# Linear operator in a given basis

Given a linear operator $$g$$ : $$R_{2}$$ $$to$$ $$R_{2}$$ maps a vector $$x$$=$$begin{pmatrix} 2\ -1 end{pmatrix}$$ and $$y$$= $$begin{pmatrix} -3\ 2 end{pmatrix}$$ to vectors $$a$$= $$g(x)$$=$$begin{pmatrix} 2\ 4 end{pmatrix}$$ and $$b$$= $$g(y)$$=$$begin{pmatrix} -1\ -2 end{pmatrix}$$

Find $$g$$ in the basis {$$x$$,$$b$$}

The approach followed by me was finding $$g$$ by using the mappings and solving a system of linear equations for $$g$$ which will be a 2×2 matrix. Then I found the basis by mapping the columns of vectors $$x$$ and $$b$$ as the following:

$$B=begin{pmatrix} 2 & -1 \ -1 & -2 end{pmatrix}$$ and $$g=begin{pmatrix} 3 & 4 \ 6 & 8 end{pmatrix}$$

and then using the below formula to calculate it in given basis {$$x$$,$$b$$} as:

$$g_{1}=B^{-1}gB$$ where $$g_{1}$$ is the linear operator in the given basis {$$x$$,$$b$$}

Can you please let me know if the approach to the above formulation is correct?