Math 21b — Determinants
OH: Tuesday, 4–5
Last Friday: Least squares and data fitting (read it!)
Topics: 2 × 2 determinants, the cross product, Sarrus’ rule, patterns.
We intruduce the determinant of a square matrix, which in a sense is a numerical measure
of invertibility. We start off by recalling examples you may already recognize.
2 × 2 determinants
Perhaps we should start with the even simpler example of a 1 × 1 determinant; a 1 × 1 matrix
is just a number a, and a matrix equation with it as the coefficient matrix is just
ax = b,
where x and b are also numbers. The solution is b/a, if that ratio is defined, and so the
number a determines whether every matrix equation with a has a (unique) solution.
2 × 2 determinants work similarly. To solve a matrix equation A~x = ~b, we again try to
form the inverse:
−1 a b
e
a b x
e
x
=
=
c d y
f
y
c d
f
d −b 1
and using our known method for finding the inverse, we see that it is ad−bc
−c a . So A
is invertible if ad − bc 6= 0, and once again, a single number determines whether matrix
equations with A have unique solutions.
Definition 1. The determinant of A = [ ac db ] is det(A) = ad − bc, also written
a b c d = ad − bc.
The cross product
In R2 , there is a trick to produce a vector orthogonal to a given one: swap the coordinates
and add a sign. For example: Given (2, 3), one orthogonal vector is (3, −2), and another is
its negative, (−3, 2), and of course any multiple would work but these at least have the same
length. In three dimensions a more complicated rule also works for this:
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Definition 2. (Cross product) The cross product of ~v = (a, b, c) and
vector ~v × w
~ = (r, s, t), where:
b c
a c a
r = s = − t = y z
x z
x
r = bz − cy
s = cx − az
w
~ = (x, y, z) is the
b , or
y
t = ay − bx.
It will remain a mystery for another day why this actually works, but you can see by
direct computation that this definition cleverly arranges minus signs so that when you take
the dot product with either ~v or w,
~ the result is zero:
(a, b, c) • (r, s, t) = abz − acy + bcx − abz + acy − bcx = 0.
So if ~u is any linear combination of ~v and w,
~ we have ~u • (~v × w)
~ = 0. Given a collection of
vectors ~u, ~v , w,
~ this quantity, the scalar triple product, measures their linear independence:
zero if they are dependent, nonzero otherwise. If these vectors are the columns of a matrix
A, then the triple product measures whether A is invertible.
Definition 3. The determinant of a 3 × 3 matrix A = [ ~u ~v w~ ] is the scalar triple product
det(A) = ~u • (~v × w).
~
Sarrus’ rule
Remembering the computation we did to show that ~v × w
~ is perpendicular to both ~v and
w,
~ we find by doing some algebra that if A = [aij ], then det(A) is a sum of products of
various entries of A with different signs. Actually, it’s much neater: in each product, one
entry from each row appears once, but the signs are harder to detect. The rule for the above
computation is as follows: duplicate the first two columns to the right of A, for example:


1 2 3
1 2 3 1 2
4 5 6 =⇒ 4 5 6 4 5
7 8 9
7 8 9 7 8
There are six diagonal lines: three down to the right, and three up to the right. Sarrus’ rule
is the following:
1. Draw all six lines;
2. For each line down to the right, multiply the three matrix entries on it;
3. For each line up to the right, multiply the entries on it and negate the result;
4. Add the six products; this is det(A).
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Patterns and upswings
Any direct attempt to generalize Sarrus’ rule to matrices with more than 3 entries per side
is doomed to failure, as the book demonstrates nicely (p. 253). Therefore, we will just give
the definition straight up:
Definition 4. Let A be an n × n matrix.
• A pattern in A is a choice of n entries, one from each row and one from each column.
• If P is a pattern, visualize it by circling the entries in the matrix.. An upswing occurs
between a pair of circled entries connected by a line with positive slope (up to the
right).
• The sign of P is 1 if the number of upswings is even, −1 if it is odd, and is denoted
sgn(P ). The product of P is the product of all the circled entries, denoted prod(P ).
The determinant of A is given by the following formula:
a11 a12 . . . a1n a21 a22 . . . a2n X
=
det(A)
=
sgn(P ) prod(P ).
..
..
..
...
.
.
.
P a pattern
an1 an2 . . . ann If this seems obtuse, don’t worry, since determinants are famously complicated. Let’s see
how it works with some simple examples:
Example 5. Say A is a 2 × 2 matrix, as above. Then there are two patterns:
a b
a b
Q:
.
P:
c d
c d
The only line segment in P is down to the right (negative slope), so sgn(P ) = 1 and
prod(P ) = ad. In Q, the line segment has positive slope, so sgn(Q) = −1 and prod(Q) = bc.
Applying the definition,
det(A) = sgn(P ) prod(P ) + sgn(Q) prod(Q) = 1 · ad + (−1) · bc = ad − bc
and this is how we defined the determinant of a 2 × 2 matrix.
Example 6. Now we move on to a 3 × 3 matrix A with columns ~u, ~v , and w.
~ Using the
previous notation ~v = (a, b, c), w
~ = (x, y, z), we take the first column to be (for example)
~u = (1, 2, 3). Then we draw the six patterns:






1 a x
1 a x
1 a x
2 b y
2 b y
2 b y
3 c z
3 c z
3 c z






1 a x
1 a x
1 a x
2 b y
2 b y
2 b y
3 c z
3 c z
3 c z
3
All the patterns on the top row have sign 1: the first has no upswings, and the other two
have two each. All the patterns on the bottom row have sign −1: the first two have one
upswing each, and the third has three. Reading off their products, the determinant is:
det(A) = bz + 2cx + 3ay − cy − 2az − 3bx = 1 · (bz − cy) + 2 · (cx − az) + 3 · (ay − bx)
= (1, 2, 3) • (a, b, c) × (x, y, z) = ~u • (~v × w)
~
which was our previous definition. As you can see, Sarrus’ rule is just a simple case of
patterns and signs.
The determinant has many useful formal properties which we will discuss next time,
along with some standard (and much more efficient) ways of computing it. One is already
available: using block matrices. You can justify the following rule straight from the definition
(see the book’s proof on p. 258):
Proposition 7. The determinant of a “block triangular” matrix in which A and C are
square can be computed by:
A B = (det A)(det D) = A 0 .
0 D
C D In particular, if A is actually triangular, then det(A) is the product of the entries on its
diagonal.
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