A}^T=I$. Then $\frac{1}{2}A\left[(A+A^T{)}^2+(A-A^T{)}^2\right]$ is equal to

Imagine you are standing at the threshold of a complex matrix problem. A square matrix $A$ with the condition $A{A}^T=I$ defines an orthogonal matrix.

This matrix preserves the geometry of space, such as rotations or reflections. Because $A$ is a square matrix, the condition $A{A}^T=I$ implies that $A^T$ is the inverse of $A$.

Consequently, the relation $A^{T}A=I$ must also hold true. Keep this in your back pocket; it is the secret weapon that will unlock the entire problem.

Respecting the Order

Now, look at the expression we need to evaluate:

In matrix algebra, multiplication is non-commutative, meaning $XY$ is rarely equal to $YX$. When we expand $(A+A^T{)}^2$, we must write it as $(A+A^T)(A+A^T)$.

Distributing this carefully, we obtain:

Since $A{A}^T=I$ and $A^{T}A=I$, the expression simplifies to:

A Satisfying Cancellation

Next, we turn our attention to the second term, $(A-A^T{)}^2$. Following the same careful expansion, we get:

Substituting our identity matrices, this simplifies to:

Now, consider the sum of these two expansions:

The $+2I$ and $-2I$ cancel out perfectly. We are left with $2A^2+2(A^T{)}^2$, or $2[A^2+(A^T{)}^2]$.

Bringing it Home

Our original expression was $\frac{1}{2}A$ multiplied by this sum. Substituting our result, we have:

Distributing the $A$, we get $A^3+A(A^T{)}^2$. We can rewrite the second term as $(A{A}^T)A^T$.

Since $A{A}^T=I$, this becomes $I{A}^T$, which is simply $A^T$. Our final result is:

You have navigated the traps of non-commutativity and emerged victorious. Remember, in JEE Advanced, complexity is often just a veil for a beautiful, underlying simplicity.