# Math Help - URGENT...DUE TODAY

1. ## URGENT...DUE TODAY

I'm a little confused about similar matrices.

Are the the following matrices similar?

$A = \begin{bmatrix} 1 & 2 & 1 \\ 0 & -2 & 1 \\ 1 & 0 & 2 \end{bmatrix} \ \$ $\ B = \begin{bmatrix} 1 & 0 & 3\\ 1 & 0 & 1 \\ 2 & 1 & 1\end{bmatrix}$

Now I know that similar matrices share a variety of properties....some of which I've already calculated.

$For A \ \Rightarrow \$ Rank = 2, Determinant = 0, Trace = 1, Char. Polynomial = $\lambda^3 - \lambda^2 - 5 \lambda$

$For B \ \Rightarrow \$ Rank = 3, Determinant = 2, Trace = 2, Char. Polynomial = $-2 + \lambda^3 - 2 \lambda^2 - 6 \lambda$

From this I would say that these two aren't similar, however, I have read other posts on this forum that make me think that the above properties aren't enough to safely conclude that the are or are not similar.

I am aware that if 2 matrices, A and B, are similar then there is a matrix P such that $B = P^{-1} AP$.

How then, do I calculate P?

Thanks in advance for any help

2. Originally Posted by Jimmy_W
I'm a little confused about similar matrices.

Are the the following matrices similar?

$A = \begin{bmatrix} 1 & 2 & 1 \\ 0 & -2 & 1 \\ 1 & 0 & 2 \end{bmatrix} \ \$ $\ B = \begin{bmatrix} 1 & 0 & 3\\ 1 & 0 & 1 \\ 2 & 1 & 1\end{bmatrix}$

Now I know that similar matrices share a variety of properties....some of which I've already calculated.

$For A \ \Rightarrow \$ Rank = 2, Determinant = 0, Trace = 1, Char. Polynomial = $\lambda^3 - \lambda^2 - 5 \lambda$

$For B \ \Rightarrow \$ Rank = 3, Determinant = 2, Trace = 2, Char. Polynomial = $-2 + \lambda^3 - 2 \lambda^2 - 6 \lambda$

From this I would say that these two aren't similar, however, I have read other posts on this forum that make me think that the above properties aren't enough to safely conclude that the are or are not similar.

I am aware that if 2 matrices, A and B, are similar then there is a matrix P such that $B = P^{-1} AP$.

How then, do I calculate P?

Thanks in advance for any help
you don't need to find P. you've shown that $\det A \neq \det B,$ which is enough to conclude that A and B are not similar.