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Question

Let P=-302056901401121201614 and A=27ω2-1-ω10-ω-ω+1

where 𝛚=(-1+i3)2, and I3 be the identity matrix of order 3.

If the determinant of the matrix P1AP-12 is αω2, then the value of α is equal to


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Solution

The value of α=36

Finding the value of α

Simplifying the required expression value using given parameters, we have

P1AP-12=αω2=(P-1AP-1)(P-1AP-1)=P-1APP-1AP-2P-1AP+I=P-1A2P-2P-1AP+P-1P=P-1(A2-2A+I)P=P-1(A-I)2P=P-1(A-I)2P=(A-I)2=27ω2-1-ω10-ω-ω+1-1000100012=17ω2-1-ω-110-ω-ω2=(1(ω2+ω+ω)-7(-ω)+ω2(ω)=(ω2+2ω-7ω+1)2ω3=1=(ω25ω+1)2(1+ω+ω2)=0(1+ω2)=-ω2αω2=(-6ω)2=36ω2ɑ=36

Hence, The value of α=36


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