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Question Number 133206 by rexford last updated on 20/Feb/21

Answered by EDWIN88 last updated on 20/Feb/21

i+j+3k= ((1),(1),(3) ) ; 3i−3j+k =  (((   3)),((−3)),((   1)) )  −4i+5j =  (((−4)),((   5)),((   0)) )  ⇔  ((1),(1),(3) ) x + (((   3)),((−3)),((   1)) ) y + (((−4)),((   5)),((   0)) ) z = λ  ((x),(y),(z) )  ⇔  (((x+3y−4z)),((x−3y+5z)),((3x+y+0z)) ) = λ ((x),(y),(z) )  ⇔  (((1       3      −4)),((1   −3        5)),((3       1        0)) )  ((x),(y),(z) ) = λ  ((x),(y),(z) )

$$\mathrm{i}+\mathrm{j}+\mathrm{3k}=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\\{\mathrm{3}}\end{pmatrix}\:;\:\mathrm{3i}−\mathrm{3j}+\mathrm{k}\:=\:\begin{pmatrix}{\:\:\:\mathrm{3}}\\{−\mathrm{3}}\\{\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$−\mathrm{4i}+\mathrm{5j}\:=\:\begin{pmatrix}{−\mathrm{4}}\\{\:\:\:\mathrm{5}}\\{\:\:\:\mathrm{0}}\end{pmatrix} \\ $$$$\Leftrightarrow\:\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\\{\mathrm{3}}\end{pmatrix}\:\mathrm{x}\:+\begin{pmatrix}{\:\:\:\mathrm{3}}\\{−\mathrm{3}}\\{\:\:\:\mathrm{1}}\end{pmatrix}\:\mathrm{y}\:+\begin{pmatrix}{−\mathrm{4}}\\{\:\:\:\mathrm{5}}\\{\:\:\:\mathrm{0}}\end{pmatrix}\:\mathrm{z}\:=\:\lambda\:\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\\{\mathrm{z}}\end{pmatrix} \\ $$$$\Leftrightarrow\:\begin{pmatrix}{\mathrm{x}+\mathrm{3y}−\mathrm{4z}}\\{\mathrm{x}−\mathrm{3y}+\mathrm{5z}}\\{\mathrm{3x}+\mathrm{y}+\mathrm{0z}}\end{pmatrix}\:=\:\lambda\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\\{\mathrm{z}}\end{pmatrix} \\ $$$$\Leftrightarrow\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{3}\:\:\:\:\:\:−\mathrm{4}}\\{\mathrm{1}\:\:\:−\mathrm{3}\:\:\:\:\:\:\:\:\mathrm{5}}\\{\mathrm{3}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\\{\mathrm{z}}\end{pmatrix}\:=\:\lambda\:\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\\{\mathrm{z}}\end{pmatrix} \\ $$$$ \\ $$

Answered by mr W last updated on 20/Feb/21

 [(1,3,(−4)),(1,(−3),5),(3,1,0) ] [(x),(y),(z) ]=λ [(x),(y),(z) ]   [((1−λ),3,(−4)),(1,(−3−λ),5),(3,1,(0−λ)) ] [(x),(y),(z) ]=0  ∣ [((1−λ),3,(−4)),(1,(−3−λ),5),(3,1,(0−λ)) ]∣=0  (1−λ)(3λ+λ^2 −5)−3(−λ−15)−4(1+9+3λ)=0  (1−λ)(λ^2 +3λ−5)+5−8λ=0  λ^2 (λ+1)=0  ⇒λ=0, −1

$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{3}}&{−\mathrm{4}}\\{\mathrm{1}}&{−\mathrm{3}}&{\mathrm{5}}\\{\mathrm{3}}&{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}\begin{bmatrix}{{x}}\\{{y}}\\{{z}}\end{bmatrix}=\lambda\begin{bmatrix}{{x}}\\{{y}}\\{{z}}\end{bmatrix} \\ $$$$\begin{bmatrix}{\mathrm{1}−\lambda}&{\mathrm{3}}&{−\mathrm{4}}\\{\mathrm{1}}&{−\mathrm{3}−\lambda}&{\mathrm{5}}\\{\mathrm{3}}&{\mathrm{1}}&{\mathrm{0}−\lambda}\end{bmatrix}\begin{bmatrix}{{x}}\\{{y}}\\{{z}}\end{bmatrix}=\mathrm{0} \\ $$$$\mid\begin{bmatrix}{\mathrm{1}−\lambda}&{\mathrm{3}}&{−\mathrm{4}}\\{\mathrm{1}}&{−\mathrm{3}−\lambda}&{\mathrm{5}}\\{\mathrm{3}}&{\mathrm{1}}&{\mathrm{0}−\lambda}\end{bmatrix}\mid=\mathrm{0} \\ $$$$\left(\mathrm{1}−\lambda\right)\left(\mathrm{3}\lambda+\lambda^{\mathrm{2}} −\mathrm{5}\right)−\mathrm{3}\left(−\lambda−\mathrm{15}\right)−\mathrm{4}\left(\mathrm{1}+\mathrm{9}+\mathrm{3}\lambda\right)=\mathrm{0} \\ $$$$\left(\mathrm{1}−\lambda\right)\left(\lambda^{\mathrm{2}} +\mathrm{3}\lambda−\mathrm{5}\right)+\mathrm{5}−\mathrm{8}\lambda=\mathrm{0} \\ $$$$\lambda^{\mathrm{2}} \left(\lambda+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\lambda=\mathrm{0},\:−\mathrm{1} \\ $$

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