Statement-1: If A is a non-singular square matrix of order n, then $|adj\, A|=|A|^{n-1}$
Statement-2: For any square matrix A of order n, $A (adj\, A) =|A| I$ and $|kA|=k|A|$
Answer & explanation
Correct answer: option 3
We know that
$A(adj\, A) =|A| I$
$∴|A (adj\, A)| = ||A|I|=| A|^n |A|$ $[∵ |kA|=k^n |A|]$
$⇒|A||adj\, A|=|A|^n$
$⇒|adj\, A|=|A|^{n-1}$
So, statement-1 is true. But, statement-2 is false. Because, $|kA|=k^n |A|$.