P=⎣⎡−1004205−38⎦⎤
p(λ)=det(P−λI)=∣∣−1−λ0042−λ05−38−λ∣∣=(−1−λ)(2−λ)(8−λ)
p(λ)=−x3+9x2−6x−16
−P3+9P2−6P−16I=0
P(−P2+9P−6I)=16I
16P−1=−P2+9P−6I=−⎣⎡10044023−3064⎦⎤+9⎣⎡−1004205−38⎦⎤−⎣⎡600060006⎦⎤
16P−1=⎣⎡−160032802232⎦⎤
P−1=161⎣⎡−160032802232⎦⎤
Let ⎣⎡x1x2x3⎦⎤=a⋅⎣⎡−100⎦⎤+b⋅⎣⎡420⎦⎤+c⋅⎣⎡5−38⎦⎤=P⎣⎡abc⎦⎤ .
Then ⎣⎡abc⎦⎤=P−1⎣⎡x1x2x3⎦⎤=⎣⎡−x1+2x2+811x321x2+163x381x3⎦⎤ .
So, ⎣⎡x1x2x3⎦⎤=(−x1+2x2+811x3)⋅⎣⎡−100⎦⎤+(21x2+163x3)⋅⎣⎡420⎦⎤+81x3⋅⎣⎡5−38⎦⎤ .
Answers:
P−1=161⎣⎡−160032802232⎦⎤
⎣⎡x1x2x3⎦⎤=(−x1+2x2+811x3)⋅⎣⎡−100⎦⎤+(21x2+163x3)⋅⎣⎡420⎦⎤+81x3⋅⎣⎡5−38⎦⎤
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