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68 CHAPTER 3. POLYNOMIAL EQUATIONS
After clearing column two,we get
⎛
⎞
a b c d
0 −eb + af −ec + ag −ed + ah
M 3 :=
0 0 (−aj + ib) (−ec + ag) + (−aj + ib) (−ed + ah) +
.
(−eb + af) (ak − ic) (−eb + af) (al − id)
⎜
⎟
⎝ 0 0 (−an + mb) (−ec + ag) + (−an + mb) (−ed + ah) + ⎠
(−eb + af) (ao − mc) (−eb + af) (ap − md)
(3.20)
The result of the next step is better contemplated than printed!
However, if we do contemplate the result printed above, we see that rows
3 and 4 contain polynomials of degree four, whereas in the “simplified” form
(3.17) we only have polynomials of degree three in the numerators. Indeed, if
we were to expand the matrix above, we would observe that rows three and
four each had a common factor of a. Similarly, if we were to (or were to get a
computer algebra system to) expand and then factor the last step, we would get
a 2 (af −eb)|M|, as in equation (3.18). Such common factors are not a coïncidence
(indeed, they cannot be, since M is the most general 4 × 4 matrix possible).
Theorem 12 (Dodgson–Bareiss [Bar68, Dod66]) 7 Consider a matrix with
entries m i,j . Let m (k)
i,j be the determinant
∣ m 1,1 m 1,2 . . . m 1,k m 1,j ∣∣∣∣∣∣∣∣ m 2,1 m 2,2 . . . m 2,k m 2,j
. . . . . . . . . . . . . . . ,
m
∣ k,1 m k,2 . . . m k,k m k,j
m i,1 m i,2 . . . m i,k m i,j
i.e. that of rows 1 . . . k and i, with columns 1 . . . k and j. In particular, the
determinant of the matrix of size n whose elements are (m i,j ) is m (n−1)
n,n and
m i,j = m (0)
i,j . Then (assuming m(−1) 0,0 = 1):
m (k)
i,j = 1
m (k−2)
k−1,k−1
∣
m (k−1)
k,k
m (k−1)
i,k
m (k−1)
k,j
m (k−1)
i,j
∣ .
Proof. By fairly tedious induction on k.
How does this relate to what we have just seen?
If we do fraction-free
7 The Oxford logician Charles Dodgson was better known as Lewis Carroll.