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# ForA = I8 1] the only eigenvalue is A = 1. ThusA -MI =Since this matrix is = 0, A = 1 is & defective eigenvalue: Thus wC Can take V2 1% V = (A - AI)vz [o] and t...

## Question

###### ForA = I8 1] the only eigenvalue is A = 1. ThusA -MI =Since this matrix is = 0, A = 1 is & defective eigenvalue: Thus wC Can take V2 1% V = (A - AI)vz [o] and the general solution is y(t) = Cie't [o] +ce' ([1]_ +t [o])Finally, the case ofA=|51 3| is very similar to the previous case The only eigenvalue is A = 4 Thus A-N=|41 Since this matrix is # 0; A = 4 is & defective eigenvalue. Thus we can take Vz = V1 = (A -A)vz = [4]and the general solution isy(t) = Ciet [4J+ce' ([93

For A = I8 1] the only eigenvalue is A = 1. Thus A -MI = Since this matrix is = 0, A = 1 is & defective eigenvalue: Thus wC Can take V2 1% V = (A - AI)vz [o] and the general solution is y(t) = Cie't [o] +ce' ([1]_ +t [o]) Finally, the case of A=|51 3| is very similar to the previous case The only eigenvalue is A = 4 Thus A-N=|41 Since this matrix is # 0; A = 4 is & defective eigenvalue. Thus we can take Vz = V1 = (A -A)vz = [4] and the general solution is y(t) = Ciet [4J+ce' ([93+'[4)

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