Page 401 - 35Linear Algebra

G.7 Determinants 401

j 0 0j
Now note that for two upper-triangular matrices U = (u ) and U = (u ),
i i
0
j
by matrix multiplication we have X = UU = (x ) is upper-triangular and
i
i
i 0i
x = u u . Also since every permutation would contain a lower diagonal entry
i i i
Q i 0
(which is 0) have det(U) = i u . Let A and A have corresponding upper-
i
0
triangular matrices U and U respectively (i.e. det(A) = det(U)), we note
0
0
that AA has a corresponding upper-triangular matrix UU , and hence we have
0
0
i 0i
det(AA ) = det(UU ) = Y u u
i i
i
! !
Y i Y 0i
= u i u i
i i
0
0
= det(U) det(U ) = det(A) det(A ).
Practice taking Determinants
Lets practice taking determinants of 2 × 2 and 3 × 3 matrices.
For 2 × 2 matrices we have a formula

a b
det = ad − bc .
c d
This formula might be easier to remember if you think about this picture.

Now we can look at three by three matrices and see a few ways to compute
the determinant. We have a similar pattern for 3 × 3 matrices. Consider the
example

 
1 2 3
det  3 1 2  = ((1 · 1 · 1) + (2 · 2 · 0) + (3 · 3 · 0)) − ((3 · 1 · 0) + (1 · 2 · 0) + (3 · 2 · 1)) = −5
0 0 1

We can draw a picture with similar diagonals to find the terms that will be
positive and the terms that will be negative.

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