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Math  /  Algebra

QuestionYou must clearly show your steps for every problem below.
1. Find all distinct (real or complex) eigenvalues of AA. Then find the basic eigenvectors of AA corresponding to each eigenvalue. A=[420−20−6−2424−4−1616]A=\left[\begin{array}{ccc} 4 & 20 & -20 \\ -6 & -24 & 24 \\ -4 & -16 & 16 \end{array}\right]

Studdy Solution

STEP 1

1. A A is a 3×3 3 \times 3 matrix.
2. We need to find eigenvalues by solving the characteristic equation.
3. Eigenvectors are found by solving (A−λI)v=0 (A - \lambda I)\mathbf{v} = \mathbf{0} for each eigenvalue λ \lambda .

STEP 2

1. Find the characteristic polynomial of matrix A A .
2. Solve the characteristic polynomial to find the eigenvalues.
3. For each eigenvalue, find the corresponding eigenvectors.

STEP 3

To find the characteristic polynomial, we need to compute the determinant of A−λI A - \lambda I , where I I is the identity matrix of the same size as A A .
The matrix A−λI A - \lambda I is:
A−λI=[4−λ20−20−6−24−λ24−4−1616−λ]A - \lambda I = \begin{bmatrix} 4 - \lambda & 20 & -20 \\ -6 & -24 - \lambda & 24 \\ -4 & -16 & 16 - \lambda \end{bmatrix}
The characteristic polynomial is given by:
det⁡(A−λI)=0\det(A - \lambda I) = 0

STEP 4

Calculate the determinant of the matrix A−λI A - \lambda I :
det⁡(A−λI)=∣4−λ20−20−6−24−λ24−4−1616−λ∣\det(A - \lambda I) = \begin{vmatrix} 4 - \lambda & 20 & -20 \\ -6 & -24 - \lambda & 24 \\ -4 & -16 & 16 - \lambda \end{vmatrix}
Expanding along the first row:
=(4−λ)∣−24−λ24−1616−λ∣−20∣−624−416−λ∣−20∣−6−24−λ−4−16∣= (4 - \lambda) \begin{vmatrix} -24 - \lambda & 24 \\ -16 & 16 - \lambda \end{vmatrix} - 20 \begin{vmatrix} -6 & 24 \\ -4 & 16 - \lambda \end{vmatrix} - 20 \begin{vmatrix} -6 & -24 - \lambda \\ -4 & -16 \end{vmatrix}
Calculate each of these 2x2 determinants and simplify to find the characteristic polynomial.

STEP 5

Solve the characteristic polynomial for λ \lambda to find the eigenvalues.

STEP 6

For each eigenvalue λ \lambda , solve the system (A−λI)v=0 (A - \lambda I)\mathbf{v} = \mathbf{0} to find the corresponding eigenvectors.
The solution will provide the distinct eigenvalues and their corresponding basic eigenvectors.

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