Given `A=[[2,0,-alpha],[5,alpha,0],[0,alpha,3]]` `For a in R-{a, b}`, `A^(-1)` exists and `A^(-1)=A^2-5bA + cI`, when `alpha = 1` . The value of `a +5b +c` is:

Let A= [[5,5 alpha, alpha],[0 ,alpha,5 alpha ],[0,0,5]] |A^(2)|=25, then |alpha|=

Let A=[[5,5 alpha,alpha],[0,alpha,5 alpha],[0,0,5]] .If |A^(2)|=25 ,then | alpha| equals

Let A=[[5,5 alpha,alpha0,alpha,5 alpha0,0,5]].IfA^(2)=25, then alpha

Let A=[(5,5 alpha, alpha),(0, alpha, 5 alpha),(0,0,5)] If |A|^(2)=25 then |alpha| equals

Let A= [[5,5alpha,alpha],[0,alpha,5alpha],[0,0,5]] . If |A^2| = 25, then alpha equals to:

Let A= [[5,5alpha,alpha],[0,alpha,5alpha],[0,0,5]] . If |A^2| = 25, then alpha equals to:

Select the correct option from the given alternatives. Let A = [[5,5alpha,alpha],[0,alpha,5alpha],[0,0,5]] , if abs(A^2) = 25 , then abs(alpha) = i] frac{1}{5} ii] 5 iii] 5^2 iv] 1