G=[24−3−5]
Finding the eigenvalues:
det((24−3−5)−λ(1001))
=(2−λ4−0(−3)−0(−5)−λ)
(2−λ)(−5−λ)−(−3)⋅4
λ2+3λ+2=0
Solving this λ2+3λ+2=0
λ=−1,λ=−2
Therefore eigenvalues are: -1 and -2
Calculating Eigenvectors for :λ=−1
=(24−3−5)−(−1)(1001)
=(24−3−5)−(−100−1)
=(34−3−4)
To solve (34−3−4)(xy)=(00), reduce the matrix
=(11) as the eigenvectors for λ=−1
Calculating Eigenvectors for :λ=−2
=(24−3−5)−(−2)(1001)
=(44−3−3)
To solve (44−3−3)(xy)=(00), reduce the matrix
=(34) as the eigenvectors for λ=−2
The Eigenvectors for (24−3−5) becomes =(11),(34)
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