Linear Algebra II - Change of Basis

[lemonsare]

Homework Statement

From Linear Algebra with applications 7th Edition by Keith Nicholson.
Chapter 9.2 Example 2.

Let T: R³ → R³ be defined by T(a,b,c) = (2a-b,b+c,c-3a).
If B₀ denotes the standard basis of R³ and B = {(1,1,0),(1,0,1),(0,1,0)}, find an invertible matrix P such that P^-1M_B0(T)P=M_B(T).

Homework Equations

The Attempt at a Solution

I know to find P, I have to first find M_B0(T) and M_B(T).

M_B0(T) is easy because I have to just find each a, b, and c as linear combinations of B₀ and the coefficients are M_B0(T).

However, I'm not sure how to find M_B(T). In the textbook, they write:

M_B(T) = [C_B(1,1,-3) C_B(2,1,-2) C_B(-1,1,0)]

I cannot figure out where they came up with these numbers (1,1,-3), (2,1,-2), and (-1,1,0).

Please help. Thank you!

 

[RUber]

$M_B(T)$ is the 3x 3 matrix where the columns are the result of T on each of the basis vectors.
T(1,1,0) = (1,1,-3).
So your matrix will look like:
$\begin{bmatrix} 1& 2& -1 \\ 1& 1& 1 \\ -3 & -2 & 0 \end{bmatrix}$

 

[lemonsare]

Oh wow thanks! That was simple.

But in the textbook $M_B(T)$ = $\begin{bmatrix} 4&4&-1\\-3&-2&0\\-3&-3&2\end{bmatrix}$

From trying to figure out where (1,1,-3) came from, I figured out that (4,-3,-3) came from the coefficients of the linear combination of each of the basis vectors. But is that wrong?