Find Matrix T with respect to the basis beta

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.

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