Question

Find the Eigen value and Eigen Vector of
A =[ 6 -2 2
-2 3 -1
2 -1 3]​

Answer 1

Answer:

Step-by-step explanation:

Given that : A = $\left[\begin{array}{ccc}6&-2&2\\-2&3&-1\\2&-1&3\end{array}\right]$

We know that |A - λI| = 0

where λ is the eigen value.

and I is the identity matrix

=>    $\left |\left[\begin{array}{ccc}6&-2&2\\-2&3&-1\\2&-1&3\end{array}\right] - \lambda \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\right| = 0$

=>    $\left |\left[\begin{array}{ccc}6&-2&2\\-2&3&-1\\2&-1&3\end{array}\right] -\left[\begin{array}{ccc} \lambda &0&0\\0& \lambda &0\\0&0& \lambda \end{array}\right]\right| = 0$

=>    $\left |\left[\begin{array}{ccc}6- \lambda&-2&2\\-2&3- \lambda&-1\\2&-1&3- \lambda\end{array}\right] \right| = 0$

On solving the determinant, we get,

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

=> (6-λ)[(3-λ)²-1] -(-2)[-6+2λ - (-2)] +2[2 - (6-2λ)] = 0

=> (6-λ)[(9+λ²-6λ-1] -(-2)[-6+2λ +2] +2[2 - 6+2λ)] = 0

=> (6-λ)[(8+λ²-6λ] +2[-4+2λ ] +2[-4+2λ)] = 0

=> 48 +6λ²- 36λ -8λ -λ³ +6λ² -8 +4λ -8 +4λ = 0

=>  λ³  -12λ²+ 36λ -32 = 0

As it is a 3 degree equation, we use hit and trail method to find the values of λ.

2³  -12*2²+ 36*2 -32 = 0

=>  λ =2

Therefore ( λ-2) is a factor of the given equation.

=> ( λ - 2) ( λ² - 10λ + 16) = 0

=> ( λ - 2) ( λ² - 8λ - 2λ + 16) = 0

=> ( λ - 2) ( λ - 8 ) (λ - 2) = 0

=> λ = 2, 8

Therefore, the eigen values are 2 and 8