# I cannot understand this change of basis problem

I know we need to get the inverse of transformation matrix first to get the coordinate relative to other frames. Explanation is in this video.

I am reading directx11 book. And the following ones are the problem and solution in this book.

1. Suppose that we have frames $$\A\$$ and $$\B\$$. Let $$\p_A=(1,-2,0)\$$ and $$\q_A=(1,2,0)\$$ represent a point and force, respectively, relative to frame $$\A\$$. Moreover, let $$\Q_B=(-6,2,0)\$$, $$\u_B=\left(\frac{1}{\sqrt 2},\frac{1}{\sqrt 2},0\right)\$$, $$\v_B=\left(-\frac{1}{\sqrt 2},\frac{1}{\sqrt 2},0\right)\$$ and $$\w_B=(0,0,1)\$$ describe frame $$\A\$$ with coordinates relative to frame $$\B\$$. Build the change of coordinate matrix that maps frame $$\A\$$ coordinates into frame $$\B\$$ coordinates, and find $$\p_B=(x,y,z)\$$ and $$\q_B=(x,y,z)\$$. Draw a picture on graph paper to verify that your answer is reasonable.

\begin{align} p_B &= \begin{bmatrix}1&-2&0&1\end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt 2}&\frac{1}{\sqrt 2}&0&0 \\ -\frac{1}{\sqrt 2}&\frac{1}{\sqrt 2}&0&0 \\ 0&0&1&0 \\ -6&2&0&1\end{bmatrix} \\ &= \begin{bmatrix}\frac{3}{\sqrt 2}-6&-\frac{1}{\sqrt 2}+2&0&1\end{bmatrix} \\ &\approx\begin{bmatrix}-3.88&1.29&0&1\end{bmatrix} \end{align}

I cannot understand how the coordinate relative to frame $$\B\$$ can be figured out by multiplying the coordinate relative to frame $$\A\$$ to the transformation matrix, not a inverse one.

Is there any material to study more or did I understand the problem in wrong way?

Note that the given $$\Q_B, u_B, v_b, w_b\$$ vectors are defined "with coordinates relative to B". That means they're already baking-in the inverse of frame B.
Whoever wrote down that $$\Q_B\$$ position vector in "coordinates relative to B" had to find the origin of A in the world, and inverse-transform that point by frame B to find the corresponding point in frame B's coordinates, and similarly for the other offset vectors.
All this problem has really done is just give you the rows of the matrix $$\M = A(B^{-1})\$$ and asked you to assemble them and use the resulting matrix to transform two vectors.
If you took the inverse of matrix $$\M\$$, you'd get $$\M^{-1} = B(A^{-1})\$$, the relative transformation from points in B back to A.