Linear Algebra

Section I. Definition 295
determines singularity will involve applying Gauss’ method to the matrix, to
show that in the end the product down the diagonal is zero if and only if the
determinant formula gives zero. This suggests our initial plan: we will look for
a family of functions with the property of being unaffected by row operations
and with the property that a determinant of an echelon form matrix is the
product of its diagonal entries. Under this plan, a proof that the functions
determine singularity would go, “Where T →···→ ˆ T is the Gaussian reduction,
the determinant of T equals the determinant of ˆ T (because the determinant is
unchanged by row operations), which is the product down the diagonal, which
is zero if and only if the matrix is singular”. In the rest of this subsection we
will test this plan on the 2×2 and 3×3 determinants that we know. We will end
up modifying the “unaffected by row operations” part, but not by much.
The first step in checking the plan is to test whether the 2 × 2 and 3 × 3
formulas are unaffected by the row operation of pivoting: if
T kρi+ρj
−→ ˆ T
then is det( ˆ T ) = det(T )? This check of the 2×2 determinant after the kρ1 + ρ2
operation
�
a
det(
ka + c
�
b
)=a(kb + d) − (ka + c)b = ad − bc
kb+ d
shows that it is indeed unchanged, and the other 2×2 pivot kρ2 + ρ1 gives the
same result. The 3×3 pivot kρ3 + ρ2 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
= aei + bfg + cdh − hfa − idb − gec
as do the other 3×3 pivot operations.
So there seems to be promise in the plan. Of course, perhaps the 4 × 4
determinant formula is affected by pivoting. We are exploring a possibility here
and we do not yet have all the facts. Nonetheless, so far, so good.
The next step is to compare det( ˆ T ) with det(T ) for the operation
T ρi↔ρj
−→ ˆ T
of swapping two rows. The 2×2 rowswapρ1↔ρ2 � �
c d
det( )=cb − ad
a b
does not yield ad − bc. This ρ1 ↔ ρ3 swap inside of a 3×3 matrix
⎛ ⎞
g h i
det( ⎝d e f⎠)
=gec + hfa + idb − bfg − cdh − aei
a b c