Let $M_a= \begin{pmatrix} 3&0&0 \\ 0&a&0 \\ 0&1&3 \end{pmatrix}$ and let $f_a:\Bbb R^3\to\Bbb R^3$ be defined by $f_a(x,y,z)=(ax,a(x+y)+2z,ax-2y+5z).$
What are the values of $a$ such that for some basis $B$ of $\Bbb R^3$, the representative matrix of $f_a$ with respect to $B$ is $M_a$?
My idea was that $M=\begin{pmatrix} a&0&0 \\ a&a&2 \\ a&-2&5 \end{pmatrix}$is the representative matrix of $f_a$ with respect to the canonical basis, and, but I'm not sure about it, we should have $M_a=(M_C^B)^{-1}\cdot M\cdot M_C^B$ where $M_C^B$ is the change of basis matrix from $B$ to the canonical basis $C$ - its columns should be "the vectors of the coordinates of $B$ with respect to $C$", but I'm having problems with this.
I tried to let the columns be $\begin{pmatrix} b_{11} \\ b_{21} \\ b_{31} \end{pmatrix}$, $\begin{pmatrix} b_{12} \\ b_{22} \\ b_{32} \end{pmatrix}$, $\begin{pmatrix} b_{13} \\ b_{23} \\ b_{33} \end{pmatrix}$, and solving for the components, but then what I find is not a basis for any $a$