Answer to Question #118586 in Linear Algebra for Nii Laryea

Question #118586
Given that M =
2 −1
−3 4
and that M2 − 6M + kI = 0, find k
(Under matrix )
1
Expert's answer
2020-05-28T19:40:14-0400
M=(2134)M=\begin{pmatrix} 2 & -1 \\ -3 & 4 \end{pmatrix}

M2=(2134)(2134)=M^2=\begin{pmatrix} 2 & -1 \\ -3 & 4 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -3 & 4 \end{pmatrix}=

=(2(2)1(3)2(1)1(4)3(2)+4(3)3(1)+4(4))==\begin{pmatrix} 2(2)-1(-3) & 2(-1)-1(4) \\ -3(2)+4(-3) & -3(-1)+4(4) \end{pmatrix}=

=(761819)=\begin{pmatrix} 7 & -6 \\ -18 & 19 \end{pmatrix}

6M=6(2134)=-6M=-6\begin{pmatrix} 2 & -1 \\ -3 & 4 \end{pmatrix}=

=(6(2)6(1)6(3)6(4))=(1261824)=\begin{pmatrix} -6(2) & -6(-1) \\ -6(-3) & -6(4) \end{pmatrix}=\begin{pmatrix} -12 & 6 \\ 18 & -24 \end{pmatrix}

M26M=(761819)+(1261824)=M^2-6M=\begin{pmatrix} 7 & -6 \\ -18 & 19 \end{pmatrix}+\begin{pmatrix} -12 & 6 \\ 18 & -24 \end{pmatrix}=

=(5005)=5(1001)=5I=\begin{pmatrix} -5 & 0 \\ 0 & -5 \end{pmatrix}=-5\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}=-5I

M26M+kI=5I+kI=0M^2-6M+kI=-5I+kI=0

k=5k=5

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Comments

Assignment Expert
30.05.20, 00:57

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Nii Laryea
29.05.20, 17:09

Thanks

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