I'm trying to solve the following Cramer's system $$AX=B \rightarrow \left\{\begin{matrix}x + y + z = 1\\ ax + by + cz = m\\ a^{2}x + b^{2}y + c^{2}z = m^{2}\\ \end{matrix}\right.$$
I tried to find the determinant of the A matrix: $$\begin{vmatrix}1 & 1 & 1\\ a & b & c\\ a^{2} & b^{2} & c^{2}\end{vmatrix}\underset{l_{2}-al_{1}, l_{3}-a^{2}l_{2}} {\rightarrow}\begin{vmatrix}1 & 1 & 1\\ 0 & b-a & c-a\\ 0 & b^{2}-a^{2} & c^{2}-a^{2}\end{vmatrix}\underset{l_{3}-(b+a)l_{2}} {\rightarrow}\begin{vmatrix}1 & 1 & 1\\ 0 & b-a & c-a\\ 0 & 0 & (c^{2}-a^{2}) - (c-a)\cdot(b+a) \end{vmatrix}$$
It seems like my determinant is wrong, because I'm getting the wrong solution to the system... Since the B matrix later substitutes a column in each of the 3 different matrices in the rule, the determinant obtained should be the same, i.e: $$\det(A_{b_{1}}) = \begin{vmatrix}1 & 1 & 1\\ m & b & c\\ m^{2} & b^{2} & c^{2}\end{vmatrix}$$ We would end up swapping a for m.
I've tried re-doing the problem over and over, but I end up getting the same wrong solution and I need help...
Note that the determinant $$ d = \begin{vmatrix}1 & 1 & 1\\ 0 & b-a & c-a\\ 0 & b^{2}-a^{2} & c^{2}-a^{2}\end{vmatrix} $$ can be easily found by cofactor expansion along the first column, you quickly get $$ \begin{split} d &= 1 \cdot \begin{vmatrix}b-a & c-a\\ b^2-a^2 & c^2-a^2\end{vmatrix} \\ &= (b-a)(c^2-a^2) - (b^2-a^2)(c-a) \\ &= (b-a)(c-a)[c+a - (b+a)] \\ &= (b-a)(c-a)(c-b). \end{split} $$