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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?

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    $\begingroup$ If you ONLY want to show that interchanging the two rows of a $2\times2$ matrix has the effect of multiplying the determinant by $-1,$ then why not just compute $\det\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]$ and $\det\left[ \begin{array}{cc} c & d \\ a & b \end{array} \right] \text{ ?} \qquad$ $\endgroup$ – Michael Hardy Jul 16 '18 at 19:13
  • $\begingroup$ Well, first you need to define what the entries of $E$ are, otherwise you cannot simplify $E_{i1}A_{1j}+E_{i2}A_{2j}$ any further. $\endgroup$ – Rahul Jul 16 '18 at 19:21
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    $\begingroup$ @Michael: Yes that is a good idea, and that is certainly the most basic route to correctly writing the proof. I just wanted to go a bit more in depth here for my own benefit. $\endgroup$ – greycatbird Jul 16 '18 at 19:29
  • $\begingroup$ @Rahul: Thank you for pointing that out - I did forget to define the matrix E. I spend so much time within the Friedberg book I sometimes take that notation for granted. $\endgroup$ – greycatbird Jul 16 '18 at 19:29
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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 $$(EA)_{ij} = \begin{bmatrix}A_{21} & A_{22} \\ A_{11} & A_{12}\end{bmatrix}_{ij} = \begin{bmatrix}c & d \\ a & b\end{bmatrix}_{ij}$$

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  • $\begingroup$ Thank you, and I think this response helped the most. I was having severe troubles with the nuts and bolts of the summation. $\endgroup$ – greycatbird Jul 16 '18 at 19:36
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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 $B$

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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 $\det \left( B \right) = \det \left( E \right)\det \left( A \right) = - \det \left( A \right)$.

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  • $\begingroup$ @ John: Thank you for the feedback. This is actually the approach I wanted to take, but in Friedberg the result you used comes well after the basic definitions I used. I wrote this proof as if my professor was grading it, and I know he would not allow theorems from later in the book to be used in earlier sections. $\endgroup$ – greycatbird Jul 16 '18 at 19:25
  • $\begingroup$ Then all you really want is the proof that the determinant of the product is the product of the determinants (this is a special case). Why don't you just follow that in general and then specialize it to this particular case? $\endgroup$ – John Polcari Jul 16 '18 at 19:30

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