# Linear Transformation from $\alpha$ to $\beta$

T: $R^3$ $\to$ $R^2$

$$[T]_{\beta\alpha} = \begin{matrix} 2 & 3 & 1 \\ 1 & 2 & 1 \\ \end{matrix}$$

$\alpha$ = {(1, -1, 1), (0, 1, 0), (1, 0, 0)}

$\beta$ = {(3, 2), (2, 1)}

Find: T((x, y, z)) for any x, y, z in $R^3$

My approach:

$[T(v)]_{\beta}$ = $[T]_{\beta\alpha}$$[v]_{\alpha} so, z\begin{pmatrix} 1 \\ -1 \\ 1 \\ \end{pmatrix} + (y+z)\begin{pmatrix} 0 \\ 1 \\ 0 \\ \end{pmatrix} +(x-z)\begin{pmatrix} 1 \\ 0 \\ 0 \\ \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} and [(x, y, z)]_{\alpha} = [(z), (y + z), (x - z)] then [T]_{\beta\alpha}[\begin{pmatrix}x\\y\\z\\\end{pmatrix}]_{\alpha}=\begin{pmatrix}x + 3y + 4z\\x + 2y + 2z\\\end{pmatrix} finally, (x + 3y + 4z)\begin{pmatrix}3\\2\\\end{pmatrix} + (x + 2y + 2z)\begin{pmatrix}2\\1\\\end{pmatrix} = \begin{pmatrix}5x + 13y + 16z\\3x + 8y + 10z\\\end{pmatrix} and T((x, y, z)) = ((5x + 13y + 16z), (3x + 8y + 10z)) But this does not look like it makes lot of sense, I think I misunderstand something. Can anyone give me a pointer? - ## 1 Answer it is correct in my humble opinion except rhe last step. finally,$$(x + 3y + 4z)\begin{pmatrix}2\\1\\\end{pmatrix} + (x + 2y + 2z)\begin{pmatrix}3\\2\\\end{pmatrix} =...$$instead of$$(x + 3y + 4z)\begin{pmatrix}3\\2\\\end{pmatrix} + (x + 2y + 2z)\begin{pmatrix}2\\1\\\end{pmatrix} = \begin{pmatrix}5x + 13y + 16z\\3x + 8y + 10z\\\end{pmatrix}$\$

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Oh, I actually made a mistake while posting it, as the order is (3, 2), (2, 1), that is why I calculated this way – tmac_balla Mar 29 '14 at 19:03