Linear Algebra - Change of basis matrices and RREF question what in the world?

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Linear Algebra - Change of basis matrices and RREF question what in the world??

Homework Statement



Suppose the linear transformation T: P3 -> P2, over R has the matrix

[tex] <br /> A = \begin{bmatrix}1&2&0&0\\0&1&2&1\\1&1&1&1 \end{bmatrix} <br /> [/tex]
relative to the standard bases of P3 and P2.

Find bases A' pf P3 and B' of P2 such that the matrix A' of T relative to A' and B' is the reduced row-echelon form for A.

Homework Equations





The Attempt at a Solution



Okay, so the relation is:
[tex] T_{B'A'} = I_{B'B}T_{BA}I_{AA'}[/tex]

Where A is the standard basis for for P3 and B is the standard basis for P2.

I also have the relation:

R = VA, where R is the RREF of A and V is an invertible matrix that maps A to R.

If I take the basis A' of A to be equal to A, then
[tex] I_{AA'} = <i>_{4x4}<br /> </i>[/tex]


and [tex] I_{B'B} = V[/tex]

Then I form an augmented matrix to find V:

[tex] \begin{bmatrix}1&2&0&0&|&1&0&0\\0&1&2&1&|&0&1&0\\1&1&1&1&|&0&0&1\end{bmatrix} [/tex]

which row reduces to

[tex] \begin{bmatrix}1&0&0&2/3&|&-1/3&2/3&4/3\\0&1&0&-1/3&|&2/3&-1/3&-2/3\\0&0&1&2/3&|&-1/3&1/3&1/3\end{bmatrix} [/tex]

Then V is the augmented part..
After this I will find the inverse of V, then find B'.. but it doesn't look right, and the final answer I got was wrong ~_~

Can someone confirm if this is correct track or if I made a mistake somewhere? Thanks.
 
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It looks like you just made an arithmetic error. V12 and V22 have the wrong sign.
 


Okay this really angers me I tried redoing it so many times and still can't get the right answer. Matrix operations really makes me angry.