Applications of the Determinant
Learning Goals: introduce applications of the determinant, such as area/volume finding, a
formula for the inverse, and Cramer’s rule.
We started by defining determinants as the volumes of n-dimensional boxes. Thus, it is
no surprise that we may use it to do just that. By its very definition, if the corners of a
parallelogram are at (0, 0), (a, b), (c, d) and (a + c, b + d), then the area of the parallelogram is
a b
(the absolute value of)
, and the same thing works in many dimensions as well. Let’s
c d
apply this to get more general areas.
x1 y1 1
1
Theorem: in the plane, a triangle with corners (x1, y1), (x2, y2) and (x3, y3) has area x2 y2 1 .
2
x 3 y3 1
Proof: The vector from (x3, y3) to (x2, y2) is (x3 – x2, y3 – y2) and similarly for the vector
from (x3, y3) to (x1, y1). So by subtracting the third row from the first and second in the
matrix (not changing its determinant!) we effectively slide the triangle along so that one
x1 ! x3 y1 ! y3 0
of its corners is at the origin. Now the determinant of x2 ! x3 y2 ! y3 0 can be
x3
y3
1
expanded along the third column and the necessary determinant gives the area.
This is the basis for the “shoelace method” for finding area of arbitrary polygons in the
plane. To find the area of a polygon whose vertices are at (x1, y1), (x2, y2), …, (xn, yn) going
around counterclockwise, form the two-column matrix
! x1 y1 $
C
#x y &
2&
B
# 2
# ! ! & where the (x1, y1) is repeated at the end. Now take
#
&
# xn yn &
D
#" x1 y1 &%
all the “downward” products, x1y2, x2y3, …, xny1 and add them.
Then subtract all the “backward” products, y1x2, y2x3, …, ynx1.
Take half of this and you have the area. This is called the
shoelace method because you can put criss-crosses down the
O
A
center of the matrix to remind you of the products you need to
take. It works because you are taking a bunch of determinants, halving them (to find areas of
triangles) and adding the results. Some of these determinants will be negative because the
corresponding triangle is traversed clockwise, but that means the area should be subtracted
anyway, as the is area not included in the figure. For instance in the figure at right, the area of
quadrilateral ABCD is that of ΔOAB plus ΔOBC, minus that areas inside ΔOCD and ΔODA.
Formula for the inverse
There is a nice way to use determinants to find a direct formula for the inverse of a matrix
when it has one. Consider the cofactor expansion along the first row. Thus we have det(A) =
a11A11 + a12A12 +  + a1nA1n along the top row. What if, instead, we used the second row of a’s
instead of the first row? What is a21A11 + a22A12 +  + a2nA1n? Well, it is a cofactor expansion
of the determinant of A, except that the first row has been replaced by the second. In other
words, we have a matrix with two identical rows, so the determinant is zero. We could do the
same thing with any other row for the a’s and also get zero. The form of the cofactor expansion
suggests dot products. We see that if a1 is the column whose entries are the cofactors of the
entries of the first row of A, we can take its dot product with that row to get the determinant,
while its dot product with every other row gives zero. That is, Aa1 = (det(A), 0, 0, …, 0). We
could do a similar thing with the cofactors of any row. Thus, if ak is a column whose entries are
the cofactors of the entries of the kth row of A, then Aak is a column which is zero except for
det(A) in the kth slot.
Consider the matrix whose columns are these a’s. That is, let B be the matrix whose ij-th
entry is the coefficient of aji (note coordinates in reverse order because the kth column of B is the
cofactors of the kth row of A). Then AB = det(A)I. If det(A) is not zero (that is, if A is invertible)
then we have proved
Theorem: if A is invertible, then A-1 = B/det(A) where B is the transposed matrix of cofactors.
!1
"1 0 2 %
" 4 2 !4 %
1$
$
'
Example: this lets us easily calculate that $ 3 2 !2 ' = $ !7 !3 8 '' .
2
$# 2 1 1 '&
$# !1 !1 2 '&
Cramer’s rule
Since we can calculate inverses, we can easily solve systems. If Ax = b, then we know
that x = A-1b. But A-1 is 1/det(A) times the transposed matrix of cofactors. So what is A-1b?
Well, any particular entry in it is the dot product of b with a row of these transposed cofactors.
That is, we have the dot product of b with a column of cofactors, like b1A1j + b2A2j +  + bnAnj.
This is like expanding the determinant of A down the jth column of A except that we don’t have
a’s, but b’s. That is, we replace the jth column of A with b and then take the determinant. If we
then divide this by det(A) we have xj.
"1 0 2 %
" 3%
$
'
Example: solving $ 3 2 !2 ' x = $$ 2 '' leads to x = 12/2 = 6, y = -19/2, z = -3/2.
$# 2 1 1 '&
$# 1 '&
Note: the book has a nice proof starting with Cramer’s rule. A[x e2 e3] = [Ax Ae2 Ae3]. Now we
are supposed to have Ax = b, and clearly Aej is the jth column of A. So the matrix on the right is
A with it’s first column replaced by b. Taking determinants of both sides gives (det(A)) (x1) =
determinant of A with its first column replaced by b. Dividing by det(A) gives Cramer’s rule,
and the same trick can appy to any column by replacing the jth column of I by x.
Now this process can be used to find A-1 by solving consecutively for Ax = e1, Ax = e2,
etc. just like we do when using Gauss-Jordan elimination.
Reading: 5.3
Problems: 5.3: 1, 3, 6, 7, 8, 12, 16, 17, 18, 22
In addition, find and explain one of the following methods of finding determinants
• Dodgson (a.k.a. Lewis Carroll) condensation mathod (a.k.a. the method of contractants)
• Chió’s pivotal condensation method
2 1 4 !3
2 !3 1 !1
Demonstrate your method of choice by finding
.
0 2 1 1
!2 2 1 !1
Extra credit: prove your method of choice works for arbitrary sized determinants.