- #1

- 5

- 0

## Main Question or Discussion Point

Im trying to figure out how to do this question. This is an example in the book i have. Im not sure how they got the answer.

Here is the example from the book:

Find the Transition Matrix P from the basis B={t+1, 2t, t-1} to B'={4t[tex]^{2}[/tex]-6t, 2t[tex]^{2}[/tex]-2, 4t} for the space R[t].

A little computataion shows that 4t[tex]^{2}[/tex]-6t:(-3, 2, -3), 2t[tex]^{2}[/tex]-2: (-1, 1, 1) and 4t:(2, 0, 2). Therefore

[tex]P=\left(\begin{array}{ccc}-3 & -1 & 2 \\ 2 & 1 & 0\\ -3 & 1 & 2\end{array}\right)[/tex]

I'd like to know how they found 4t[tex]^{2}[/tex]-6t to be(-3, 2, -3), 2t[tex]^{2}[/tex]-2 to be (-1, 1, 1) and 4t to be(2, 0, 2). Any help is apprectated.

Here is the example from the book:

Find the Transition Matrix P from the basis B={t+1, 2t, t-1} to B'={4t[tex]^{2}[/tex]-6t, 2t[tex]^{2}[/tex]-2, 4t} for the space R[t].

A little computataion shows that 4t[tex]^{2}[/tex]-6t:(-3, 2, -3), 2t[tex]^{2}[/tex]-2: (-1, 1, 1) and 4t:(2, 0, 2). Therefore

[tex]P=\left(\begin{array}{ccc}-3 & -1 & 2 \\ 2 & 1 & 0\\ -3 & 1 & 2\end{array}\right)[/tex]

I'd like to know how they found 4t[tex]^{2}[/tex]-6t to be(-3, 2, -3), 2t[tex]^{2}[/tex]-2 to be (-1, 1, 1) and 4t to be(2, 0, 2). Any help is apprectated.