Question

find eigen values and eigen vectors 1 2 3 2 4 6 3 6 9​

Answer 1

Answer: One eigen value is 0 and corresponding eigen vectors are $\left[\begin{array}{ccc}-2\\1\\0\end{array}\right] and \left[\begin{array}{ccc}-3\\0\\1\end{array}\right]$  and other eigen value is 14 and corresponding eigen vector is $\left[\begin{array}{ccc}1\\2\\3\end{array}\right]$

Step-by-step explanation:

Let given matrix be named A

$A = \left[\begin{array}{ccc}1&2&3\\2&4&6\\3&6&9\end{array}\right]$

Let λ be eigen value of A

∴ det ( A -  λI ) = 0

$\lambda I = \left[\begin{array}{ccc}\lambda&0&0\\0&\lambda&0\\0&0&\lambda\end{array}\right]$

$A- \lambda I = \left[\begin{array}{ccc}1-\lambda&2&3\\2&4-\lambda&6\\3&6&9-\lambda\end{array}\right]$

det ( A -  λI )  = (1- λ)[(4- λ)(9- λ) - 36] -2[2(9- λ) - 18] + 3[12 - 3(4 - λ)]

 0  =  (1- λ)[ λ² -13λ ]  -2[-2λ] + 3[3λ]

 0   =  (1- λ)[ λ² -13λ ]  + 4λ + 9λ

 0   =  (1- λ)[ λ² -13λ ] + 13λ

 0   =  λ(1- λ)[ λ -13 ]  + 13λ

 0  =   λ [ (1 - λ)(λ -13) + 13 ]

 0  =    λ [ -λ² + 14λ -13 + 13 ]

 0  =    λ [ -λ² + 14λ ]

 0  =    λ² [ -λ + 14 ]

λ = 0 , 0, 14

Finding eigen vectors corresponding to λ = 0 :

Av = λv

Av = 0

$\left[\begin{array}{ccc}1&2&3\\2&4&6\\3&6&9\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = 0$

$x + 2y + 3z = 0$        

$2x + 4y + 6z = 0$

$x + 2y + 3z = 0$        

$3x + 6y + 9z = 0$

$x + 2y + 3z = 0$        

We get only 1 unique equation. Hence, two parameters out of x y and z are independent.

Choose y and z as independent.

y = r and z =s

x + 2y + 3z = 0

x + 2r + 3s = 0

x  =  -2r - 3s

$v = \left[\begin{array}{ccc}-2r-3s\\r\\s\end{array}\right]$

$v = \left[\begin{array}{ccc}-2r\\r\\0\end{array}\right] + \left[\begin{array}{ccc}-3s\\0\\s\end{array}\right]$

$v = r\left[\begin{array}{ccc}-2\\1\\0\end{array}\right] + s\left[\begin{array}{ccc}-3\\0\\1\end{array}\right]$

Eigen vectors corresponding to λ = 0  are :

$\left[\begin{array}{ccc}-2\\1\\0\end{array}\right] and \left[\begin{array}{ccc}-3\\0\\1\end{array}\right]$

Finding eigen vectors corresponding to λ = 14 :

Av = λv

Av = 14v

$\left[\begin{array}{ccc}1&2&3\\2&4&6\\3&6&9\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = 14\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$

$x + 2y + 3z = 14x$

$-13x + 2y + 3z = 0$       eq(1)

$2x + 4y + 6z = 14y$

$2x - 10y + 6z = 0$         eq(2)

$3x + 6y + 9z = 14z$

$3x + 6y -5z = 0$        eq(3)

eq(1) can be written as  -(3eq(3) + 2eq(2))

eq(1) =  -(3eq(3) + 2eq(2))

We get 2 unique equations. Hence, one parameter is independent.

Choose z as independent.

z = s

2x - 10y + 6z = 0

2x - 10y + 6s = 0

x  - 5y  = - 3s                 eq(i)

3x + 6y - 5z = 0

3x + 6y = 5s

x + 2y  =  5s/3             eq(ii)

eq(ii) - eq(i)

7y  = 14s/3

y  = 2s/3

x = 5y - 3s

x  = 10s/3 - 3s

x = s/3

$v = \left[\begin{array}{ccc}\frac{s}{3} \\\frac{2s}{3} \\s\end{array}\right]$

$v = \frac{s}{3} \left[\begin{array}{ccc}1\\2\\3\end{array}\right]$

Eigen vectors corresponding to λ = 14 is :

$\left[\begin{array}{ccc}1\\2\\3\end{array}\right]$

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