Find a matrix $P$ such that $[\overrightarrow{\rm x}]_C$ = $P[\overrightarrow{\rm x}]_B$ for all x in $\mathbb{R}^2$ Let $B$ be the basis of $\mathbb{R}^2$ consisting of the vectors 

${\begin{bmatrix}
5\\ 
-2
\end{bmatrix}
, 
\begin{bmatrix}
2\\ 
3
\end{bmatrix}}$

and let $C$ be the basis consisting of 

${\begin{bmatrix}
-2\\ 
3
\end{bmatrix} , 
\begin{bmatrix}
1\\ 
-2
\end{bmatrix}}$



I tried doing $ P =C*B^{-1}$ in which I got

$\begin{bmatrix}
\frac{-4}{19} &amp; \frac{9}{19}\\ 
\frac{5}{19} &amp; \frac{-16}{19}
\end{bmatrix}$

but when I went to check it, it was wrong.
Note that for any $x \in \mathbb R^2$, we can write $x =  x_1(5,-2) + x_2(2,3) = (5x_1 + 2x_2, 3x_2 - 2x_1)$, and it can also be written as $x = x_3(-2,3) + x_4(1,-2) =(x_4-2x_3,3x_3-2x_4)$.

Equating these, we get:$5x_1+2x_2 = x_4-2x_3, 3x_2-2x_1 = 3x_3-2x_4$. Now, solve for $x_3,x_4$, we get :
$$
x_3 = -8x_1-7x_2 , x_4 = -11x_1-12x_2
$$ 
and therefore,
$$
\begin{pmatrix} x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix} -8 \quad -7 \\ -11 \quad -12\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}
$$
where $x_3,x_4$ is the representation of $x$ in basis $C$, and $x_1,x_2$ is the representation of $x$ in basis $B$, both as column vectors.
Since you are given $xC = PxB$

**The Answer is: $P=C^{-1} xB$ **