Suppose we're working within the vector space of $2 \times 2$ matrices over $\mathbb{R}$. For any matrix, $$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix},$$ it should be the case that, irrespective of whether or not $A$ is invertible, there exists a matrix $B$ such that $$AB = BA = \det A \cdot I,$$ and $B$ is the matrix of cofactors.
In this lecture I am watching, the professor argues that the matrix of cofactors of the above matrix $A$ is given by $$\begin{pmatrix} d & -b \\ -c & a\end{pmatrix}.$$ Though this should be correct because the result gives $ad - bc$ along the main diagonal and zeroes everywhere else, which is a scalar multiple, $\det A$, of $I$, I am for some reason not able to rederive the result. If I take $A$ and compute the matrix of minors, I get: \begin{align*} \text{minor}(1,1) & = \det(d) = d \\ \text{minor}(1,2) & = \det(c) = c \\ \text{minor}(2,1) & = \det(b) = b \\ \text{minor}(2,2) & = \det(a).= a. \end{align*} The matrix of cofactors is, hence, $$\begin{pmatrix} d & -c \\ -b & a \end{pmatrix}.$$ This doesn't match, and I cannot figure out what I have done wrong, even though I am sure it is probably rather obvious.
Could someone take a look at this? In addition, I would be very interested in truly understanding the underlying purpose of these computations. This isn't the way I originally learned determinants, but I am sure there is more meaning to it than brute-force computation.
