8$ Vectors and geometry in three dimensions
containing aik have been covered or removed. Thus for the determinant
As, we have
(2.53) 1Q 11 a22 a23
' (^412) = a21 a23, 1422 = all a13
1 a32 ass Ia31 ass asl ass
and for the determinant A4 we have
-411 =
a22 a23 a24
a32 a33 a34
a42 a43 a44
, 412=
a23 a23 a24
a33 a33 a34
a41 a43 a44
, -423 =
all a12
asi a32
a41 a42
a14
a34
a44
While it is possible to give more gruesome definitions, the number A3
can be defined by the formula
(2.54)
or
As = a11A11 - a121412 + aisA13
a22 a23
A3=a11I - a12
a32 asi
a21 - a23
a31 asi
- a13 a21 a22a31
a32
which makes sense because we know how to evaluate determinants of
order 2. The above formulas give the expansion of As in terms of the
elements of the first row. It can be proved that the same number As is
furnished by expansions in terms of another row or any column. Thus
As = -a23421 + a22422 - a23423
As = a31A31 - a32432 4- a33433
AS = a11A11 - a21421 a314s1
As = -a12412 + a22422 - a32432
As = a13A13 - a231423 + a33433.
In these expansions, the sign of the term involving ail, is plus whenever
the sum (or difference) of the subscripts is an even number like 0, ±2,
±4, and is minus when the sum (or difference) is odd like ±1,
±3, ±5,. To put this in other words, we can say that we get
a plus sign whenever aik lies on the main diagonal (running from the upper
left corner to the lower right corner), and that we get a change in sign
whenever we move one step right, left, down, or up.
Progress to determinants of order 4 and more is now easy. The
expansion of A4 in terms of the elements of the first row is
(2.55) A4 = all411 - a12A12 + a13 d1a - a141414
and A4 has seven more expansions, in terms of the otherrows and the
columns, all of which yield the same number A4. The usefulness of
the possibility of expanding a determinant of order 4 interms of elements