M317 – Algebra 2
Unit 3
Worksheet 2
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Date
Teacher
Period
2 x 2 and 3 x 3 Determinants are easy to evaluate by using simple operations of the diagonal products, but
these methods are only applicable to 2 x 2 and 3 x 3 matrices. All determinants can be evaluated by a
method called expansion by minors.
a b c
Given:
−2 3 5
notice the elements in the top row.
1 −4 6
" a " is in row 1 column 1
The minor of " a " is a 2 x 2 determinant which is created by covering up row 1 and column 1, therefore
the minor of " a " is
3 5
"b " is in row 1 column 2.
−4 6
The minor of "b " is a 2 x 2 determinant which is created by covering up row 1 and column 2, therefore
the minor of "b " is
−2 5
1 6
.
The minor of " c " is a 2 x 2 determinant which is created by covering up row_______ and
column_______, therefore the minor of " c " is
.
To expand the determinant by minors we alternate the signs of the values of a, b, c and the product of
their respective minors.
SO . . . . . . .
a b c
−2 3 5
=a
1 −4 6
3 5
−4 6
−b
−2 5
1 6
+c
−2
3
1 −4
by elevating each minor we get
a ( 38 ) − b ( −17 ) + c ( 5 )
or
38a + 17b + 5c
M317 – Algebra 2
Unit 3 – Worksheet 2
10/09/2008
Here is a more realistic problem:
Evaluate
( −2 )
−2 − 3 4
1 2 − 1 by expansion of minors
5 0 3
2 −1
1 −1
1 2
− ( −3)
+ ( 4)
0
3
5 3
5 0
= ( −2 )( 6 ) + ( 3)( 8 ) + ( 4 )( −10 )
= −12 + 24 − 40
= −28
Here are 2 problems for you to try:
4 3 −2
1) 1 − 1 3 =
−2 − 2 − 3
−1 3 3
2) 0 2 0 =
11− 1 3
Find the determinants using the diagonal method:
−2 3 1
3) 0 4 −3 =
2 5 −1
M317 – Algebra 2
0 −4 0
4) 2 −1 1 =
3 −2 5
Unit 3 – Worksheet 2
10/09/2008