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Are the matrices $A = \begin{bmatrix}1&2&-3\\1&-1&2\\0&3&5\end{bmatrix}$ and $B = \begin{bmatrix}1&-1&4\\1&1&5\\1&1&6\end{bmatrix}$ similar?

I know that similar matrices have the same determinant, the same rank, and the same characteristic polynomial (and therefore, the same eigenvalues).

What I have tried:

$det(A) = -30$

$det(B) = 2$

$rank(A) = 3$

$rank(B) = 3$

How do I show the characteristic polynomial for $A$ and $B$?

Also since $det(A) \ne det(B)$, we already know that matrices $A$ and $B$ are not similar so we don't need to prove that using $B = P^{-1}AP$, right?

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Similar matrices have the same eigenvalues, determinants are product of eigenvalues so similar matrices have same determinants. Thus, different determinants $\implies$ not similar, so you don't need to do any more work to prove they aren't similar.

You can find the characteristic polynomial from the definition -- the characteristic polynomial of $A$ is $det(A - \lambda I)$ [ or depending on who you ask, sometimes $det(\lambda I - A)$; this is negative of the other definition if $A$ has an odd number of rows] where $I$ is the identity matrix of the same size as $A$, and the variable is $\lambda$.

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Easiest property is that similar matrices have the same trace, and not in your case.

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