Let T: R^2 --> R^3 be defined by T(a1,a2)=(a1-a2,a1,2a1+a2) Let beta be the standard basis for R^2 and gamma={(1,1,0),(0,1,1),(2,2,3)}.
compute [T]_beta to gamma
"beta be the standard basis for R^2" => T(1,0) = (1,1,2) and T(0,1) = (-1,0,1)
The question essentially asks:
We want to write (a1-a2,a1,2a1+a2) as a linear combination of gamma. Let the coefficients of this linear combination be x,y,z in that order.
Then given (a1,a2) in R^2, we want to find a matrix that maps it to T(a1,a2) which is written as a linear combination of gamma.
Thus:
T(a1,a2) = (a1-a2,a1,2a1+a2) = x(1,1,0)+y(0,1,1)+z(2,2,3)
But observe that:
$\displaystyle (a1-a2,a1,2a1+a2) = \begin{pmatrix}1 & -1 \\ 1 & 0 \\ 2 & 1\end{pmatrix} \begin{pmatrix}a1\\ a2\end{pmatrix}$
$\displaystyle x(1,1,0)+y(0,1,1)+z(2,2,3) = \begin{pmatrix}1 & 0 & 2 \\ 1 & 1 & 2 \\ 0 & 1 & 3\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}$
So:
$\displaystyle \begin{pmatrix}1 & -1 \\ 1 & 0 \\ 2 & 1\end{pmatrix} \begin{pmatrix}a1\\ a2\end{pmatrix} = \begin{pmatrix}1 & 0 & 2 \\ 1 & 1 & 2 \\ 0 & 1 & 3\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}$
This means,
$\displaystyle \begin{pmatrix}1 & 0 & 2 \\ 1 & 1 & 2 \\ 0 & 1 & 3\end{pmatrix}^{-1}\begin{pmatrix}1 & -1 \\ 1 & 0 \\ 2 & 1\end{pmatrix} \begin{pmatrix}a1\\ a2\end{pmatrix} = \begin{pmatrix}x\\ y\\ z\end{pmatrix}$
Thus the matrix $\displaystyle \begin{pmatrix}1 & 0 & 2 \\ 1 & 1 & 2 \\ 0 & 1 & 3\end{pmatrix}^{-1}\begin{pmatrix}1 & -1 \\ 1 & 0 \\ 2 & 1\end{pmatrix}$ maps (a1,a2) to (x,y,z)
I will let you compute that matrix....