Let A and B be non-singular square matrices of order 3. Consider the following statements, each claimed to hold for all such A and B:
(A) \(\mathrm{adj}(A^{T}) = (\mathrm{adj}\,A)^{T}\)
(B) \(\mathrm{adj}(A^{-1}) = (\mathrm{adj}\,A)^{-1}\)
(C) \(\mathrm{adj}(A + B) = \mathrm{adj}\,A + \mathrm{adj}\,B\)
(D) \(|\mathrm{adj}\,A| = |A|\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (B) only
For a non-singular matrix, \(\mathrm{adj}\,A = |A|\,A^{-1}\).
\(\mathrm{adj}(A^{T}) = |A^{T}|\,(A^{T})^{-1} = |A|\,(A^{-1})^{T} = (\mathrm{adj}\,A)^{T}\)
\(\displaystyle \mathrm{adj}(A^{-1}) = |A^{-1}|\,A = \frac{A}{|A|}\)
\(\displaystyle (\mathrm{adj}\,A)^{-1} = \left(|A|\,A^{-1}\right)^{-1} = \frac{A}{|A|}\)
For order n, \(|\mathrm{adj}\,A| = |A|^{\,n - 1}\), so here \(|\mathrm{adj}\,A| = |A|^{2}\).
So the adjoint goes with the transpose and with the inverse, it does not split over a sum, and \(|\mathrm{adj}\,A| = |A|^{2}\).
Checking the statements
(A) \(\mathrm{adj}(A^{T}) = (\mathrm{adj}\,A)^{T}\) — Correct. Both equal \(|A|\,(A^{-1})^{T}\).
(B) \(\mathrm{adj}(A^{-1}) = (\mathrm{adj}\,A)^{-1}\) — Correct. Both equal \(\displaystyle\frac{A}{|A|}\).
(C) \(\mathrm{adj}(A + B) = \mathrm{adj}\,A + \mathrm{adj}\,B\) — Incorrect. For A = B = I: \(\mathrm{adj}(2I) = 4I\), but \(\mathrm{adj}\,I + \mathrm{adj}\,I = 2I\).
(D) \(|\mathrm{adj}\,A| = |A|\) — Incorrect. For order 3, \(|\mathrm{adj}\,A| = |A|^{2}\); for A = 2I it is 64, not 8.