Given the polynomials $p_1,p_2,p_3,v_1,v_2,v_3 \in P_2$: \begin{gather*} p_1(t) := t^2 − 2t + 5, \qquad p_2(t) := 2t^2 − 3t, \qquad p_3(t) := t + 1, \\ v_1(t) := t^2 + 4t − 3, \qquad v_2(t) := t − 1, \qquad v_3(t) := 1. \end{gather*} find the matrix of transition from base $P=\{p_1,p_2,p_3\}$ to base $V = \{v_1,v_2,v_3\}$.
I have found the matrix: $$ M_1 := \begin{pmatrix} 1 & -6 & 2 \\ 2 & - 11 & -5 \\ 0 & 1 & 2 \end{pmatrix} $$ and I thought that this is the answer because \begin{gather*} p_1(t) = v_1(t) − 6 v_2(t) + 2 v_3(t), \qquad p_2(t) = 2 v_1(t) − 11 v_2(t) − 5 v_3(t), \quad p_3(t) = 0 \cdot v_1(t) + 1 \cdot v_2(t) + 2 v_3(t). \end{gather*} but it turns out that I should find matrix like this: $$ M_2 := \begin{pmatrix} 1 & 2 & 0 \\ -6 & -11 & 1 \\ 2 & -5 & 2 \end{pmatrix} $$ and after that to find a matrix of transition i should find inverse matrix from that matrix so i would get matrix like this: $$ \frac{1}{11} \begin{pmatrix} -17 & -4 & 2 \\ 14 & 2 & -1 \\ 52 & 9 & 1 \end{pmatrix} $$ Can you please explain why
Why I had to find the matrix $M_2$ and not $M_1$, which is actually transposed matrix $M_2$.
Why I had to find the inverse matrix $M_2$, (why is the inverse matrix of $M_2$ matrix of transition).
Your $M_1$ is the correct matrix if you have a polynomial expressed in terms of the $v$s and you want it expressed in terms of the $p$s and if your polynomials are column vectors. If I have a polynomial $av_1+bv_2+cv_3$ and express it as a column matrix $\begin {pmatrix}a\\b\\c \end{pmatrix}$ I can premultiply it by $M_1$ and get the polynomial expressed in terms of the $p$s. You needed to transpose it because you presumably use row vectors, so the polynomial would be expressed by the matrix $\begin {pmatrix}a&b&c \end{pmatrix}$ and postmultiplied by $M_2$, giving a new row vector.
The reason you needed to invert it is that you were asked to start with $p$s and get $v$s. If you had written $v_1=$an expression of $p$'s you would have gotten the right matrix at the start.