Solve for Eigenvector: Characteristic Equation of Matrix A with Solution Method

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SUMMARY

The characteristic equation of the matrix A, defined as A = \(\begin{pmatrix} 4 & 0 & 1 \\ -2 & 1 & 0 \\ -2 & 0 & 1 \end{pmatrix}\), is accurately derived as \(L^3 - 6L^2 + 11L - 6 = 0\). This result is obtained by calculating the determinant of the matrix \(LI - A\). The equation can also be expressed as \(x^3 - 6x^2 + 11x - 6 = 0\). For formatting mathematical expressions, LaTeX is recommended for clarity and precision.

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johnnyboy2005
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hey, not sure how this works on this website but was just wondering if someone can tell me if I'm doing this right...

find characteristic equation of matrix

A=l 4 0 1 l
l -2 1 0l
l -2 0 1l

i found it to be (L=lambda)...

L^3 - 6L^2 + 11L - 6 = 0


only because i solved the matrix det(LI - A) and found that

=(L-4)[(L-1)(L-1)] -1[0 - (2)(L-1)] ...have i got this right? thanks for the help
 
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Yes, that's right. Or just say:

[tex]x^3-6x^2+11x-6=0[/tex]

We use a formatting language called LaTex to format it in traditional form:

[tex] M=\left(<br /> \begin{array}{ccc} 4 & 0 & 1 \\<br /> -2 & 1 & 0 \\<br /> -2 & 0 & 1<br /> \end{array}<br /> \right)[/tex]

If you choose to post frequently here you may wish to learn it. Start by looking at the "General Physics" forum, top thread titled "Introducing LaTex".
 

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