Practicing Success
Let A and B be two non-singular, square matrices of same order, and A. (AB)–1 = B–1.A–1 B. (A + B)–1 = B–1 + A–1 C. adj. A = |A|.A–1 D. det(A–1) = [det A]–1 Choose the correct answer from the options given below |
A and B only B and C only B and D only A, C and D only |
A, C and D only |
A. (AB)–1 = B–1.A–1 B. (A + B)–1 ≠ B–1 + A–1 C. adj. A = |A|1 $adj(A) = |A|\frac{1}{A}⇒|A|.A^{-1}$ D. det(A–1) = [det A]–1 ∴ A, C, & D options are correct. So option 4th is correct. |