4.0 MATRICES AND SYSTEMS OF LINEAR EQUATIONS 4.2 DETERMINANT OF MATRICES http://portal.kmpp.matrik.edu.my/mod/ resource/view.php?id=2380
LEARNING OUTCOMES At the end of this topic, students should be able to : (a) Find the minors and cofactors of a matrix (b) Find the determinant of a matrix.
Determinant of a 2x2 matrix a b A c d ad bc E.g. 3 5 1 1 A 3 5 8 YOUR PREVIOUS KNOWLEDGE http://portal.kmpp.matrik.edu.my/ mod/resource/view.php?id=2382
MINOR Let be × matrix , = × The minor, , is the determinant of − × ( − ) matrix obtained by deleting the row and column of A 11 12 13 21 22 23 31 32 33 a a a A a a a a a a m21 a a12 33 12 13 32 33 a a a a http://portal.kmpp.matrik.edu.my/ mod/resource/view.php?id=2384 a a13 32
= − + The cofactor, , of the element is COFACTOR
EXAMPLE 1 = − find the minors and cofactors of and .
= = () – () = 4 2 4 3 SOLUTION 1 2 1 3 4 2 1 4 3A = = − (−)() = 1 1 3 2 1 2 1 3 4 2 1 4 3A
The cofactor, , of the element = − + = − + = = − = = − + = = − = −
11 12 13 21 22 23 31 32 33 minor matrix : m m mM m m mm m m NOTE:
= − + 11 12 13 21 22 23 31 32 33 m m mC m m mm m m 1 1 1 2 1 3 11 12 13 2 1 2 2 2 3 21 22 23 3 1 3 2 3 3 31 32 33 Cofactor matrix, 1 m 1 1 1 1 1 1 1 1 m mC m m mm m m
http://portal.kmpp.matrik.edu.my/mod EXAMPLE 2 /resource/view.php?id=2386 = , find all the minors of A. Hence, form the and
= = − ()= 1 1 0 3 3 1 2 3 3 1 2 0 SOLUTION 1 2 0 3 1 1 2 0 3 A = = − () = = = − () = −
= = − () = 2 0 0 3 1 0 2 3 1 2 2 0 1 2 03 1 12 0 3A = = − () = = = − () = −
= = − () = 2 0 1 1 1 0 3 1 1 2 3 1 1 2 0 3 1 1 2 0 3A = = − () = = = − () = −
3 7 2 minor 6 3 4 2 1 5 A 3 7 2 cofactor 6 3 4 2 1 5 3 7 2 6 3 4 2 1 5 A
COFACTOR EXPANSIONS Let be × matrix with entries . || = = = + +. . . +. Theorem http://portal.kmpp.matrik.edu.my/mod/resource /view.php?id=2388 cofactor expansion along the row. • For any = 1,2, … , , we have
COFACTOR EXPANSIONS Let be × matrix with entries . • For any = 1,2, … , , we have || = = = + +. . . + . Theorem cofactor expansion along the .
By expanding along the first row. A a m a m a m 11 11 12 12 13 13 = − + 11 11 12 12 13 13 A a c a c a c Elements in 1st row: , , If 3 3 A aij 11 12 13 21 22 23 31 32 33 a a a A a a a a a a
EXAMPLE 3 = − − − , find by expanding along the first column.
2 3 4 2 7 1 4 1 4(3)( 1) (1)( 1) (5)( 1) 1 10 1 10 2 7 = (− + ) + (−)(− + ) + (− + ) = − 5 1 10 1 2 7 3 1 4 A Expansion along 1st column, 11 11 21 21 31 31 |A| a c a c a c = − + + SOLUTION
= − − − by using expansion of the cofactors. EXAMPLE 4 Choose the row or column that has the most zeroes. When = , =
= + + (−)() = − 5 2 1 (0) (0) (1)( 1) 0 5 Choose second row 2 1 3 0 0 1 0 5 4 A A a c a c a c 21 21 22 22 23 23 SOLUTION
1. = 2. = 3. = where =scalar & =order of matrix A 4. If the corresponding elements are equal in any 2 row or column in the matrix , then = 0 5. If any 2 rows or column interchange in the matrix , then the new determinant is − . Properties of Determinant http://portal.kmpp.matrik.edu.my/ mod/resource/view.php?id=2391 http://portal.kmpp.matrik.edu.my/ mod/resource/view.php?id=2393 http://portal.kmpp.matrik.edu.my/mod/ resource/view.php?id=2394 http://portal.kmpp.matrik.edu.my/mod/ resource/view.php?id=2396
= − . By using the properties of determinant, find EXAMPLE 5 = . if = & = − . − = .
if = & = . − = .
41 2 3 104x y A w z SOLUTION = = = = (−) = − = − − = − = =
41 2 3 104x y A w z SOLUTION if = & = = − = [1st & 3rd row have the same entries] − = − − Interchan ge of 2 nd & 3rd column = − = −
CONCLUSION Minor and Cofactor i n and j n A a c i j i j 1,2,..., 1,2,..., ; Expansion of the cofactors The minor, , is the determinant of − × ( − ) matrix obtained by deleting the row and column of A The cofactor, , of the element is = − +
Exercise: . = − , find the minors and cofactors of . . = − − − , find by expanding along the second column. Answers: 10 ; −10 Answer: −28
Solution 1 : = = 1 2 1 3 4 2 1 4 3A 2 1 4 3 = − − () = − + = = − = −
Solution 2 : 5 1 10 1 2 7 3 1 4 A 5 1 3 1 (7)( 1) 5 10 3 4 ( 2)( 1) 1 10 1 4 (1)( 1) 3 4 5 = (−)(−) + (−)() + (−)() = – – A a c a c a c 21 21 22 22 23 23 a) Expansion along 2nd row; = −