derivation of formula to determine determinants Please explain the derivation of formula to determine determinant.
e.g., to calculate determinant of  why do we first multiply $a_{11}$ and $a_{22}$? Why not $a_{11}$ and $a_{21}$?  Also why do we then take the difference of the cross products , and not the sum?
 A: Ultimately, the determinant represents a volume.  To be specific, the determinant of an $n \times n$ matrix is the (signed) volume of the $n$-dimensional parallelepiped spanned by its row vectors.  For a $2 \times 2$ and perhaps even a $3 \times 3$ matrix, it is relatively simple to check that the determinant formula does indeed yield this volume.
Keeping this definition in mind, let's look at a $2 \times 2$ determinant with row vectors (a, b) and (c, d).

Can you see why the area of this parallelogram will be $ad - bc$?  If you're stuck, this graphic might help:

A: Consider the system $$\eqalign{ax+by=r\cr cx+dy=s\cr}$$ Multiply the first equation by $d$, the second, by $b$, and subtract the second from the first to get $$(ad-bc)x=rd-bs$$ and thus $$x={rd-bs\over ad-bc}$$ provided $ad-bc\ne0$. This is as good a reason as any for defining $$\det\pmatrix{a&b\cr c&d\cr}=ad-bc,\qquad\det\pmatrix{r&b\cr s&d\cr}=rd-bs$$
A: You have got it the wrong way. Actually you should first have asked question why do we multiply rows with columns in matrices? Why not rows with rows? What is the use of determinants? How do you think you will derive the formula? We don't even know what a determinant is!
The answer is that the whole topic matrices and determinants were developed to help us in other topics. We defined them this way. You are taught that$$\det\pmatrix{a&b\cr c&d\cr}=ab-bc$$
Better truth is $$ad-bc=\det\pmatrix{a&b\cr c&d\cr}$$
The term $ad-bc$ appears in so many contexts that we write it in this way.
A: Here is a way to visualize the $2\times 2$ case in a similar fashion as this other answer, but using an interactive visualization (sorry about the intrusive watermark, if anyone knows a good free gif recorder for iMac feel free to suggest):

(this was generated with Mathematica, see bottom for the code building up the interactive version of this visualization)
Denoting with $(a_1,a_2)$ the coordinates of the first vector and with $(b_1,b_2)$ those of the second vector, we can see that:


*

*The big black dashed rectangle has sizes $a_1\times b_2$ (and thus area $a_1 b_2$).

*The smaller rectangle delimited by the blue dashed line has dimensions $(a_2 b_1/b_2)\times b_2$ (and thus area $a_2 b_1$).


It then easily follows that the area of the parallelepiped is given by $a_1 b_2-a_2 b_1$.
Equivalently, one can use the parallelepiped obtained after only the first translation above, which is seen to have area given by
$$\left( a_1-a_2\frac{b_1}{b_2}\right) b_2=a_1 b_2-a_2 b_1.$$

Run the following code on a Mathematica notebook to get the interactive visualization from which the gif above is made:
DynamicModule[{a1 = 3, a2 = 1, b1 = 1, b2 = 2}, Manipulate[
  Graphics[{
    {Green, Polygon@{
       {0, 0},
       {a1 - a2 b1/b2, a2 - a2/a1 (a2 b1/b2)},
       {a1 - a2 b1/b2, b2},
       {b1, b2}
       }}
    , {Purple, Translate[#, {b1, b2} t] &@Polygon@{
        {0, 0}, {a1 - a2 b1/b2, 0}, {a1 - a2 b1/b2, 
         a2 - a2/a1 (a2 b1/b2)}
        }}
    , {Purple, 
     Translate[#, {b1, b2} t - {a1 - a2 b1/b2, 0} t2] &@Polygon@{
        {a1 - a2 b1/b2, 0}, {a1, a2}, {a1 - a2 b1/b2, 
         a2 - a2/a1 (a2 b1/b2)}
        }}
    , {Orange, Translate[#, -{a1 - a2 b1/b2, 0} t2] &@Polygon@{
        {a1 - a2 b1/b2, a2 - a2/a1 (a2 b1/b2)},
        {a1, a2},
        {a1 + b1 (b2 - a2)/b2, b2},
        {a1 - a2 b1/b2, b2}
        }}
    , {EdgeForm@{Dashed, Thick}, FaceForm@None, Opacity@0.2, Rectangle[
      {0, 0}, {a1, b2}
      ]}
    , {EdgeForm[{Blue, Dashed}], FaceForm@None, Opacity@0.2, Rectangle[
      {a1 - b1 a2/b2, 0}, {a1, b2}
      ], Dashed, Blue, Opacity@1, 
     Line@{{a1 - b1 a2/b2, 0}, {a1 - b1 a2/b2, -0.5}}}
    , Locator@Dynamic@{a1, a2}, Locator@Dynamic@{b1, b2}
    , With[{hx1 = -0.3}, {Arrowheads@{-.04, .04}, 
      Arrow@{{0, hx1}, {a1, hx1}}}]
    , With[{hx2 = -0.5}, {Arrowheads@{-.04, .04}, 
      Arrow@{{0, hx2}, {a1 - b1 a2/b2, hx2}}}]
    }, Frame -> True, PlotRange -> ({#, #} &@{-1, 5})],
  {{t, 1}, 0, 1, 0.01, Appearance -> "Labeled"},
  {{t2, 0}, 0, 1, 0.01, Appearance -> "Labeled"}, 
  ControlPlacement -> Right
  ]]

