a) Find the coordinate vector of the vector $v = (1, 2, 3)^{\top}$ relative to this basis
b) Let $\mathfrak{B}' = ((0,1,0)^{\top} , (1,0,1)^{\top}, (2,1,0)^{\top})$. Find the matrix P that describes the transition from $\mathfrak{B}$ to $\mathfrak{B}'$
a) Finding the coordinate vector relative to this basis means solving the system $$\begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \\ 0 & 2 & 1 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$$
After solving the system https://matrixcalc.org/slu.html#solve-using-Gaussian-elimination%28%7B%7B1,2,3,1%7D,%7B2,1,1,2%7D,%7B0,2,1,3%7D%7D%29
We get the vector $x = \begin{bmatrix} 4/7 \\ 15/7 \\ -9/7 \end{bmatrix}$
b) If I understand correctly $P \cdot \mathfrak{B} = \mathfrak{B}'$
This means $P = \mathfrak{B}' \cdot \mathfrak{B}^{-1}$
So we need to find the inverse of $\mathfrak{B}$. To do so, we set up an indentity matrix next to it, and then apply row operations to $\mathfrak{B}$ until we get the identity matrix. We apply the same row operations to the identity matrix, and that will give the inverse of $\mathfrak{B}$
We get https://matrixcalc.org/#%7B%7B1,2,3%7D,%7B2,1,1%7D,%7B0,2,1%7D%7D%5E%28-1%29
$\mathfrak{B}^{-1} = 1/7 \cdot \begin{bmatrix} -1 & 4 & -1 \\ -2 & 1 & 5 \\ 4 & -2 & -3 \end{bmatrix}$
Thus $P = \mathfrak{B}' \cdot \mathfrak{B}^{-1} = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \cdot 1/7 \cdot \begin{bmatrix} -1 & 4 & -1 \\ -2 & 1 & 5 \\ 4 & -2 & -3 \end{bmatrix} = 1/7 \begin{bmatrix} 6 & -3 & -1 \\ 3 & 2 & -4 \\ -2 & 1 & 5 \end{bmatrix}$
Is this correct ? I'm quite new to the subject, so getting your feedback would be awesome. Thank you so much !