Solution: λ1=1, λ2=−4, λ3=6, λ4=−6det(λI−A)=λn+c1λn−1+c2λn−2+....+cnHere n=4,∴ det(λI−A)=λ4+c1λ3+c2λ2+c3λ+c4Forλ=1,det(I−A)=(1)4+c1(1)3+c2(1)2+c3(1)+c4 =1+c1+c2+c3+c4.........................................(I)Forλ=−4,det(−4I−A)=(−4)4+c1(−4)3+c2(−4)2+c3(−4)+c4 =256−64c1+16c2−4c3+c4.........................................(II)Forλ=6,det(6I−A)=(6)4+c1(6)3+c2(6)2+c3(6)+c4 =1296+216c1+36c2+6c3+c4.........................................(III)Forλ=−6,det(−6I−A)=(−6)4+c1(−6)3+c2(−6)2+c3(−6)+c4 =1296−216c1+36c2−6c3+c4.........................................(IV)Re−write the equation (I),(II),(III) and (IV),c4+c1+c2+c3=−1..........................................................(I)∗c4−64c1+16c2−4c3=−256..............................................(II)∗c4+216c1+36c2+6c3=−1296...........................................(III)∗c4−216c1+36c2−6c3=−1296...........................................(IV)∗Solving (I)∗,(II)∗,(III)∗ and (IV)∗,we get ,c4=144,c1=3,c2=−40 and c3=−108∴det(λI−A) becomes,det(λI−A)=λ4+3λ3−40λ2−108λ+144For λ=0det(−A)=c4=144 or det(A)=(−1)4.144=144∴det(A)=144Note:Their are several methods to solve system of linear equations.
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