If A=[α22α] and |A3|=125, then the value of α is [2004]
(3)
A=[α22α] and |A3|=125⇒|A3|=125
∴ |A|=α2-4
Now, |A|3=125
⇒(α2-4)3=125=53
⇒α2-4=5
⇒α=±3