Evaluate a Determinant of any Order

General method for Calculating of Determinants

• The "criss-cross" method discussed earlier can only be used on 2x2 and 3x3 determinants
• A general method for calculating determinants known as the Method of Cofactors is completely general and can be used to evaluate a determinant of any order (2x2, 3x3, 4x4, 5x5,...)

First, some definitions:

• Element - each individual entry or number in a determinant is referred to as an element. We often use the notation a_ij to designate the element in the "i^th" row and "j^th" column.

ex.

a11	a12	a13
a21	a22	a23
a31	a32	a33

a₂₁ --> means element in second row, and first column.

• Minor - the minor of an element is the determinant that is left when the row and column containing the element are crossed out.

ex. referring to previous example
The minor of 
a₁₃ =

a21	a22
a31	a32

The minor of
a₂₂ =

a11	a13
a31	a33

The minor of
a₂₃ =

a11	a12
a31	a32

in total there would be 9 minors for 3x3 determinant

• Cofactor - The cofactor of an element is equal to the minor of the element multiplied by +1 or -1, depending on the elements position. In general,

Cofactor of a_ij = (-1)^i+j . (minor of a_ij)

Rather than using the fancy formula above to generate the appropriate (+1) or (-1) we will oftenuse the checker board pattern of signs shown below:
always "+" in top left corner

+	-	+
-	+	-
+	-	+

example:
Given

a11	a12	a13
a21	a22	a23
a31	a32	a33

==>

+ a11	- a12	+ a13
- a21	+ a22	- a23
+ a31	- a32	+ a33

Cofactor of

a₁₃ = (+1) . minor of a₁₃
a₁₃ = (+1) .

a21	a22
a31	a32

Cofactor of

a₃₃ = (+1) . minor of a₃₃
a₃₃ = (+1) .

a11	a12
a21	a22

Cofactor of

a₃₂ = (-1) . minor of a₃₂
a₃₂ = (-1) .

a11	a13
a21	a23

example:
Given

4	-2	6
-4	3	1
8	5	2

==>

+ 4	- -2	+ 6
-4	3	1
8	5	2

Cofactor of  8

a₃₁ = (+1) . minor of a₃₁
a₃₁ = (+1) .

-2	6
3	1

= ... = -20


Cofactor of 5

a₃₂ = (-1) . minor of a₃₂
a₃₂ = (-1) .

4	6
-4	1

= ... = -28



Cofactor of  -4

a₂₁ = (-1) . minor of a₂₁
a₂₁ = (-1) .

-2	6
5	2

= ... = 34



General Rule for evaluating a determinant (method of Cofactors)

Pick any row or any column of the determinant and multiply each element in the row or column by its corresponding cofactor and then sum up these products. The result obtained is the value of the determinant!!


Example:
Evaluate the determinant below using the method of cofactors.

∆ =

4	3	2
6	-2	5
1	-4	-8

Solution:

Expanding using cofactor method with 1st row
∆ =

+ 4	- 3	+ 2
6	-2	5
1	-4	-8

∆ = 4 (+1) .

-2	5
-4	-8

+ 3 (-1) .

6	5
1	-8

+ 2 (+1) .

6	-2
1	-4

∆ = 4 (16 - (-20)) - 3 (-48 - 5) + 2 (-24 - (-2))
∆ = 4 (36) - 3 (-53) + 2 (-22) 
∆ = 259

Note:
Redo same ∆ calculation using 1st column
∆ =

+    4	3	2
-      6	-2	5
+     1	-4	-8

∆ = 4 (+1) .

-2	5
-4	-8

+ 6 (-1) .

3	2
-4	-8

+ 1 (+1) .

3	2
-2	5

∆ = 4 (16 - (-20)) - 6 (-24 - (-81)) + 1 (15 - (-4))
∆ = 4 (36) - 3 (-53) + 2 (-22)
∆ = 259
ie. same answer.


Example:
Evaluate the 4x4 determinant shown using the method of cofactors. Which row(s) or column(s) would be most convenient to use?

∆ =

0	5	0   0
2	1	4   4
0 0	3 8	2   0   1   1

 Solution:
Expanding using cofactor method with 1st column

∆ =

+ 0	5	0 0
-  2	1	4 4
+ 0 -  0	3 8	2 0 1 1

∆ = 0 (+1) .

1	4	4
3	2	0
8	1	1

+ 2 (-1) .

5	0	0
3	2	0
8	1	1

+ 0 (-1) .

5	0	0
1	4	4
3	2	0

∆ = - 2 .

5	0	0
3	2	0
8	1	1

Always "+" in top left position

∆ = -2 [ 0. (+1) .

3	2
8	1

+ 0 (-1) .

5	0
5	1

+ 1 (+1) .

5	0
3	2

]

∆ = -2 . [0 + 0 + 1 (10 - 0)]
∆ = -20


Some Useful Simplification Rules for Determinants

Property 1:

• Pick any row of determinant. You may add / subtract any multiple of another row to the given row and the value of the determinant will remain the same. ---> useful for introducing zeros...

Example:

∆ =

-2	0	2
2	4	3
4	-4	4

<--- row 2 + 1 . row 1
<--- row 3 + 2 . row 1



∆ =

-2	0	2
0	4	5
0	-4	8

Note: brought in zeros... good stuff!

Example:
∆ =

1	2	3      4
5	6	7      8
9 13	10 14	11    12 15    16

 <--- row 2 - row 1
<--- row 4 - row 3



∆ =

1	2	3     4
4	4	4     4
9 4	10 4	11   12 4     4

 <--- row 4 - row 2


∆ =

1	2	3     4
4	4	4    4
9 0	10 0	11 12 0     0

expanding using row 4...
\ ∆ = 0




Property 2:

• Interchanging two rows (or, in fact, two columns) of a determinant changes the sign of the determinant. 

∆ =

a	b
c	d

= a d - b c ; 

∆₁ =

c	d
a	b

= b c - a d = - (a d - b c) = -∆

Example:
∆ =

2	4
8	6

=
-1 .
 

8	6
2	4

Note: not very useful property

Property 3:

• If every element in a row (or column) has a common factor, then you can pull the common factor out in front and multiply it by the remaining determinant. ---> Useful for reducing arithmetic...
  

∆ =

a	b
c	d

= a d - b c ;

∆₁ =

ka	kb
c	d

= k a d - k b c = k (a d - b c) = k∆

Examples:
∆ =

200	400	600
1	7	9
-3	6	9

= 200 .

1	2	3
1	7	9
-3	6	9

 ∆ =  (200) (3) . 

1	2	1
1	7	3
-3	6	3

= (200) (3) (3) .

1	2	1
1	7	3
-1	2	1

= 1800 .

1	2	1
1	7	3
-1	2	1

Example:
Solve the following system by usind Cramer's Rule.
  2 x + 3 y +    z = 2
-   x + 2 y + 3 z = -1
-3 x -  3 y +    z = 0

Solution:
∆ =

2	3	1	2	3
-1	2	3	-1	2
-3	-3	1	-3	-3

∆ = (2) (1) (1) + (3) (3) (-3) + (1) (-1) (-3)
    - (1) (2) (-3) - (2) (3) (-3) - (3) (-1) (1)
\ ∆ = 7  <--- System Determinant

∆_x =

2	3	1	2	3
-1	2	3	-1	2
0	-3	1	0	-3

∆_x = (2) (2) (1) + (3) (3) (0) + (1) (-1) (-3)
- (1) (2) (0) - (2) (3) (-3) - (3) (-1) (1)
\ ∆_x = 28
\ x = ∆_x / ∆ = 28 / 7 = 4
\ x = 4


∆_y =

2	2	1	2	2
-1	-1	3	-1	-1
-3	0	1	-3	0

∆_y = 2 (-1) .

-1	3
-3	1

+ 1 (-1) .

2	1
-3	1

+ 0 (-1) .

2	1
-1	3

∆_y = -2 (-1 - (-9)) - 1 (2 - (-3)) + 0
\ ∆_y = -21
\ y = ∆_y / ∆ = -21 / 7 = -3
\ y = -3



∆_z =

2	3	2
-1	2	-1
-3	-3	0

<--- row 1 + 2 row 2



∆_z =

0	7	0
-1	2	-1
-3	-3	0

∆_z = 7 (-1) .

-1	-1
-3	0

∆_z = -7 (0 - 3) = 21
\ ∆_z = 21
\ z = ∆_z / ∆ = 21 / 7 = 3
\ z = 3