Question

If A.x=\lambda x,where A=\begin{vmatrix}2&2&-2\\1&3&1\\1&2&2\end{vmatrix},determine the eigen values of the matrix A, and an eigen vector corresponding to each eigen value. If \lambda=2,what is b

Answer 1

Answer on Question # 40994 – Math-Linear Algebra

Question:

If $Ax = \lambda x$, where $A = \begin{pmatrix} 2 & 2 & -2 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{pmatrix}$, determine the eigen values of the matrix $A$, and an eigen vector corresponding to each eigen value. If $\lambda = 2$, what is $b$.

Solution:

Step 1:

The given matrix is $A = \begin{pmatrix} 2 & 2 & -2 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{pmatrix}$.

Step 2:

The characteristic equation is $|A - \lambda I| = 0$

$$\text{i.e. } \left| \begin{array}{ccc} 2 - \lambda & 2 & -2 \\ 1 & 3 - \lambda & 1 \\ 1 & 2 & 2 - \lambda \end{array} \right| = -\lambda^3 + 7 * \lambda^2 - 14 * \lambda + 8 = 0.$$

Solving the above determinant we get

$$\lambda_1 = 1, \lambda_2 = 2 \text{ and } \lambda_3 = 4 \text{ are eigen values}.$$

Step 3:

For every eigen value let's find eigen vector.

To find eigen vectors take $(A - \lambda_i I) X = 0$, i.e. $\begin{pmatrix} 2 - \lambda_i & 2 & -2 \\ 1 & 3 - \lambda_i & 1 \\ 1 & 2 & 2 - \lambda_i \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$.

Firstly take $\lambda_1 = 1$.

Using Gaussian elimination we get

$$\left( \begin{array}{ccc} 1 & 2 & -2 \\ 1 & 2 & 1 \\ 1 & 2 & 1 \end{array} \right) \sim \left( \begin{array}{ccc} 1 & 2 & -2 \\ 0 & 0 & 3 \\ 0 & 0 & 3 \end{array} \right) \sim \left( \begin{array}{ccc} 1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{array} \right)$$

So, corresponding eigen vector is $b_1 = \begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix}$.

Now, let's take $\lambda_2 = 2$

Using Gaussian elimination we get

$$\left( \begin{array}{ccc} 0 & 2 & -2 \\ 1 & 1 & 1 \\ 1 & 2 & 0 \end{array} \right) \sim \left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 2 & -2 \\ 1 & 2 & 0 \end{array} \right) \sim \left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 2 & -2 \\ 0 & 1 & -1 \end{array} \right) \sim \left( \begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{array} \right)$$

So, corresponding eigen vector is $b_2 = \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}$.

Now, let's take $\lambda_3 = 4$

Using Gaussian elimination we get

$$\left( \begin{array}{cccc} -2 & 2 & -2 & 0 \\ 1 & -1 & 1 & 0 \\ 1 & 2 & -2 & 0 \end{array} \right) \sim \left( \begin{array}{cccc} 1 & -1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 3 & -3 & 0 \end{array} \right) \sim \left( \begin{array}{cccc} 1 & -1 & 1 & 0 \\ 0 & 3 & -3 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \sim \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right)$$

So, corresponding eigen vector is $b_{3} = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$ .

**Answer:**

Eigen values $\lambda_1 = 1, \lambda_2 = 2, \lambda_3 = 4$ .

And corresponding eigen vectors $b_{1} = \begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix}$ , $b_{2} = \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}$ , $b_{3} = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$ .

For eigen value $\lambda = 2$ the corresponding eigen vector is $b = \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}$ .