adj(kA)=k^n(n-1)det(A)^(n-1) HOW TO PROVE ????

Question

A is non singular square matrix of order n and k is a non zero scaler

Answer 1

We can use the adjugate matrix relation: A adj(A) = I_n det(A)

 

Then we have:

 

kA adj(kA) = I_n det(kA)

 

kA adj(kA) = kⁿ I_n det(A)

 

adj(kA) = k^n-1 A^-1 det(A)

 

det(adj(kA)) = det(k^n-1 A^-1 det(A))

 

det(adj(kA)) = k^n(n-1) [det(A)]ⁿ det(A^-1)

 

det(adj(kA)) = k^n(n-1) [det(A)]ⁿ / det(A)

 

det(adj(kA)) = k^n(n-1) [det(A)]^n-1