Prove that the output of the function equals the determinant Let $δ$ : $M_{2×2}$($F$) $→$ $F$ be a function with the following three properties.
($i$) $δ$ is a linear function of each row of the matrix when the other row is held fixed.
($ii$) If the two rows of $A$ ∈ $M_{2×2}$($F$) are identical, then $δ(A)$ =$ 0$.
($iii$) If $I$ is the $2 × 2$ identity matrix, then $δ(I)$ = $1$.
Prove that $δ(A)$ = det($A$) for all $A$ ∈ $M_{2×2}$$(F)$.
What I have so far: Suppose $A$ ∈ $M_{2×2}$($F$) and let $A$= $\ \left( \begin{array}{ccc} a&b\\
c&d \\
\end{array} \right)\ $
where $a$, $b$, $c$, $d$ are scalars. We know that $\delta$$\ \left( \begin{array}{ccc} 1&0\\
0&1\\
\end{array} \right)\ $ = $1$. We also know that If $A$=$\ \left( \begin{array}{ccc} 
d&e\\
d&e \\
\end{array} \right)\ $, where $d$ and $e$ are scalars, then $\delta$($A$)=$0$.
I am not sure what else to do from here...
 A: i think you can do this. let $r_1, r_2$ be two row vectors. 
first we can show that $$det \pmatrix{r_1\\r_2} = -\pmatrix{r_2\\r_1}$$
 by using the fact that $$det \pmatrix{r_1 + r_2\\r_1 + r_2} =0 $$
consider the determinant of a product two matrices: 
$$\begin{align}det \pmatrix{a & b\\c&d} \pmatrix{r_1\\r_2} &= det \pmatrix{ar_1+br_2\\cr_1+dr_2}\\
 &= det\pmatrix{a r_1\\cr_1} + det\pmatrix{a r_1\\dr_2} + det\pmatrix{b r_2\\cr_1}+ det\pmatrix{b r_2\\dr_2} \\
&=ad \, det\pmatrix{r_1\\r_2}+ bc\,det\pmatrix{r_2\\r_1}\\
&=(ad - bc)det\pmatrix{r_1\\r_2} \end{align}$$
now choose $$r_1 = (1,0)^T, r_2 = (0,1)^T$$ and you get $$det \pmatrix{a & b\\c&d} = ad - bc. $$
A: Write $\delta(\begin{smallmatrix}a&b\\c&d\end{smallmatrix})=f([a,b],[c,d])$ to simplify, and so that $f$ is an actual bilinear function. Then using linearity twice, and then property (ii) one gets:
$$
\begin{align}
  f ([a,b],[c,d]) &= af([1,0],[c,d])+bf([0,1],[c,d])
\\&=a(cf([1,0],[1,0])+df([1,0],[0,1]]))+b(df([0,1],[1,0])+df([0,1],[0,1]))
\\&=adf([1,0],[0,1])+bcf([0,1],[1,0]).
 \end{align}
$$
Now $f([1,0],[0,1])=1~$ by property (iii), and it to simplify the result to $ad-bc=\det(\begin{smallmatrix}a&b\\c&d\end{smallmatrix})$, it remains to see that one has $f([0,1],[1,0])=-1~$ as well. Applying property (ii) and then the equation above for the special case $[a,b]=[c,d]=[1,1]~$ gives
$$
  0=f([1,1],[1,1])=f([1,0],[0,1])+f([0,1],[1,0])=1+f([0,1],[1,0]),
$$
so indeed $f([0,1],[1,0])=-1~$, and we are done.

The final step is somewhat ad hoc for the $2\times2$ case; for more general determinants it would be best to prove in general that $f(\ldots,x,\ldots,y,\ldots)=-f(\ldots,y,\ldots,x,\ldots)~$ for any pair of arguments.
