If A is square matrix of order 3 and A.(Adj.(A)) = 10I, then the value of \(\frac{1}{25}\)|Adj.(A)is

4

Consider the following identity $A(adjA)=∣A∣I$. Thus comparing it with the above equation gives us $∣A∣=10$

∣adjA∣=∣A∣ⁿ⁻¹ where n is the order of the square matrix.

Here 'n' is 3

$∴ \frac{1}{25}∣adjA∣=\frac{1}{25}∣A∣^{3−1}⇒\frac{1}{25}∣A∣^2=10^2=\frac{1}{25}×100=4$

So option 4 is correct.