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А linear map has this construction in ${e_1, e_2, e_3}$ basis \begin{pmatrix}1&2&3\\-1&0&3\\2&1&5\end{pmatrix}

Find the matrix of A linear map in ${e_2,e_1,e_3}$ basis.

I have calculated the transform matrix from first basis to the second, and got \begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix} And I know that A linear map in second basis is $1/T * A * T$, where T is the transform matrix and A is the representation of linear map. (First matrix in the question). but my Transform matrix's Determinant is 0. Any suggestions?

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    $\begingroup$ The determinant is $-1$ not $0$. $\endgroup$ – wgrenard Mar 29 '18 at 19:19
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Hint:

Observe that $T^2=I_3$, for a couple of obvious reasons.

Unrelated: don't use fractional notation for the inverse of a matrix. This has nothing to do with rational numbers.

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Oh, how inattentive it was :D, So after calculations, I got \begin{pmatrix}0&-1&3\\2&1&3\\1&2&5\end{pmatrix}. It's the right answer yes? Thanks

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You can solve this without doing any explicit matrix multiplication or inversion. The target ordered basis is just the standard basis with the first two vectors swapped, so coordinates in the latter basis are obtained by swapping the first to coordinates relative to the standard basis. Recall that the columns of a transformation matrix are the images of the basis vectors, so the new matrix is obtained by swapping the first two columns and first two rows of the original matrix.

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