Proof: Inverse of a Matrix

Question

I know that to find the inverse of a matrix, I need to divide 1 by the determinant of the matrix followed by multiplying it by the adjugate of the matrix. However, what is the proof for this? I know the proof involves using a vector of sorts which results in a system of equations that can be solved to result in the above equation.

Answer 1

We let $A(i;j)$ be the matrix you get by crossing out $A$'s $i$-th row and $j$-th column.

Let $X$ be the adjugate of $A$. That means, that: $$X_{i,k} = (-1)^{i+k}\det(A(k;i))$$

We want to prove, that: $XA = \det(A)I$. Then the result follows, if $A$ is invertible (that is: $\det(A) \neq 0$). We have:

$$(XA)_{i,j} = \sum_{k=1}^n X_{i,k} A_{k,j} = \sum_{k=1}^n (-1)^{i+k}A_{k,j}\det(A(k;i))$$

If $i=j$, then $(XA)_{i,i} = \det(A)$ by the Laplace-Formula for determinants. So the diagonal entries of $XA$ are all $\det(A)$.

Suppose $i\neq j$. Then:

$$(XA)_{i,j} = \sum_{k=1}^n (-1)^{i+k}A'_{k,i}\det(A'(k;i)) = \det(A')$$

where $A'$ results from replacing the $i$-th column of $A$ with it's $j$-th column, again by the Laplace-Formula for determinants and noticing that $A'_{k,i} = A_{k,j}$ and $\det(A(k;i))=\det(A'(k;i))$. Because $A'$ has the same column twice, we have $\det(A')=0$. So all non-diagonal entries of $XA$ are $0$.

Therefore: $XA = \det(A)I$.