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Let T : R^3 -> R^3 be the linear transformation

(a) The matrix of T with respect to the standard basis alpha of R^3

(b) The matrix of T with respect to the basis beta

$$ T\begin{pmatrix} x \\ y \\ z \\ \end{pmatrix}= \begin{pmatrix} 5x & + & z \\ 3x & +2y & -3z \\ 5x \\ \end{pmatrix} $$ For the standard basis alpha and the basis beta given by $$ \begin{matrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & 1 \\ \end{matrix} $$ (a) $$ \begin{pmatrix} T \end{pmatrix} = \begin{pmatrix} 5 & 0 & 1 \\ 3 & 2 & -3 \\ 5 & 0 & 0 \\ \end{pmatrix} $$ (b)

$$ T\begin{pmatrix} 1 \\ 1 \\ 2 \\ \end{pmatrix}= \begin{pmatrix} 7 \\ -1 \\ 5 \\ \end{pmatrix} $$

This is in my textbook i can't seem to figure out where the 7,-1,5 vector comes from. This is in terms of beta. Any help is really appreciated it.

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  • $\begingroup$ Just apply the first formula for vector (1,1,2) and you get (7,-1,5) $\endgroup$ – Andrei Nov 12 '18 at 21:23
  • $\begingroup$ Wow ok, thank you.. I think im retarded. $\endgroup$ – Bryce Blake Nov 12 '18 at 21:48

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