Chapter 6.7
Determinants
• In this chapter all matrices are square; for
example: 1x1 (what is a 1x1 matrix, in
fact?), 2x2, 3x3
• Our goal is to introduce a new concept, the
determinant, which is only defined for
square matrices, yet any size square
matrices
• The determinant is, first of all, just a
number; and, since we want to have a
natural definition for it, we say that for 1x1
matrices the determinant is EXACTLY that
number:
det(a11 )  a11  a11
• Examples:
det(3)  3  3; det( 5)   5  5
• Notation: we use either the keyword “det”
in front of the matrix OR we replace the
brackets (or parenthesis, if applicable) with
vertical bars
• Careful: for 1x1 matrices DO NOT confuse
the vertical bars notation with the absolute
value notation (the context will tell you
whether the number is assumed to be a
matrix, in which case it’s determinant, or a
number, in which case it’s absolute value)
• What about bigger matrices? The idea is to
define a bigger matrix’ determinant based
on smaller (sub-)matrices’ determinant
• Take a 2x2 matrix; a smaller matrix is a 1x1
matrix, whose determinant we know right
now; hence we can define the 2x2 matrix
determinant
– Choose a row (usually the first one)
 a11
a
 21
a12    this one!

a22 
– take the first entry in the row, and remove from
the matrix the column corresponding to it and
the chosen row; what’s left is a matrix of order
2-1=1
* # 
# a 
22 

– take the determinant of this matrix (we know
how to compute it!); we call this the minor and
it’s usually denoted by capital letter with the
initial entry indices (in our case 11)
A11  a22  a22
– The last thing we do for the 11 index is to build
the cofactor, which is the minor multiplied by
(-1) to the power (sum of indices), in our case
1+1=2; the usual notation for cofactor is c with
original indices
11
c11  (1) A11  a22
– Take now the second entry and build its
cofactor: remove the second column and the
first row; what’s left is a 1x1 matrix, namely
 # *
  a21 
a

 21 #
– compute the minor
A12  a21
– Get the cofactor:
1 2
c12  (1)
A12  a21
– we exhausted the row, and now we construct
the sum:
a11
a12
a21
a22

 a11c11  a12 c12  a11a22  a12 a21
• For the 2x2 matrix we get, in fact, an easy
to remember formula: product of the first
diagonal (NW-SE) minus product of the
second diagonal (NE-SW)
• Example:
1 5 
3  2  1 (2)  5  3  2  15  17


• Things get a bit more complicated as we go
to 3x3 matrices; we use the VERY SAME
TECHNIQUE, though
– take the matrix
 a11
a
 21
 a31
a12
a22
a32
a13 

a23 
a33 
– Choose a row - again, usually we take the first
row, so let’s go with that one
– take first entry and compute its cofactor:
remove the first column and the first row, and
we get the following matrix:
* #
# a
22

 # a32
#

a23 
a33 
– We compute this smaller (2x2) matrix’
determinant (because we know now how!) and
so we get the minor of index 11:
a22
A11 
a32
a23
 a22 a33  a23 a32
a33
– finally, the cofactor:
11
c11  (1) A11  (a22 a33  a23a32 )
– Next entry’s cofactor: (less talk, just
computation)
#
a
 21
a31
A12 
#
a21

# a23    
a31

# a33 
*
a21 a23
a31
a33
a23 
a33 
 a21a33  a23 a31
– So, the cofactor is:
c12  (1)12 A12  (a21a33  a23a31 )
– and, for the last entry of this row, voila the
cofactor:
1 3
c13  (1)
A13  (1)
4
a21 a22
a31
a32
 (a21a32  a22 a31 )
– Hence the determinant for our 3x3 matrix is
a11
a12
a13
a21
a22
a23  a11c11  a12 c12  a13c13 
a31
a32
a33
 a11 (a22 a33  a23 a32 )  a12 (a21a33  a23 a31 )  a13 (a21a32  a22 a31 )
• There actually is a way of remembering
even this formula, reminiscent of the 2x2
matrix’ determinant; again, we have a first
diagonal (NW-SE), but also 2 “first halfdiagonals”; the product of entries on each
gets added; we have a second diagonal (NESW) and 2 “second half-diagonals”, and the
product of entries on each, respectively, gets
substracted (for alternate description please
read at the bottom of page 280)
• Example:
1 2 3 
 4 5 6 


7 8 9
 1 5  9  2  6  7  4  8  3 
 3  5  7  2  4  9  6  8 1 
 45  84  96  105 72  48 
0
• One interesting issue: as mentioned, you
could choose any row you want (and, in
fact, if you “look sidewise”, you could do a
very similar thing for columns!); but the
process is pretty complex, right? So how
can we be so sure we get the same number
all the time? Well, be assured - it really
works, regardless of row (or of column, in
fact)
• Why would you choose a different row? For
example, you have a lot of 0 (zero) entries
in that row; since you got to multiply those
entries with the respective cofactors, 0 times
anything is 0! So we don’t need to compute
those cofactors!
• Example:
1 2 3
0 4 0


 4 3 3
– if you choose the first row, you need to
compute all three cofactors; but if you choose
the second row, you only need compute the
second entry’s cofactor!
1 # 3 
# * #   1 3   A  1 3  3  12  9
22
4 3


4 3


4 # 3
– the determinant is hence (I‘m mentioning the
other 2 cofactors, but, as you see, they get
multiplied by 0, so we don’t care what values
they have):
1 2 3
0 4 0  0  c21  4  c22  0  c23  4  (1) 2 2 A22  4  (9)  36
4 3 3
• As an exercise, try to prove that, by
choosing the second row in a 2x2 matrix
you get the same number (you can work
with an actual matrix, or try the general
form, with a’s)
• Complicated as it was for 3x3 matrices, you
can see now how complicated it could get
even further (4x4, 5x5 and so on); still the
method still works, and the moment you
know how to compute a 3x3 matrix’
determinant, you can compute the 4x4
matrix’ one; the moment you know how to
compute a 4x4 matrix’ determinant, you can
compute the 5x5 matrix’ one
• But … computing a determinant is not
always a hideously long and intricate task,
because the way we compute this number
leaves a few backdoors open
• For instance:
– 1. If each of the entries in a certain row (or
column) of the matrix is 0, then its determinant
is 0 (remember the example above? What if the
4 was also a 0?)
– 2. If two rows (or columns) are identical, then
the determinant is 0
1 2 3
1 2 3 0
2 1 5
– 3. If the matrix is upper/lower triangular (in
particular, if it is diagonal) then the determinant
is equal to the product of the main (first)
diagonal entries
1 2 3
0 1 2  1 1  2  2
0 0 2
– 4. If one modifies the matrix by adding a
multiple of one row to another row (same for
columns), the determinant doesn’t change here you see the connection to elementary
matrix operations!
1 3 5
1
3
5
1 3 5
4 3 5  4 1 3  3 5  5  3 0 0 
3 2 1
3
2
1
3 2 1
 3  (1)
2 1
3 5

 3  (3  10)  21
2 1
– 5. If one interchanges two rows (or columns)
the determinant changes sign
1 2
3 4

3 4
1 2
– 6. If one multiplies the entries of a row (or
column) by a number, the determinant gets
multiplied by that number
3 1 3  2
1 2
3
3
5
3 5
– 7. If one multiplies the matrix by a scalar
(which, if you remember, means multiplying all
elements by that number - all rows, that is, and
see 6.) then the determinant gets multiplied by
that number as many times as many rows it has
(its size; for a 2x2 matrix, twice; 3x3 matrix,
thrice; and so on)
3
1 3
2 1
2
2

2 7
2 7 
 2 6 


 4 14 


– 8. If one multiplies two matrices, the
determinant of the product is the product of the
determinants
A B  A  B
– 9. The determinant of the transpose of a mtrix
equal the determinant of the original matrix
A  A
T
1 3 5
1 0 4
0 1 23 1 3
4 3 2 5 2 2
• One last thing: as you can see, it’s very
convenient to have an upper triangular
matrix (or lower, or diagonal); on the other
hand, when reducing a matrix (a square
matrix, that is) that’s exactly what we get
• So … why not use this? As you can see, one
elementary operation doesn’t change the
determinant (adding a multiple of a row to
another one; probably the most important
one; see 4.); a second one only changes its
sign (interchanging two rows; see 5.); the
last one multiplies the determinant by a
controllable quantity (multiplying by that
quantity a certain row; see 6.)
• Here’s an example:
2 3 0
3
4
2 3 0
9  31
1
4
6
8
2
8
3
1
– think of this as “factoring out the 3 out of the
second row”
2 3 0
31
4
2
8
1
2
3
R2  2 R1
1
2
3
3  3 2  3 0   3 0  7  6 
R3  4 R1
1
4 8 1
0 0  11
– (we now have an upper triangular matrix, so we
can stop here - or, if you can waste time, NOT
DURING THE EXAM! you can continue
reducing the matrix by “factoring out” the -7
out of the second row etc)
 3  (1 (7)  (11))  231
• This method is called triangulation (for
obvious reasons!)
• It’s especially useful for higher order
matrices (4x4, 5x5, etc) since we don’t have
to compute many complex cofactors, but
rather use simple elementary operations;
both methods take time, though (in general,
expect to waste a lot of time when
computing a large matrix, with few 0 …)