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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