In matrix from, the system is
The coefficient matrix has eigenvalues "\\lambda" such that
"\\begin{vmatrix} \n 3-\\lambda & -1 & -1 \\\\\n 1 & 1-\\lambda & -1 \\\\\n 1 & -1 & 1-\\lambda \\\\\n \\end{vmatrix} \n=-(\\lambda-2)^2(\\lambda-1)=0 \\implies 2_{(2)}, \\lambda=1"
(where the subscript denotes multiplicity of the eigenvalue).
"\\lambda =2" :
"\\begin{bmatrix}1 &-1 & -1 \\\\ 1 & -1 & -1 \\\\ 1 &-1 & -1 \\end{bmatrix} \\vec n_1=\\vec 0\\\\\n\\implies n_{1,1}-n_{1,2}-n_{1,3}=0 \\implies n_{1,1}=n_{1,2}+n_{1,3}"
By picking "n_{1,1}=1" , we can then set "n_{1,2}=1" and "n_{1,3}=0" , and vice versa, to find two corresponding eigenvectors,
"\\vec n_1=\\begin{bmatrix}1 \\\\ 1 \\\\ 0 \\end{bmatrix}, \\vec n_2=\\begin{bmatrix}1 \\\\ 0 \\\\ 1 \\end{bmatrix}"
"\\lambda=1" :
"\\begin{bmatrix}2 &-1 & -1 \\\\ 1 & 0 & -1 \\\\ 1 &-1 & 0 \\end{bmatrix} \\vec n_3=\\vec 0\\\\\n\\implies 2n_{3,1}-n_{3,2}-n_{3,3}=0 \\implies 2n_{3,1}=n_{3,2}+n_{3,3}"
We obtain "\\vec n_3" independent of "\\vec n_1, \\vec n_2" by picking "\\vec n_{3,2}=\\vec n_{3,3}=1", so that the third corresponding eigenvector is;
"\\vec n_3=\\begin{bmatrix}1 \\\\ 1 \\\\ 1 \\end{bmatrix}"
Then the general solution to this system is
"\\begin{bmatrix}x \\\\ y \\\\ z \\end{bmatrix}=c_1e^{2t} \\vec n_1+ c_2te^{2t} \\vec n_2+ c_3e^{t} \\vec n_3\\\\\n\\begin{cases}\n x=c_1e^{2t} + c_2te^{2t} + c_3e^{t}\\\\ \n y=c_1e^{2t} + c_3e^{t}\\\\ \n z=c_2te^{2t} + c_3e^{t}\\\\ \n\\end{cases}"
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