Animated Solution for Mathematics - Matrices and Determinants: If A, B, and $(adj(A^{-1})+adj(B^{-1}))$ are non-singular matrices of same order, then the inverse of $A(adj(A^{-1})+adj(B^{-1}){)}^{-1}B$, is equal to

Unraveling the Complexity

Welcome, fellow traveler on the path to JEE Advanced mastery. Today, we are not just solving a matrix problem; we are choreographing a dance.

When you look at an expression like $X=A(adj(A^{-1})+adj(B^{-1}){)}^{-1}B$, it is natural to feel a momentary surge of anxiety. It looks like a tangled knot of operators, inverses, and adjoints.

But I want you to take a deep breath. In the world of linear algebra, complexity is often just a mask for elegance. Our goal is to peel back these layers, one by one, until the truth reveals itself.

The Reversal Law

Imagine you are standing before a locked door, and to open it, you must perform a sequence of actions in reverse order. Matrix inversion is exactly like that. We are given the matrix $X=A(adj(A^{-1})+adj(B^{-1}){)}^{-1}B$. We need to find $X^{-1}$.

Many students instinctively try to distribute the inverse, but we must respect the Reversal Law of Inverses. If we have a product of matrices $PQR$, the inverse is $(PQR{)}^{-1}=R^{-1}{Q}^{-1}{P}^{-1}$.

Here, our '$P$' is $A$, our '$Q$' is the entire bracket $(adj(A^{-1})+adj(B^{-1}){)}^{-1}$, and our '$R$' is $B$. Applying the law, we get:

Notice how the $B$ jumped to the front as $B^{-1}$, and the $A$ moved to the back as $A^{-1}$. The middle term, however, has become an inverse of an inverse. This is where the beauty begins.

The Vanishing Act

Look closely at that middle term: $[(adj(A^{-1})+adj(B^{-1}){)}^{-1}{]}^{-1}$. In the realm of matrices, the inverse of an inverse is simply the original matrix.

Just as the negative of a negative is a positive, $(M^{-1}{)}^{-1}=M$. This property is our greatest ally. It allows us to strip away the outer inverse, leaving us with a much cleaner expression:

Suddenly, the 'scary' part of the problem has vanished. We are left with a simple product of matrices. But we still have these adjoint terms. How do we handle them?

The Adjoint Transformation

This is the heart of the problem. We need to bridge the gap between the adjoint and the original matrices. Recall the fundamental identity: $adj(M)=|M|M^{-1}$.

Let us apply it to $M=A^{-1}$:

We know two things: first, the determinant of an inverse is the reciprocal of the determinant, so $|A^{-1}|=\frac{1}{|A|}$. Second, the inverse of an inverse is the original matrix, so $(A^{-1}{)}^{-1}=A$. Substituting these, we get:

By the exact same logic, $adj(B^{-1})=\frac{1}{|B|}B$. We have successfully converted abstract adjoints into simple scalar multiples of our original matrices. Let us substitute these back into our equation for $X^{-1}$:

The Final Assembly

Now, we distribute. We take $B^{-1}$ from the left and $A^{-1}$ from the right, applying them to both terms inside the bracket. Remember, matrix multiplication is not commutative, so we must be precise with our placement:

Look at the products formed! In the first term, we have $A{A}^{-1}$, which is the identity matrix $I$. In the second term, we have $B^{-1}B$, which is also $I$.

Since any matrix multiplied by the identity remains unchanged, our expression collapses beautifully:

We are almost there. To match the options provided, we need a common denominator. Let's combine these terms using $|A||B|$ as the common denominator:

Finally, recognize that $|B|B^{-1}=adj(B)$ and $|A|A^{-1}=adj(A)$. Also, $|A||B|=|AB|$. Substituting these, we arrive at the elegant conclusion:

This matches our target option perfectly. You see? What started as an intimidating wall of symbols was merely a series of logical steps waiting to be taken. Trust the properties, respect the order of operations, and the math will always guide you home.