4.5 Determinants
Finding the area of a triangle
using a matrix
A Determinants is a number
The matrix must be square, meaning its
dimension can be a 2 X 2, 3 X 3, 4 X 4
and so on.
The determinants is used to solve system of
equations or find the area of a triangle.
Lets find the determinant of a 2 X 2
Let
2 3
A

5 8
The determinant of A is
written as
| A |
2 3
5 8
How to find the Determinant
Multiply the top left and bottom right, then
subtract it from the multiplication of the top
right and the bottom left.
| A |
2 3
5 8
 2 *8  5*3
| A | = 16 – 15 = 1
Find the Determinant
6 
 7
B

10

12


Find the Determinant
6 
 7
B

10

12


(-7 * - 12) – (6 * 10) =
Find the Determinant
6 
 7
B

10

12


(-7 * - 12) – (6 * 10) = 84 – 60 = 24
What about a 3 X 3
One method is expansion by minors.
1
0
1
2
1
3
4 2 3
1
0
1
2
1
3
4 2 3
The top row are the minors.
1, 0, -1
1 3
1
The minor of 1 is  2  3
You cut out the numbers
in the row and column
the minor is in
What about a 3 X 3
1
0
1
2
1
3
The minor of 1 is
1
4 2 3
1
0
1
2
1
3
4 2 3
1
0
1
2
1
3
4 2 3
The minor of 0 is
0
1
3
2 3
2
3
4 3
The minor of - 1 is  1
2
1
4 2
Using the minor to find the
determinant
Since we are using the top row.
We subtract the second minor from the first
the add the last minor.
+
1
1
3
2 3
-
0
2
3
4 3
+
1
2
1
4 2
Using the minor to find the
determinant
Since we are using the top row.
We subtract the second minor from the first
the add the last minor.
+
1
1
3
2 3
-
0
2
3
4 3
+
1
2
1
4 2
Find the determinants of the 2 X 2
Using the minor to find the
determinant
Since we are using the top row.
We subtract the second minor from the first the
add the last minor.
+
1
1
3
2 3
2
3
- 04 3 +
1
2
1
4 2
1((-1*-3)-(3*-2)) – 0((2*- 3)-(3*4)) + (-1)((2*-2)-(-1 * 4))
1(3 – (-6)) – 0(-6 – 12) + (-1)(-4 – (-4))
1(9) -0(- 18) + (-1)(0) = 9 – 0 + 0 = 9
Using the minor to find the
determinant
Since we are using the top row.
We subtract the second minor from the first the
add the last minor.
+
1
1
3
2 3
2
3
- 04 3 +
1
2
1
4 2
1((-1*-3)-(3*-2)) – 0((2*- 3)-(3*4)) + (-1)((2*-2)-(-1 * 4))
1(3 – (-6)) – 0(-6 – 12) + (-1)(-4 – (-4))
1(9) -0(- 18) + (-1)(0) = 9 – 0 + 0 = 9
Using the minor to find the
determinant
Since we are using the top row.
We subtract the second minor from the first the
add the last minor.
+
1
1
3
2 3
2
3
- 04 3 +
1
2
1
4 2
1((-1*-3)-(3*-2)) – 0((2*- 3)-(3*4)) + (-1)((2*-2)-(-1 * 4))
1(3 – (-6)) – 0(-6 – 12) + (-1)(-4 – (-4))
1(9) -0(- 18) + (-1)(0) = 9 – 0 + 0 = 9
Lets look at that with just variables
a b
c
d
f  a
e
g h
So
i
e
f
h
i
b
d
f
g
i
c
d
e
g h
a(ei  hf )  b(di  gf )  c(dh  eg )
Another way
Take the matrix's last two columns and place
them outside the matrix.
a b
c
d
e
f
g
h
i
a
b
c a b
becomes d e f d e
g
h
i g h
We are going to multiply down at an
angle, one way then the other way
Multiply at from top to bottom left to right,
then add. Repeat going from right to left
then add. To finish subtract the first
multiplied sum from the last.
a
b
c a b
d
e
f d e
g
h
i g h
( a * e * i  b * f * g  c * d * h )  (c * e * g  a * f * h  b * d * i )
Lets use this method
A
1
0
1 1
2
1
3
0
2 1
4 2 3 4 2
(1* -1 * -3 + 0 * 3 * 4 + -1 * 2 * - 2) –
(-1* -1 * 4 + 1* 3 * - 2 + 0 * 2 * -3)
(3 + 0 + 4) – (4 - 6 + 0) = 7 – ( - 2) = 9
To find the area of a triangle
Find the vertices of the triangle.
Place them in
(0,4)
a 3 X 3 matrix
Making the last
column
4 0 1
(4,0)
All 1’s
0
2
4
1
2 1
(2,2)
Find the determinants
4
0
1
0
4
1
2
2 1
-4(4 - (-2)) - 0(0 - 2) + 1(0 - 8) = -32
Then multiply + ½ or - ½
whatever makes it a positive number.
- ½ * -32 = 16
With variables: Area of a Triangle
With points (A,B) , (C, D) and (E,F)
Homework
Page 186 – 187
# 15 – 37 odd, 43
Homework
Page 186 – 187
# 16 – 38 even, 44