If \( A \) is a non-singular matrix of order \( n \) , then \( adj(adj\ A) \) is equal to :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( |A|^{n-2}A \)
For any square matrix B, \( B\,(\mathrm{adj}\,B)=|B|\,I \). Taking \( B=\mathrm{adj}\,A \) and using \( |\mathrm{adj}\,A|=|A|^{n-1} \):
\( \displaystyle (\mathrm{adj}\,A)\,\mathrm{adj}(\mathrm{adj}\,A)=|A|^{n-1}\,I \)
Multiplying both sides on the left by A, and using \( A\,(\mathrm{adj}\,A)=|A|\,I \):
\( \displaystyle |A|\,\mathrm{adj}(\mathrm{adj}\,A)=|A|^{n-1}\,A \)
Since A is non-singular, \( |A|\ne 0 \), so dividing by |A|:
\( \displaystyle \mathrm{adj}(\mathrm{adj}\,A)=|A|^{n-2}\,A \)
Why the other options are incorrect
Option (1) \( |A|^{n-1} \) — Incorrect. This is \( |\mathrm{adj}\,A| \), a number, not a matrix.
Option (3) \( |A|^{n-1}A \) — Incorrect. The power of |A| is n − 2, not n − 1.
Option (4) \( |A|^{(n-1)^2} \) — Incorrect. This is the determinant \( |\mathrm{adj}(\mathrm{adj}\,A)| \), a number, not the matrix itself.