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May 14, 2024
Solution by Steps
step 1
We are given a 3x3 matrix A=(30amp;−8amp;−83amp;16amp;−236amp;−4amp;8) A = \begin{pmatrix} 30 & -8 & -8 \\ 3 & 16 & -23 \\ 6 & -4 & 8 \end{pmatrix} . We need to find its eigenvalues and corresponding eigenvectors
step 2
To find the eigenvalues, we solve the characteristic equation det⁡(A−λI)=0 \det(A - \lambda I) = 0
step 3
The characteristic polynomial is det⁡(30−λamp;−8amp;−83amp;16−λamp;−236amp;−4amp;8−λ)=0 \det \begin{pmatrix} 30 - \lambda & -8 & -8 \\ 3 & 16 - \lambda & -23 \\ 6 & -4 & 8 - \lambda \end{pmatrix} = 0
step 4
Solving the determinant, we get the characteristic polynomial −λ3+54λ2−104λ+480=0 -\lambda^3 + 54\lambda^2 - 104\lambda + 480 = 0
step 5
The roots of this polynomial are the eigenvalues. Solving, we find the eigenvalues to be λ1=2 \lambda_1 = 2 , λ2=20 \lambda_2 = 20 , and λ3=30 \lambda_3 = 30
step 6
For λ1=2 \lambda_1 = 2 , we solve (A−2I)v⃗=0 (A - 2I)\vec{v} = 0 to find the eigenvector. This gives us v⃗1=(131) \vec{v}_1 = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}
step 7
For λ2=20 \lambda_2 = 20 , we solve (A−20I)v⃗=0 (A - 20I)\vec{v} = 0 to find the eigenvector. This gives us v⃗2=(403) \vec{v}_2 = \begin{pmatrix} 4 \\ 0 \\ 3 \end{pmatrix}
step 8
For λ3=30 \lambda_3 = 30 , we solve (A−30I)v⃗=0 (A - 30I)\vec{v} = 0 to find the eigenvector. This gives us v⃗3=(3−15) \vec{v}_3 = \begin{pmatrix} 3 \\ -1 \\ 5 \end{pmatrix}
Answer
The eigenvalues of the matrix A A are 2,20, 2, 20, and 30 30 . The corresponding eigenvectors are v⃗1=(131) \vec{v}_1 = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} , v⃗2=(403) \vec{v}_2 = \begin{pmatrix} 4 \\ 0 \\ 3 \end{pmatrix} , and v⃗3=(3−15) \vec{v}_3 = \begin{pmatrix} 3 \\ -1 \\ 5 \end{pmatrix} .
Key Concept
Eigenvalues and Eigenvectors
Explanation
Eigenvalues are the special set of scalars associated with a linear system of equations (i.e., a matrix equation) that provide insights into the system's properties. Eigenvectors are the vectors that correspond to these eigenvalues and indicate the directions in which the linear transformation acts by stretching or compressing.
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