Discrete systems of equations217In each of these instructions the last line is a check that undertaking the matrixmultiplication does indeed lead to the Jordan form of the matrix. In each packagewe get the Jordan formJ=±1003²However, the transition matrix in each package on the face of it looks different.More speciFcally,MathematicaV=±−11²MapleV=³1212−1212´Butthesearefundamentallythesame.Wenotedthiswhenderivingtheeigenvectorsin the previous section. We arbitrarily chose values forvr1orvr2(along with thevalues associated with the eigenvalues). InMaple, consider the Frst column,which is the Frst eigenvector. Settingvr2=1, means multiplying the Frst term by−2, which gives a value forvr1=−1. Similarly, settingvs1=1inMaple, convertsvs2also to the value of unity. Hence, the two matrices are identical. In each casethe last instruction veriFes thatV−1AV=J.
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