Prove the formula for the inverse of a matrix 
Assuming that the matrix $A = ||a_{ij}||_{1 \leq i,j \leq n}$ is invertible, write down the explicit formula for the inverse matrix $A^{-1} = ||b_{ij}||_{1 \leq i,j \leq n}$. Prove that this formula gives the inverse matrix. 

I wrote down the formula to be that every element of the inverse matrix is given by 
$$b_{ij} = \frac{1}{\det(A)} \cdot A_{ji}$$
where $A_{ji}$ is the algebraic complement of the element at row $j$ column $i$. Now I'm a little stuck on how to prove this. Would I be able to do it by induction? I.e show it works for some matrix where $n=2$ (this would be my base case), then somehow use that to show it for the $n= k$ and then $n=k+1$ which would be an $n=3$ matrix, something like this?
 A: It is straight consequence of Cramer's rule:
j-th column $v$ of inverse matrix should satisfy equation:
$$A v = e_j$$
Where $e_j$ is j-th vector of canonical basis.
So apply Cramer's rule to this equation and compare it with what you want to prove.
A: With the correct formula for $b_{i,j}$, just write down the coefficient $\sum_jA_{i,j}b_{j,k}$ of the matrix product. This gives $\frac1{\det A}$ times the Laplace expansion along row$~k$ of the determinant of the matrix obtained from$~A$ by replacing that row$~k$ by row$~i$ of $A$: the first factor $A_{i,j}$ of the summand traverses row$~i$ of$~A$, while the second factor $b_{j,k}$ traverses the cofactors for row$~k$ of$~A$ (due to the transposition in the formula); we copied row$~i$ to the position of row$~k$ in order to get a proper Laplace expansion of a matrix. Now if $i\neq k$ that new matrix has two equal rows, so the Laplace expansion is zero. But when $i=k$ we just get a Laplace expansion for an unmodified $A$, whose value is of course $\det A$, which cancels against the factor $\frac1{\det A}$. The conclusion is that
$$
  \sum_jA_{i,j}b_{j,k} = \delta_{i,k}
$$
in other words the product of $A$ and the proposed candidate for $A^{-1}$ is indeed the identity matrix. It is well known that for square matrices $AB=I$ implies $BA=I$ so one really has the inverse matrix; alternatively, the proof for the matrix product in the opposite order is quite similar, but using Laplace expansion by columns.
