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Let A= [[5,5 alpha, alpha],[0 ,alpha,5 a...

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

A

`5^(2)`

B

1

C

`1//5`

D

5

Text Solution

Verified by Experts

The correct Answer is:
C
`A = [[5, 5alpha ,alpha ],[0,alpha,alpha5],[0,0,5]]rArr abs(AcdotA)= abs(A) abs(A)= (25alpha)^(2) = 25`
`rArr alpha^(2) = 1/25`
`rArr alpha = pm1/5`
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