If A, B and (adj(A⁻¹) + adj(B⁻¹)) Are Non-Singular Matrices of the Same Order, Then the Inverse of A(adj(A⁻¹) + adj(B⁻¹))⁻¹B Is Equal To

JEE Main 2025 — 23 January, Morning Shift. Previous Year Question.

Problem

If $A,B$, and $\left(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})\right)$ are non-singular matrices of the same order, then the inverse of $A\left(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})\right)^{-1}B$, is equal to:

Key insight. The reversal rule for inverses, $(PQR)^{-1}=R^{-1}Q^{-1}P^{-1}$, immediately turns “inverse of a product” into a product of inverses in reverse order — and once that’s applied, the identity $M\cdot\operatorname{adj}(M)=\det(M)\cdot I$ makes the adjoint terms collapse almost entirely, leaving only scalar determinants behind.

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Approach

Let $X=\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})$. Apply the reversal rule to find the inverse of $AX^{-1}B$, which introduces $X$ itself (since $(X^{-1})^{-1}=X$). Expand $X$ back into its two adjoint terms, and use $M\cdot\operatorname{adj}(M)=\det(M)I$ (and its variant $\operatorname{adj}(M)\cdot M=\det(M)I$) to simplify each resulting term into a scalar multiple of a single matrix.

Solution

Step 1 — Apply the reversal rule for inverses

Let $X=\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})$. Using $(PQR)^{-1}=R^{-1}Q^{-1}P^{-1}$:

$$\left(AX^{-1}B\right)^{-1} = B^{-1}\left(X^{-1}\right)^{-1}A^{-1} = B^{-1}XA^{-1}$$

Step 2 — Expand X

$$B^{-1}XA^{-1} = B^{-1}\left[\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})\right]A^{-1}$$

$$= B^{-1}\operatorname{adj}(A^{-1})A^{-1} + B^{-1}\operatorname{adj}(B^{-1})A^{-1}$$

Step 3 — Simplify the first term using adj(M)·M = det(M)I

$$\operatorname{adj}(A^{-1})\cdot A^{-1} = \det(A^{-1})\cdot I = \frac{1}{\det A}I$$

So the first term becomes:

$$B^{-1}\cdot\frac{1}{\det A}I = \frac{1}{\det A}B^{-1}$$

Step 4 — Simplify the second term using M·adj(M) = det(M)I

$$B^{-1}\cdot\operatorname{adj}(B^{-1}) = \det(B^{-1})\cdot I = \frac{1}{\det B}I$$

So the second term becomes:

$$\frac{1}{\det B}I\cdot A^{-1} = \frac{1}{\det B}A^{-1}$$

Step 5 — Combine both terms

$$\left(AX^{-1}B\right)^{-1} = \frac{1}{\det A}B^{-1}+\frac{1}{\det B}A^{-1}$$

Step 6 — Rewrite B⁻¹ and A⁻¹ using B⁻¹=adj(B)/det(B) and A⁻¹=adj(A)/det(A)

$$= \frac{1}{\det A}\cdot\frac{\operatorname{adj}(B)}{\det B} + \frac{1}{\det B}\cdot\frac{\operatorname{adj}(A)}{\det A}$$

$$= \frac{\operatorname{adj}(B)+\operatorname{adj}(A)}{\det A\cdot\det B} = \frac{1}{|AB|}\left(\operatorname{adj}(A)+\operatorname{adj}(B)\right)$$

Answer

Common mistakes

• Getting the order wrong when reversing $(AX^{-1}B)^{-1}$. The reversal rule flips both the order and inverts each factor — the correct result is $B^{-1}(X^{-1})^{-1}A^{-1}$, not $A^{-1}(X^{-1})^{-1}B^{-1}$ or any other permutation.
• Mixing up $M\cdot\operatorname{adj}(M)$ and $\operatorname{adj}(M)\cdot M$. Both equal $\det(M)\cdot I$, but which side of the product they appear on in the working matters for correctly matching terms — here, $\operatorname{adj}(A^{-1})A^{-1}$ has the adjoint on the left, while $B^{-1}\operatorname{adj}(B^{-1})$ has it on the right, and both simplify correctly to $\det(\cdot)I$ regardless of order, but tracking this carefully avoids confusion.

Practise next

• If $P,Q$, and $\left(\operatorname{adj}(P^{-1})-\operatorname{adj}(Q^{-1})\right)$ are non-singular matrices of the same order, find the inverse of $P\left(\operatorname{adj}(P^{-1})-\operatorname{adj}(Q^{-1})\right)^{-1}Q$, using the same reversal-and-adjoint-identity approach.