Answer to Question #118586 in Linear Algebra for Nii Laryea
Given that M =
2 −1
−3 4
and that M2 − 6M + kI = 0, find k
(Under matrix )
1
2020-05-28T19:40:14-0400
M=(2−3−14)
M2=(2−3−14)(2−3−14)=
=(2(2)−1(−3)−3(2)+4(−3)2(−1)−1(4)−3(−1)+4(4))=
=(7−18−619)
−6M=−6(2−3−14)=
=(−6(2)−6(−3)−6(−1)−6(4))=(−12186−24)
M2−6M=(7−18−619)+(−12186−24)=
=(−500−5)=−5(1001)=−5I
M2−6M+kI=−5I+kI=0
k=5
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