logistic_guy
Full Member
- Joined
- Apr 17, 2024
- Messages
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here is the question
Solve \(\displaystyle \bold{X}' = \begin{bmatrix}1 & -2 & 2 \\-2 & 1 & -2 \\2 & -2 & 1 \end{bmatrix} \bold{X}\).
my attemb
\(\displaystyle \bold{Z} = \bold{A} - \lambda\bold{I} = \begin{bmatrix}1 & -2 & 2 \\-2 & 1 & -2 \\2 & -2 & 1 \end{bmatrix} - \lambda \begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix} = \begin{bmatrix}1 & -2 & 2 \\-2 & 1 & -2 \\2 & -2 & 1 \end{bmatrix} + \begin{bmatrix}-\lambda & 0 & 0 \\0 & -\lambda & 0 \\0 & 0 & -\lambda \end{bmatrix} = \begin{bmatrix}1 - \lambda & -2 & 2 \\-2 & 1 - \lambda & -2 \\2 & -2 & 1 - \lambda \end{bmatrix}\)
det\(\displaystyle (\bold{Z}) = (1 - \lambda)[(1 - \lambda)(1 - \lambda) - (-2)(-2)] - -2[(-2)(1 - \lambda) - (-2)(2)] + 2[(-2)(-2) - (1 - \lambda)(2)]\)
\(\displaystyle = (1 - \lambda)[1 - \lambda - \lambda + \lambda^2 - 4] + 2[-2 + 2\lambda + 4] + 2[4 - 2 + 2\lambda]\)
\(\displaystyle = (1 - \lambda)[-2\lambda + \lambda^2 - 3] + 2[2\lambda + 2] + 2[2 + 2\lambda]\)
\(\displaystyle = -2\lambda + \lambda^2 - 3 + 2\lambda^2 - \lambda^3 + 3\lambda + 4\lambda + 4 + 4 + 4\lambda\)
\(\displaystyle = 9\lambda + 3\lambda^2 - \lambda^3 + 5 = 0\)
this mean
\(\displaystyle \lambda^3 - 3\lambda^2 - 9\lambda - 5 = 0\)
i'm strugle to solve degree \(\displaystyle 3\) so i learn rational root theorem
this give me possible rational roots \(\displaystyle \pm 1\) and \(\displaystyle \pm 5\)
testing one by one
\(\displaystyle \lambda(1) = 1^3 - 3(1)^2 - 9(1) - 5 = 1 - 3 - 9 - 5 = -16 \neq 0\)
\(\displaystyle \lambda(-1) = (-1)^3 - 3(-1)^2 - 9(-1) - 5 = -1 - 3 + 9 - 5 = 0\)
so \(\displaystyle (\lambda + 1)\) is factor
\(\displaystyle \frac{\lambda^3 - 3\lambda^2 - 9\lambda - 5}{\lambda + 1} = \lambda^2 - 4\lambda - 5\) i don't know how to write long division steps in latex
i've get \(\displaystyle (\lambda + 1)(\lambda^2 - 4\lambda - 5) = (\lambda + 1)(\lambda + 1)(\lambda - 5) = (\lambda + 1)^2(\lambda - 5) = 0\)
this mean \(\displaystyle \lambda_1 = \lambda_2 = -1\) and \(\displaystyle \lambda_3 = 5\)
i get the same problem as before\(\displaystyle \bold{X_1} = \bold{X_2}\)
this give me only two solutions \(\displaystyle \bold{X_1}\) and \(\displaystyle \bold{X_3}\)but i need three solutions
Solve \(\displaystyle \bold{X}' = \begin{bmatrix}1 & -2 & 2 \\-2 & 1 & -2 \\2 & -2 & 1 \end{bmatrix} \bold{X}\).
my attemb
\(\displaystyle \bold{Z} = \bold{A} - \lambda\bold{I} = \begin{bmatrix}1 & -2 & 2 \\-2 & 1 & -2 \\2 & -2 & 1 \end{bmatrix} - \lambda \begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix} = \begin{bmatrix}1 & -2 & 2 \\-2 & 1 & -2 \\2 & -2 & 1 \end{bmatrix} + \begin{bmatrix}-\lambda & 0 & 0 \\0 & -\lambda & 0 \\0 & 0 & -\lambda \end{bmatrix} = \begin{bmatrix}1 - \lambda & -2 & 2 \\-2 & 1 - \lambda & -2 \\2 & -2 & 1 - \lambda \end{bmatrix}\)
det\(\displaystyle (\bold{Z}) = (1 - \lambda)[(1 - \lambda)(1 - \lambda) - (-2)(-2)] - -2[(-2)(1 - \lambda) - (-2)(2)] + 2[(-2)(-2) - (1 - \lambda)(2)]\)
\(\displaystyle = (1 - \lambda)[1 - \lambda - \lambda + \lambda^2 - 4] + 2[-2 + 2\lambda + 4] + 2[4 - 2 + 2\lambda]\)
\(\displaystyle = (1 - \lambda)[-2\lambda + \lambda^2 - 3] + 2[2\lambda + 2] + 2[2 + 2\lambda]\)
\(\displaystyle = -2\lambda + \lambda^2 - 3 + 2\lambda^2 - \lambda^3 + 3\lambda + 4\lambda + 4 + 4 + 4\lambda\)
\(\displaystyle = 9\lambda + 3\lambda^2 - \lambda^3 + 5 = 0\)
this mean
\(\displaystyle \lambda^3 - 3\lambda^2 - 9\lambda - 5 = 0\)
i'm strugle to solve degree \(\displaystyle 3\) so i learn rational root theorem
this give me possible rational roots \(\displaystyle \pm 1\) and \(\displaystyle \pm 5\)
testing one by one
\(\displaystyle \lambda(1) = 1^3 - 3(1)^2 - 9(1) - 5 = 1 - 3 - 9 - 5 = -16 \neq 0\)
\(\displaystyle \lambda(-1) = (-1)^3 - 3(-1)^2 - 9(-1) - 5 = -1 - 3 + 9 - 5 = 0\)
so \(\displaystyle (\lambda + 1)\) is factor
\(\displaystyle \frac{\lambda^3 - 3\lambda^2 - 9\lambda - 5}{\lambda + 1} = \lambda^2 - 4\lambda - 5\) i don't know how to write long division steps in latex
i've get \(\displaystyle (\lambda + 1)(\lambda^2 - 4\lambda - 5) = (\lambda + 1)(\lambda + 1)(\lambda - 5) = (\lambda + 1)^2(\lambda - 5) = 0\)
this mean \(\displaystyle \lambda_1 = \lambda_2 = -1\) and \(\displaystyle \lambda_3 = 5\)
i get the same problem as before\(\displaystyle \bold{X_1} = \bold{X_2}\)
this give me only two solutions \(\displaystyle \bold{X_1}\) and \(\displaystyle \bold{X_3}\)but i need three solutions