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Diberikan masalah nilai awal \\(\\mathbf{x}'=\\begin{bmatrix} -2 & 1 \\\\ -5 & 4  \\end{bmatrix}\\mathbf{x}\\), \\(\\mathbf{x}_0=\\begin{bmatrix} 1 \\\\ 5 \\end{bmatrix}\\). 

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Jika solusi khusus masalah nilai awal tersebut adalah \\(\\mathbf{x}(t)=\\begin{bmatrix} ae^{bt} \\\\ ce^{bt} \\end{bmatrix}\\), maka \\(a+b+c=\\ldots\\).

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