Suppose we have an nxn matrix A, and $A\underline{x} = \underline{p}$ prove that $x_i = \dfrac{\det(A_i)}{\det(A)}$ where $A_i$ is the matrix obtained from A by replacing the $i^{th}$ column by $\underline{p}$
The question hints to multiply the equation by $adj A$
Following the hint:
$adj(A) A \underline{x} = adj A \underline{p}$
$$adj A \underline{p} = \sum_{j=1}^n adj A p_j = \sum_{j=1}^np_jC_{ji}$$ where $C_{ji}$ is a cofactor of A, from here I just concluded that $\sum_{j=1}^np_jC_{ji} = \det(A_i)$ so
$$adj(A) A \underline{x} = \det{A_i} \implies \dfrac{adj(A)}{\det(A)}A \underline{x} = \dfrac{\det(A_i)}{\det(A)} $$ so using $\dfrac{adj(A)}{\det(A)} = A^{-1}$ we get $\underline{x} = \dfrac{\det(A_i)}{\det(A)} $
Now to be honest, I'm pretty bad with matrix proofs which is why I ended up with $$\underline{x} = \dfrac{\det(A_i)}{\det(A)} $$
instead of
$$x_i = \dfrac{\det(A_i)}{\det(A)} $$
Could someone help me clean this proof up, and how could I justify $\sum_{j=1}^np_jC_{ji} = \det(A_i)$?
thank you