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Question
Math
Posted 10 months ago
Given a matrix AA, where
A=[224135234]
A=\left[\begin{array}{lll}
2 & 2 & 4 \\
1 & 3 & 5 \\
2 & 3 & 4
\end{array}\right]


Tasks:
1. Find the Eigenvalues of AA : Determine the eigenvalues of matrix AA.
2. Find the Eigenvectors Corresponding to Each Eigenvalue: For each eigenvalue found in Task 1, find the corresponding eigenvectors.
3. Diagonalize AA, if Possible: If matrix AA is diagonalizable, find a matrix PP consisting of the eigenvectors of AA and a diagonal matrix DD such that A=PDP1A=P D P^{-1}. If AA is not diagonalizable, explain why.
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Answer from Sia
Posted 10 months ago
Solution by Steps
step 1
To find the eigenvalues of matrix A A , we calculate the characteristic polynomial and solve for the values of λ \lambda that satisfy the equation det(AλI)=0 \det(A - \lambda I) = 0
step 2
The eigenvalues calculated from the asksia-ll calculator are λ18.80916 \lambda_1 \approx 8.80916 , λ20.926209 \lambda_2 \approx 0.926209 , and λ30.735373 \lambda_3 \approx -0.735373
step 3
To find the eigenvectors corresponding to each eigenvalue, we solve the system (AλiI)v=0 (A - \lambda_i I) \mathbf{v} = \mathbf{0} for each λi \lambda_i
step 4
The eigenvectors calculated from the asksia-ll calculator are v1(0.885,1.01305,1) \mathbf{v}_1 \approx (0.885, 1.01305, 1) , v2(7.51633,6.03548,1) \mathbf{v}_2 \approx (7.51633, -6.03548, 1) , and v3(0.601327,1.17757,1) \mathbf{v}_3 \approx (-0.601327, -1.17757, 1)
step 5
To diagonalize matrix A A , we form matrix P P with the eigenvectors as columns and matrix D D as the diagonal matrix with the corresponding eigenvalues
step 6
The diagonalization of matrix A A is given by A=PDP1 A = PDP^{-1} , where P P and D D are as follows: P(0.6013277.516330.8851.177576.035481.01305111) P \approx \begin{pmatrix} -0.601327 & 7.51633 & 0.885 \\ -1.17757 & -6.03548 & 1.01305 \\ 1 & 1 & 1 \end{pmatrix} , D(0.7353730000.9262090008.80916) D \approx \begin{pmatrix} -0.735373 & 0 & 0 \\ 0 & 0.926209 & 0 \\ 0 & 0 & 8.80916 \end{pmatrix} , and P1(0.2819050.2652190.5181680.08761390.05944550.01731670.1942920.3246650.499149) P^{-1} \approx \begin{pmatrix} -0.281905 & -0.265219 & 0.518168 \\ 0.0876139 & -0.0594455 & -0.0173167 \\ 0.194292 & 0.324665 & 0.499149 \end{pmatrix}
Answer
The matrix A A is diagonalizable, and the diagonalization is given by A=PDP1 A = PDP^{-1} with matrices P P , D D , and P1 P^{-1} as calculated above.
Key Concept
Diagonalization of a matrix
Explanation
Diagonalization involves finding a matrix P P of eigenvectors and a diagonal matrix D D of eigenvalues such that A=PDP1 A = PDP^{-1} . This process simplifies matrix operations and is possible when matrix A A has a full set of linearly independent eigenvectors.

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