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Question: Find eigen values and eigen vectors of following matrices.
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1. $A=$ $\left[ {\begin{array}{cc} 3 & -1 & 1\\ -1 & 5 & -1 \\ 1 & -1 & 3 \\ \end{array} } \right]$

1. $A=$ $\left[ {\begin{array}{cc} 6 & -2 & 2\\ -2 & 3 & -1 \\ 2 & -1 & 3 \\ \end{array} } \right]$

Subject: Applied Mathematics 4

Topic: Matrices

Difficulty: Medium

m4m(64) • 82 views
 modified 12 weeks ago  • written 12 weeks ago by
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1. Characteristic equation of A is $\mid A-\lambda I \mid = 0$

$\lambda^3-(3+5+3) \lambda^2 + [(15+15+9)-(1+1+1)]\lambda - \mid A \mid =0$

$\lambda^3-11\lambda^2+36\lambda-36=0$

$\lambda=6,3,2$ are eigen values of A

Now to find eigen vector $X\neq0$ such that $[A-\lambda I]=0$

$\left[ {\begin{array}{cc} 3-\lambda & -1 & 1 \\ -1 & 5-\lambda & -1 \\ 1 & -1 & 3-\lambda \end{array} } \right]$ $\left[ {\begin{array}{cc} x \\ y \\ z \end{array} } \right]$= $\left[ {\begin{array}{cc} 0 \\ 0 \\ 0 \end{array} } \right]$ (1)

Put $\lambda=6$ in (1)

$\left[ {\begin{array}{cc} -3 & -1 & 1 \\ -1 & -1 & -1 \\ 1 & -1 & 3 \end{array} } \right]$ $\left[ {\begin{array}{cc} x \\ y \\ z \end{array} } \right]$= $\left[ {\begin{array}{cc} 0 \\ 0 \\ 0 \end{array} } \right]$

Choose any two distinct rows. Let us choose 2nd and 3rd row, we get,

$-x-y-z=0$

$x-y-3z=0$

By Cramer's rule,

$\frac{x}{\mid {\begin{array}{cc} -1 & 1 \\ -1 & 3 \\ \end{array} } \mid}$= $\frac{-y}{\mid {\begin{array}{cc} -1 & 1 \\ 1 & 3 \\ \end{array} } \mid}$= $\frac{z}{\mid {\begin{array}{cc} -1 & -1 \\ 1 & -1 \\ \end{array} } \mid}$

$\frac{x}{2}=\frac{-y}{4}=\frac{z}{2}$

$X_1=$ $\left[ {\begin{array}{cc} 2 \\ -4 \\ 2 \end{array} } \right]$ or $\left[ {\begin{array}{cc} 1 \\ -2 \\ 1 \end{array} } \right]$

Put $\lambda=3$ in (1) we get,

$\left[ {\begin{array}{cc} 0 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 0 \end{array} } \right]$$\left[ {\begin{array}{cc} x \ y \ z \end{array} } \right]= \left[ {\begin{array}{cc} 0 \\ 0 \\ 0 \end{array} } \right] From 2nd and 3rd row as they are distinct -x+2y-z=0 x-y+0=0 By Cramer's rule, \frac{x}{\mid {\begin{array}{cc} 2 & -1 \\ -1 & 0 \\ \end{array} } \mid}= \frac{x}{\mid {\begin{array}{cc} -1 & 1 \\ 1 & 0 \\ \end{array} } \mid}= \frac{x}{\mid {\begin{array}{cc} -1 & 2 \\ 1 & -1 \\ \end{array} } \mid} \frac{x}{-1}=\frac{-y}{1}=\frac{z}{-1} X_2= \left[ {\begin{array}{cc} -1 \\ -1 \\ -1 \end{array} } \right] or \left[ {\begin{array}{cc} 1 \\ 1 \\ 1 \end{array} } \right] Put \lambda=2 in (1), we get \left[ {\begin{array}{cc} 1 & -1 & 1\\ -1 & 3 & -1 \\ 1 & -1 & 1 \end{array} } \right] \left[ {\begin{array}{cc} x\\ y \\ z \end{array} } \right] = \left[ {\begin{array}{cc} 0\\ 0 \\ 0 \end{array} } \right] -x+3y-z=0 x-y+z=0 \frac{x}{\mid {\begin{array}{cc} 3 & -1 \\ -1 & 1\\ \end{array} } \mid} = \frac{-y}{\mid {\begin{array}{cc} -1 & 1 \\ 1 & 1 \\ \end{array} } \mid}= \frac{z}{\mid {\begin{array}{cc} -1 & 3 \\ 1 & -1 \\ \end{array} } \mid} \frac{x}{2}=\frac{-y}{0}=\frac{z}{-2} X_3= \left[ {\begin{array}{cc} 2\\ 0 \\ -2 \end{array} } \right] or \left[ {\begin{array}{cc} 1\\ 0 \\ -1 \end{array} } \right] Therefore,\lambda=6,3,2 are eigen values of A and corresponding eigen vectors are X_1= \left[ {\begin{array}{cc} 1\\ -2 \\ 1 \end{array} } \right], X_2= \left[ {\begin{array}{cc} 1\\ 1\\ 1 \end{array} } \right] , X_3= \left[ {\begin{array}{cc} 2\\ 0 \\ -2 \end{array} } \right] 1. Characteristic equation of A is \mid A-\lambda I \mid = 0 \lambda^3-(6+3+3) \lambda^2 + [(18+18+9)-(4+4+1)]\lambda - \mid A \mid =0 \lambda^3-12\lambda^2+36\lambda-2=0 \lambda=8,2,2 are eigen values of A Now to find eigen vector X\neq0 such that [A-\lambda I]=0 \left[ {\begin{array}{cc} 6-\lambda & -2 & 2 \\ -2 & 3-\lambda & -1 \\ 2 & -1 & 3-\lambda \end{array} } \right] \left[ {\begin{array}{cc} x \\ y \\ z \end{array} } \right]= \left[ {\begin{array}{cc} 0 \\ 0 \\ 0 \end{array} } \right] (1) Put \lambda=8 in (1) \left[ {\begin{array}{cc} -2 & -2 & 2 \\ -2 & -5 & -1 \\ 2 & -1 & -5 \end{array} } \right] \left[ {\begin{array}{cc} x \\ y \\ z \end{array} } \right]= \left[ {\begin{array}{cc} 0 \\ 0 \\ 0 \end{array} } \right] Choose any two distinct rows. Let us choose 2nd and 3rd row, we get, -2x-5y-z=0 2x-y-5z=0 By Cramer's rule, \frac{x}{\mid {\begin{array}{cc} -5 & -1 \\ -1 & -5 \\ \end{array} } \mid}= \frac{-y}{\mid {\begin{array}{cc} -2 & -1 \\ 2 & -5 \\ \end{array} } \mid}= \frac{z}{\mid {\begin{array}{cc} 2 & -5 \\ 2 & -1 \\ \end{array} } \mid} \frac{x}{24}=\frac{-y}{12}=\frac{z}{12} X_1= \left[ {\begin{array}{cc} 24 \\ -12 \\ 12 \end{array} } \right] or \left[ {\begin{array}{cc} 2 \\ -1 \\ 1 \end{array} } \right] Put \lambda=2 in (1) we get, \left[ {\begin{array}{cc} 4 & -2 & 2 \\ -2 & 1 & -1 \\ 2 & -1 & 1 \end{array} } \right]$$ \left[ {\begin{array}{cc} x \ y \ z \end{array} } \right]$=$\left[ {\begin{array}{cc} 0 \\ 0 \\ 0 \end{array} } \right]$

Here, all rows are the same if we reduce it. Therefore, we will get only one equation.

$2x-y+z=0$

$x=y-2x$

Put $x=0$ and $x=1$ $y=1$ $z=1$ and $y=0$ and $z=-2$

$X_2=$ $\left[ {\begin{array}{cc} 0 \\ 1 \\ 1 \end{array} } \right]$ and $X_3=$ $\left[ {\begin{array}{cc} 1\\ 0 \\ -2 \end{array} } \right]$

Therefore, $\lambda=8,2,2$ are eigen values of A and corresponding eigen vectors are $X_1$= $\left[ {\begin{array}{cc} 2 \\ -1 \\ 1 \end{array} } \right]$, $X_2=$ $\left[ {\begin{array}{cc} 0 \\ 1 \\ 1 \end{array} } \right]$ and $X_3=$ $\left[ {\begin{array}{cc} 1\\ 0 \\ -2 \end{array} } \right]$