Verifying change of basis calculation by example, my lack of understanding/mistake or Author's?

Question

a. I got the change of coordinates from B to A, P = $\begin{pmatrix}6&0&-6\\ -1&4&0\\ 1&1&3\end{pmatrix}$

b. However, I tried to use the formula for change of basis $P^{-1}(x)=(x)_B$ and got the following:

$\begin{pmatrix}6&0&-6\\ -1&4&0\\ 1&1&3\end{pmatrix}^{-1}\cdot \begin{pmatrix}1\\ -4\\ 4\end{pmatrix}=\begin{pmatrix}\frac{22}{17}\\ -\frac{23}{34}\\ \frac{115}{102}\end{pmatrix}$

Which according to the answer $\begin{pmatrix}14\\ \:-12\\ \:6\end{pmatrix}$, mine is incorrect.

I tried to work backward from their answer, by doing the below and also could not get the answer:

$\begin{pmatrix}6&0&-6\\ \:\:\:-1&4&0\\ \:\:\:1&1&3\end{pmatrix}^{-1}\vec{v\:}=\begin{pmatrix}14\\ \:-12\\ \:6\end{pmatrix} \rightarrow \frac{1}{102}\begin{pmatrix}12&-6&24\\ 3&24&6\\ -5&-6&24\end{pmatrix}\vec{v}= \begin{pmatrix}14\\ \:-12\\ \:6\end{pmatrix}$

$\begin{pmatrix}12&-6&24\\ 3&24&6\\ -5&-6&24\end{pmatrix}\vec{v}= 102 \begin{pmatrix}14\\ \:-12\\ \:6\end{pmatrix} \rightarrow \vec{v}=\begin{pmatrix}48\\ \:-62\\ \:20\end{pmatrix}$

Qn: Did I make a mistake or did the author make a mistake?

Answer 1

Actually, the matrix $P$ that defines the change of coordinates from $B$ to $A$ is obtained as $$ P = \left[ \begin{array}{ccc} 6 & -1 & 1 \\ 0 & 4 & 1 \\ -6 & 0 & 3 \\ \end{array} \right] $$

Define $$ v = \left[ \begin{array}{c} 1 \\ -4 \\ 4 \\ \end{array} \right] $$

Note that $$ P v = \left[ \begin{array}{c} 14 \\ -12 \\ 6 \\ \end{array} \right] $$

Answer 2

The issue was that:

1. I was reading P values horizontally instead of vertically.
2. I inverted what was already the right transformation to use

---

Corrections and clarifications thanks to Dr. Sundar's prompt:

a. I got the change of coordinates transformation from B to A, $P_{B\rightarrow A}=\begin{pmatrix}6&-1&1\\ 0&4&1\\ -6&0&3\end{pmatrix}$

b. I should have used the change of basis $P_{B\rightarrow A}(x)=(x)_A$. Because inverting P would have been $P_{B\rightarrow A}^{-1}=P_{A\rightarrow B}$