Matrices – Inverse of Matric – 23 January 2025 (Shift 1)

Solution:

Let $E = [A(adj(A^{-1})+adj(B^{-1}))^{-1}B]^{-1}$.
Using reversal law $(XYZ)^{-1} = Z^{-1}Y^{-1}X^{-1}$:
$$E = B^{-1} \left[ (adj(A^{-1}) + adj(B^{-1}))^{-1} \right]^{-1} A^{-1}$$
$$E = B^{-1} (adj(A^{-1}) + adj(B^{-1})) A^{-1}$$
Using property $adj(P^{-1}) = |P^{-1}|P = \frac{P}{|P|}$:
$$adj(A^{-1}) = \frac{A}{|A|} \quad \text{and} \quad adj(B^{-1}) = \frac{B}{|B|}$$
Substitute these values:
$$E = B^{-1} \left( \frac{A}{|A|} + \frac{B}{|B|} \right) A^{-1}$$
$$E = \frac{B^{-1}AA^{-1}}{|A|} + \frac{B^{-1}BA^{-1}}{|B|} = \frac{B^{-1}}{|A|} + \frac{A^{-1}}{|B|}$$
Using formula $X^{-1} = \frac{adj(X)}{|X|}$:
$$E = \frac{adj(B)}{|B||A|} + \frac{adj(A)}{|A||B|} = \frac{1}{|AB|}(adj(B)+adj(A))$$
Ans. (3)