Difference between revisions of "Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 2 (a)/Solution 1"

From UBC Wiki
Jump to: navigation, search
m
m
 
Line 20: Line 20:
  
  
we obtain <math display="block">\begin{bmatrix}1&0&3\\0&1&-4\\0&0&0 \end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix}=0.</math>  
+
we obtain <math display="block">\begin{bmatrix}1&0&3\\0&1&-4\\0&0&0 \end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix}=\mathbf{0}.</math>  
  
 
Thus <math display="inline">x_1+3x_3=0</math> and <math display="inline">x_2-4x_3=0</math>; in vector form we get <math display="block">\begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix}=\begin{bmatrix}-3x_3\\4x_3\\x_3 \end{bmatrix}=x_3\begin{bmatrix}-3\\4\\1 \end{bmatrix},</math> where <math>x_3 \in \mathbb{R}</math> is free. So an eigenvector corresponding to <math display="inline">\lambda_1=3</math> is <math display="block">\color{blue}\begin{bmatrix}-3\\4\\1 \end{bmatrix}.</math>
 
Thus <math display="inline">x_1+3x_3=0</math> and <math display="inline">x_2-4x_3=0</math>; in vector form we get <math display="block">\begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix}=\begin{bmatrix}-3x_3\\4x_3\\x_3 \end{bmatrix}=x_3\begin{bmatrix}-3\\4\\1 \end{bmatrix},</math> where <math>x_3 \in \mathbb{R}</math> is free. So an eigenvector corresponding to <math display="inline">\lambda_1=3</math> is <math display="block">\color{blue}\begin{bmatrix}-3\\4\\1 \end{bmatrix}.</math>

Latest revision as of 20:37, 8 March 2018

Since an eigenvector of of eigenvalue satisfies , i.e., , all we need to do is solve the following system.

By applying Gaussian elimination,


we obtain

Thus and ; in vector form we get where is free. So an eigenvector corresponding to is