Linear Algebra, 2020a

Section I. Definition 327
A good strategy for solving problems is to explore which properties the
solution must have, and then search for something with those properties. So we
shall start by asking what properties we’d like the determinant formulas to have.
At this point, our main way to decide whether a matrix is singular or not is
to do Gaussian reduction and then check whether the diagonal of the echelon
form matrix has any zeroes, that is, whether the product down the diagonal
is zero. So we could guess that whatever determinant formula we find, the proof
that it is right may involve applying Gauss’s Method to the matrix to show that
in the end the product down the diagonal is zero if and only if our formula gives
zero.
This suggests a plan: we will look for a family of determinant formulas that
are unaffected by row operations and such that the determinant of an echelon
form matrix is the product of its diagonal entries. In the rest of this subsection
we will test this plan against the 2×2 and 3×3 formulas. In the end we will
have to modify the “unaffected by row operations” part, but not by much.
First we check whether the 2×2 and 3×3 formulas are unaffected by the row
operation of combining: if
T
kρ i +ρ j
−→
ˆT
then is det(ˆT) =det(T)? This check of the 2×2 determinant after the kρ 1 + ρ 2
operation
(
)
a b
det(
)=a(kb + d)−(ka + c)b = ad − bc
ka + c kb+ d
shows that it is indeed unchanged, and the other 2×2 combination kρ 2 +ρ 1 gives
the same result. Likewise, the 3×3 combination kρ 3 + ρ 2 leaves the determinant
unchanged
⎛
⎞
a b c
⎜
⎟
det( ⎝kg + d kh+ e ki+ f⎠) =a(kh + e)i + b(ki + f)g + c(kg + d)h
g h i − h(ki + f)a − i(kg + d)b − g(kh + e)c
as do the other 3×3 row combination operations.
= aei + bfg + cdh − hfa − idb − gec
So there seems to be promise in the plan. Of course, perhaps if we had
worked out the 4×4 determinant formula and tested it then we might have found
that it is affected by row combinations. This is an exploration and we do not
yet have all the facts. Nonetheless, so far, so good.
Next we compare det(ˆT) with det(T) for row swaps. Here we hit a snag: the