Issue with Matrix Multiplication using the Formal Definition I am writing a formal proof to show that if $B$ is the matrix obtained by interchanging the rows of a $2\times2$ matrix $A$, then $\det(B)=-\det(A)$.  My reasoning and proof are coming along nicely but I hit a bump in the road that highlighted to me a gap in my knowledge - that is, I guess I do not completely understand the definition of matrix multiplication.  Note I went the rigorous route here only because I wanted to prove to myself I fully understood matrix multiplication... and I don't.  My proof thus far is:

Let $E$ be the elementary matrix obtained by performing a type 1 elementary row operation on $I_2$.  By Theorem 3.1 (Friedberg), $B=EA$.  Note
$$\det(A) =\det\begin{pmatrix}
    a & b \\
    c & d
    \end{pmatrix}=ad-bc$$ By the definition of matrix multiplication,
\begin{align}
& B_{ij}=(EA)_{ij} \\[10pt]
= {} & \sum_{k=1}^2 E_{ik}A_{kj} \text{ for } 1\le i\le2\text{, }1\le j\le2 \\[10pt]
= {} & E_{i1}A_{1j}+E_{i2}A_{2j}\text{ for } 1\le i\le2\text{, }1\le j\le2 \\[10pt]
\vdots\,\,\, \\[10pt]
= {} & \begin{pmatrix}
    c & d \\
    a & b
    \end{pmatrix}_{ij}
\end{align}
If $B=EA=\begin{pmatrix}
    c & d \\
    a & b \\
    \end{pmatrix}$, then by the definition of a determinant of a $2\times2$ matrix,
\begin{align}
\det(B) & =\det(EA)=bc-ad \\[10pt]
& =-(ad-bc) \\[10pt]
& =-\det(A)
\end{align}

My issue is, how do I formally express the steps where I put my "$\cdots$"?  That is, the column and row vector multiplication and their sum? Maybe I'm wrong, but I feel most sources don't fully explain all the steps of matrix multiplication and just resort to hand-waving.  The way I think about it - the column and row vectors I will be multiplying in my proof are actually just $2\times1$ and $1\times2$ matrices, respectively.  I know they result in a $2\times2$ matrix, but how?  And why?
 A: If I understood correctly, the matrix $E = \begin{bmatrix}E_{11} & E_{12} \\ E_{21} & E_{22}\end{bmatrix}$ is given by
$$E = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$$
so for $1\le i,j\le 2$ we have$$(EA)_{ij} = \sum_{k=1}^2E_{ik}A_{kj} =  E_{i1}A_{1j}+E_{i2}A_{2j}$$
If $i= 1$ then $$(EA)_{1j} = E_{11}A_{1j}+E_{12}A_{2j} = 0 \cdot A_{1j}+1\cdot A_{2j} = A_{2j}$$
If $i= 2$ then $$(EA)_{2j} = E_{21}A_{1j}+E_{22}A_{2j} = 1 \cdot A_{1j}+0\cdot A_{2j} = A_{1j}$$
So $$(EA)_{ij} = \begin{bmatrix}A_{21} & A_{22} \\ A_{11} & A_{12}\end{bmatrix}_{ij} = \begin{bmatrix}c & d \\ a & b\end{bmatrix}_{ij}$$
A: Notice that since $E$ is obtained from switching the rows on the identity matrix, then the elements of $E$ are either $0$ or $1$.  So, for your sum you will have something like $$B_{ij} = E_{i1} A_{1j} + E_{i2} A_{2j} = 0(A_{1j}) + 1(A_{2j}) = A_{2j}$$
Since you are only considering $2 \times 2$ matrices, then you only need to check a few options.
Using this you can determine the elements of the matrix $B$
A: Don't see a real issue here.  A simple way to show your desired result is $B = EA$ with
$$E = \left[ {\begin{array}{*{20}{c}}
0&1\\
1&0
\end{array}} \right]\quad  \Rightarrow \quad \det \left( E \right) =  - 1$$
so that $\det \left( B \right) = \det \left( E \right)\det \left( A \right) =  - \det \left( A \right)$.
