Question $\mathbf{1 6}$ to $\mathbf{2 0}$ refer to the vector equation $\mathbf{A} \cdot \mathbf{x}=\lambda \mathbf{x}$ . For the coefficient matrix $\mathbf{A}$ given in each case, determine the eigenvalues and an eigenvector corresponding to each eigenvalue: $16 \quad \mathbf{A}=\left(\begin{array}{rrr}2 & 1 & 1 \\ 1 & 3 & 2 \\ -1 & 1 & 2\end{array}\right)$ $17 \mathbf{A}=\left(\begin{array}{lll}1 & 2 & 2 \\ 1 & 3 & 1 \\ 2 & 2 & 1\end{array}\right)$ $18 \quad \mathbf{A}=\left(\begin{array}{rrr}2 & 0 & 1 \\ -1 & 4 & -1 \\ -1 & 2 & 0\end{array}\right)$ $19 \mathbf{A}=\left(\begin{array}{rrr}1 & -4 & -2 \\ 0 & 3 & 1 \\ 1 & 2 & 4\end{array}\right)$ $20 \quad \mathbf{A}=\left(\begin{array}{lll}3 & 0 & 3 \\ 0 & 3 & 3 \\ 2 & 3 & 1\end{array}\right)$

Studdy Solution

STEP 1

1. We are given a set of matrices and need to find their eigenvalues and corresponding eigenvectors.
2. The eigenvalue problem is defined by the equation $\mathbf{A} \cdot \mathbf{x} = \lambda \mathbf{x}$ .
3. To find eigenvalues, we will solve the characteristic equation $\det(\mathbf{A} - \lambda \mathbf{I}) = 0$ .
4. For each eigenvalue, we will find a corresponding eigenvector by solving $(\mathbf{A} - \lambda \mathbf{I}) \mathbf{x} = \mathbf{0}$ .

STEP 2

1. Determine the eigenvalues for each matrix.
2. Find an eigenvector corresponding to each eigenvalue.

STEP 3

For each matrix $\mathbf{A}$ , calculate the characteristic polynomial by finding the determinant of $\mathbf{A} - \lambda \mathbf{I}$ , where $\mathbf{I}$ is the identity matrix of the same size as $\mathbf{A}$ .
Example for Matrix 16: $\mathbf{A} = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 3 & 2 \\ -1 & 1 & 2 \end{pmatrix}$ Calculate $\det(\mathbf{A} - \lambda \mathbf{I})$ : $\det\begin{pmatrix} 2-\lambda & 1 & 1 \\ 1 & 3-\lambda & 2 \\ -1 & 1 & 2-\lambda \end{pmatrix}$

STEP 4

Solve the characteristic polynomial equation $\det(\mathbf{A} - \lambda \mathbf{I}) = 0$ to find the eigenvalues $\lambda$ .
Example for Matrix 16: Solve: $\det\begin{pmatrix} 2-\lambda & 1 & 1 \\ 1 & 3-\lambda & 2 \\ -1 & 1 & 2-\lambda \end{pmatrix} = 0$

STEP 5

For each eigenvalue $\lambda$ found, substitute it back into the equation $(\mathbf{A} - \lambda \mathbf{I}) \mathbf{x} = \mathbf{0}$ and solve for the eigenvector $\mathbf{x}$ .
Example for Matrix 16: Assume $\lambda_1$ is an eigenvalue, solve: $\begin{pmatrix} 2-\lambda_1 & 1 & 1 \\ 1 & 3-\lambda_1 & 2 \\ -1 & 1 & 2-\lambda_1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$
Repeat the above steps for each matrix (17, 18, 19, 20) to find their eigenvalues and corresponding eigenvectors.

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