Linear Algebra

304 Chapter 4. Determinants
conflict: here are two Gauss’ method reductions of the same matrix, the first
without any row swap
� � � �
1 2 −3ρ1+ρ2 1 2
−→
3 4
0 −2
and the second with a swap.
� �
1 2 ρ1↔ρ2
−→
3 4
� � � �
3 4 −(1/3)ρ1+ρ2 3 4
−→
1 2
0 2/3
Following Definition 2.1 gives that both calculations yield the determinant −2
since in the second one we keep track of the fact that the row swap changes
the sign of the result of multiplying down the diagonal. But if we follow the
supposition and change the second condition then the two calculations yield
different values, −2 and 2. That is, under the supposition the outcome would not
be well-defined — no function exists that satisfies the changed second condition
along with the other three.
Of course, observing that Definition 2.1 does the right thing in this one
instance is not enough; what we will do in the rest of this section is to show
that there is never a conflict. The natural way to try this would be to define
the determinant function with: “The value of the function is the result of doing
Gauss’ method, keeping track of row swaps, and finishing by multiplying down
the diagonal”. (Since Gauss’ method allows for some variation, such as a choice
of which row to use when swapping, we would have to fix an explicit algorithm.)
Then we would be done if we verified that this way of computing the determinant
satisfies the four properties. For instance, if T and ˆ T are related by a row swap
then we would need to show that this algorithm returns determinants that are
negatives of each other. However, how to verify this is not evident. So the
development below will not proceed in this way. Instead, in this subsection we
will define a different way to compute the value of a determinant, a formula,
and we will use this way to prove that the conditions are satisfied.
The formula that we shall use is based on an insight gotten from property (2)
of the definition of determinants. This property shows that determinants are
not linear.
3.1 Example For this matrix det(2A) �= 2· det(A).
� �
2 1
A =
−1 3
Instead, the scalar comes out of each of the two rows.
�
�
� 4
�−2 �
2�
�
6�
=2·
�
�
� 2
�−2 �
1�
�
6�
=4·
�
�
� 2
�−1 �
1�
�
3�
Since scalars come out a row at a time, we might guess that determinants
are linear a row at a time.