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

Answered by EDWIN88 last updated on 20/Feb/21

$$\mathrm{i}+\mathrm{j}+3\mathrm{k}=\left(\begin{array}{c}1\\ 1\\ 3\end{array}\right);3\mathrm{i}-3\mathrm{j}+\mathrm{k}=\left(\begin{array}{c}3\\ -3\\ 1\end{array}\right)$$$$-4\mathrm{i}+5\mathrm{j}=\left(\begin{array}{c}-4\\ 5\\ 0\end{array}\right)$$$$\iff \left(\begin{array}{c}1\\ 1\\ 3\end{array}\right)\mathrm{x}+\left(\begin{array}{c}3\\ -3\\ 1\end{array}\right)\mathrm{y}+\left(\begin{array}{c}-4\\ 5\\ 0\end{array}\right)\mathrm{z}=\lambda \left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right)$$$$\iff \left(\begin{array}{c}\mathrm{x}+3\mathrm{y}-4\mathrm{z}\\ \mathrm{x}-3\mathrm{y}+5\mathrm{z}\\ 3\mathrm{x}+\mathrm{y}+0\mathrm{z}\end{array}\right)=\lambda \left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right)$$$$\iff \left(\begin{array}{c}13-4\\ 1-35\\ 310\end{array}\right)\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right)=\lambda \left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right)$$$$$$

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

$$\left[\begin{array}{ccc}1& 3& -4\\ 1& -3& 5\\ 3& 1& 0\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\lambda \left[\begin{array}{c}x\\ y\\ z\end{array}\right]$$$$\left[\begin{array}{ccc}1-\lambda & 3& -4\\ 1& -3-\lambda & 5\\ 3& 1& 0-\lambda \end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=0$$$$\mid \left[\begin{array}{ccc}1-\lambda & 3& -4\\ 1& -3-\lambda & 5\\ 3& 1& 0-\lambda \end{array}\right]\mid =0$$$$(1-\lambda )(3\lambda +{\lambda }^2-5)-3(-\lambda -15)-4(1+9+3\lambda )=0$$$$(1-\lambda )({\lambda }^2+3\lambda -5)+5-8\lambda =0$$$${\lambda }^2(\lambda +1)=0$$$$\Rightarrow \lambda =0,-1$$

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