Question

If A.x = λx, where A = {{{{211232−212}}}}, determine the eigen values of the matrix A, and an eigen vector corresponding to each eigen value. If λ = 4, what is C?

(A) {2,3,0}

(B) {2,1,0}

(C) {-2,1,1}

(D) {3,2,6}

Answer 1

Let find eigen values. For that we need the next formula:

$$\det (A - \lambda I) = 0$$

Where $I$ is the identity matrix

$$\left| \begin{array}{ccc} 2 - \lambda & 1 & 1 \\ 2 & 3 - \lambda & 2 \\ -2 & 1 & 2 - \lambda \end{array} \right| = (2 - \lambda)^2 (3 - \lambda) - 4 + 2 + 2(3 - \lambda) - 2(2 - \lambda) - 2(2 - \lambda)$$ $$= -\lambda^3 + 7\lambda^2 - 14\lambda + 8$$ $$- \lambda^3 + 7\lambda^2 - 14\lambda + 8 = 0$$ $$\lambda^3 - 7\lambda^2 + 14\lambda - 8 = 0$$ $$(\lambda - 1)(\lambda - 2)(\lambda - 4) = 0$$ $$\left| \begin{array}{l} \lambda - 1 = 0 \\ \lambda - 2 = 0 \Rightarrow \end{array} \right| \begin{array}{l} \lambda_1 = 1 \\ \lambda_2 = 2 \\ \lambda_3 = 4 \end{array}$$

Now we will find the eigen vectors corresponding to each eigen value

1. For $\lambda_1 = 1$

$$\left( \begin{array}{ccc} A - \lambda_1 & \alpha = 0 \\ 1 & 1 & 1 \\ 2 & 2 & 2 \\ -2 & 1 & 1 \end{array} \right) \left| \begin{array}{l} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{array} \right| = 0$$

Having the system of three equations

$$\left\{ \begin{array}{l} \alpha_1 + \alpha_2 + \alpha_3 = 0 \\ 2\alpha_1 + 2\alpha_2 + 2\alpha_3 = 0 \\ -2\alpha_1 + \alpha_2 + \alpha_3 = 0 \end{array} \right.$$

Solve the system by the Gauss method

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

Having the next equations

$$\left\{ \begin{array}{l} \alpha_1 + \alpha_2 + \alpha_3 = 0 \\ \alpha_2 + \alpha_3 = 0 \end{array} \right.$$ $$\left\{ \begin{array}{l} \alpha_1 + \alpha_2 = -\alpha_3 \\ \alpha_2 = -\alpha_3 \end{array} \right.$$

Let $\alpha_3 = t$, then $\alpha_2 = -t$ and $\alpha_1 = 0$. That means that our eigen vector has the following form:

$$\alpha = \left( \begin{array}{c} 0 \\ -t \\ t \end{array} \right) \text{ or } \alpha = \left( \begin{array}{c} 0 \\ -1 \\ 1 \end{array} \right)$$

2. For $\lambda_2 = 2$

Similar to the previous

$$\left[ \begin{array}{ccc} 0 & 1 & 1 \\ 2 & 1 & 2 \\ -2 & 1 & 0 \end{array} \right] \left[ \begin{array}{c} \beta_1 \\ \beta_2 \\ \beta_3 \end{array} \right] = 0$$ $$\begin{array}{l} \left( \begin{array}{cccc} 0 & 1 & 1 & 0 \\ 2 & 1 & 2 & 0 \\ -2 & 1 & 0 & 0 \end{array} \right) \sim \left( \begin{array}{cccc} 0 & 1 & 1 & 0 \\ 2 & 1 & 2 & 0 \\ 0 & 2 & 2 & 0 \end{array} \right) \sim \left( \begin{array}{cccc} 2 & 1 & 2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \sim \left( \begin{array}{cccc} 1 & 1/2 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \\ \left\{ \begin{array}{l} \beta_1 + \frac{1}{2} \beta_2 + \beta_3 = 0 \\ \beta_2 + \beta_3 = 0 \end{array} \right. \\ \left\{ \begin{array}{l} \beta_1 + \frac{1}{2} \beta_2 = -\beta_3 \\ \beta_2 = -\beta_3 \end{array} \right. \\ \end{array}$$

Let $\beta_3 = t \Rightarrow \beta_2 = -t \Rightarrow \beta_1 = -\frac{1}{2} t \Rightarrow \beta = \begin{pmatrix} -\frac{1}{2} t \\ -t \\ t \end{pmatrix}$ or $\beta = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}$

3. For $\lambda_3 = 4$

Similar to the previous

$$\begin{array}{l} \left[ \begin{array}{ccc} -2 & 1 & 1 \\ 2 & -1 & 2 \\ -2 & 1 & -2 \end{array} \right] * \left[ \begin{array}{c} \gamma_1 \\ \gamma_2 \\ \gamma_3 \end{array} \right] = 0 \\ \left( \begin{array}{cccc} -2 & 1 & 1 & 0 \\ 2 & -1 & 2 & 0 \\ -2 & 1 & -2 & 0 \end{array} \right) \sim \left( \begin{array}{cccc} 0 & 0 & 3 & 0 \\ 2 & -1 & 2 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \sim \left( \begin{array}{cccc} 1 & -\frac{1}{2} & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \\ \left\{ \begin{array}{l} \gamma_1 - \frac{1}{2} \gamma_2 + \gamma_3 = 0 \\ \gamma_3 = 0 \end{array} \right. \\ \left\{ \begin{array}{l} \gamma_1 = \frac{1}{2} \gamma_2 \\ \gamma_3 = 0 \end{array} \right. \\ \end{array}$$

Let $\gamma_2 = t \Rightarrow \gamma_1 = \frac{1}{2} t \Rightarrow \gamma = \begin{pmatrix} \frac{1}{2} t \\ t \\ 0 \end{pmatrix}$ or $\gamma = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$

Answer:

1) $\lambda_1 = 1$; $\alpha = \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}$

2) $\lambda_2 = 2$; $\beta = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}$

3) $\lambda_3 = 4$; $\gamma = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$