We will denote this also by x the above formula x x

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. We will denote this also by ˙ X . )
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The above formula ˙ X ( x ) = AX ( x ) is true for all x , so in particular when x = 0 we have A = ˙ X (0) X - 1 (0) which gives a way to recover the matrix A if we are only given a fundamental matrix solution X . I Example Does the matrix X = e 2 x e - x 2 e 2 x - e - x arise as a fundamental matrix solution for some system ˙ y = A y ? If so, find A . Solution If X was a fundamental matrix solution, then X (0) would be invertible. So we first check if this is the case. We compute X (0) = 1 1 2 - 1 Since det X (0) = - 3 6 = 0 , X (0) is invertible, so we compute X - 1 (0) = 1 / 3 1 / 3 2 / 3 - 1 / 3 We also compute ˙ X = 2 e 2 x - e - x 4 e 2 x e - x so we have ˙ X (0) = 2 - 1 4 1 . Then by the above formula X is a fundamental matrix solution for the system with coefficient matrix A = ˙ X (0) X - 1 (0) = 2 - 1 4 1 1 / 3 1 / 3 2 / 3 - 1 / 3 = 0 1 2 1 .
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