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The matrix of Linear operator A in the basis B=$\{ e_1,e_2,e_3 \}$ is $$\begin{pmatrix} 1 & 2 & 3 \\ 4 & 0 & 5 \\ -1 & 3 & 1 \end{pmatrix}$$ i should find the matrix of this linear operator in the basis E=$\{ e_2, e_1, e_3 \}$

I found the change matrix P from B to E $$\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

thus when i calculate $P^{-1}AP$, it is $$\begin{pmatrix} 0 & 4 & 5 \\ 2 & 1 & 3 \\ 3 & -1 & 1 \end{pmatrix}$$

So the matrix of Linear Operator A in the basis E turned to be matrix with the change of the first and second rows and columns.

Is this right? Did i find the right matrix ?

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Yes, that is correct. A 3 by 3 matrix will map basis vector $e_1$ to its first column, $e_2$ to its second column, and $e_3$ to its third column. Changing the order of the first two basis vectors swaps the first and second rows and columns.

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