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Question

If A=[α22α] and |A3|=125, then α is equal to


Solution

A=[α22α][α22α]=[α2+42α+2α2α+2α4+α2]
=[α2+44α4αα2+4]
A2A=[α2+44α4αα2+4][α22α]
=[(α2+4)α+8α2α2+8+4α24α2+(α2+4)28α+α(α2+4)]
=[α3+12α6α2+86α2+8α3+12α]
Now,R1R1+R2[α3+6α2+12α+8α3+6α2+12α+86α2+8α3+12α](α+2)3[116α2+8α3+12α]
(α+2)3[α3+12α6α28](α+2)3.(α2)3Now,|A3|=125125=53=(3+2)3(31)3So,α=±3.

 

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