Edit complete, but it doesn't seem as though I can change the title.
The latex arrows next to the 'P' aren't showing up for me but they're supposed to be left arrows

1. The problem statement, all variables and given/known data
Let B and C be bases of R^2. Find the change of basis matrices [itex]P_{B \leftarrow C}[/itex] and [itex]P_{C\leftarrow B}[/itex]

To find the change of basis matrix from C to B [itex]P_{B \leftarrow C}[/itex] , I followed the steps in Lay's Linear Algebra book and found the coordinates of the B vectors relative to C.

But in the solutions, this is the change of basis matrix for going from C to B, i.e. [itex]P_{C\leftarrow B}[/itex]. Have I misinterpreted something? The steps I'm referring to are based on this theorem on page 273 of the 3rd edition of the text by Lay:

If you want a matrix like this:
[tex] \pmatrix{7/4 & -2/4 \\ 1/4 & 2/4}[/tex]
use the "\ p matrix" command (remove spaces!) To get the above I just entered "\ p matrix{ 7/4 & -2/4 \\ 1/4 & 2/4}". In LaTeX/TeX, the '&' separates elements in the same row and the "\\" starts a new row.

Anyway, you are encountering the classic confusion between the matrix that changes the basis vs. the matrix that changes the ccomponents. So, if {u1, u2} and {v1,v2} are two bases (each ui or vi is a two-dimensional vector), then if
[tex] \pmatrix{u_1\\u_2} = \pmatrix{a&b\\c&d} \pmatrix{v_1\\v_2} [/tex]
and if ##x = x_1 u_1 + x_2 u_2 = y_1 v_1 + y_2 v_2## we have
[tex] (y_1 \;y_2) = (x_1 \; x_2) \pmatrix{a &b\\c &d}[/tex]