Section 9 The Lplce Expnsion In the lst section, we defined the determinnt of (3 3) mtrix A 12 to be 22 12 21 22 2231 22 12 21. In this section, we introduce generl formul for computing determinnts. Rewriting 22 12 23 31 21 32 22 31 23 32 12 21 22 23 32 12 23 31 12 21 21 32 22 31 ( ) ( ) ( ) 22 23 32 12 23 31 21 21 32 22 31 ( ) ( ) ( ) 22 23 32 12 21 23 31 21 32 22 31 note tht the terms outside the brckets re the terms long the first row of the mtrix 12 The term in brckets ssocited with is the determinnt of the (2 2) mtrix fter deleting the 1 st row nd 1 st column of A : ( ) 22 23 32 12 21 22 23 31 32 The term in brckets ssocited with 12 is the determinnt of the (2 2) mtrix fter deleting the 1 st row nd 2 nd column of A : ( ) 21 23 31 12 21 22 23 31 32 The term in brckets ssocited with is the determinnt of the (2 2) mtrix fter deleting the 1 st row nd 3 nd column of A : ( ) 21 32 22 31 12. Mtrix Algebr Notes Anthony Ty 9-1
There is the mtter of the minus sign in front of 12. This cn be chieved by multiplying into ech term ( 1) i j. Therefore, we cn write 12 12 12 12 ( 1) 21 22 23 31 32 12 ( 1) 12 21 22 23 31 32 ( 1) (1,1)th Minor of A (1,2)th Minor of A (1,3)th Minor of A (1,1)th Cofctor of A (1,2)th Cofctor of A (1,3)th Cofctor of A Some nmes hve been introduced in the formul bove: the determinnt of the mtrix fter removing the i th row nd j th column is clled the ( i, j )th minor of A ; including the sign ( 1) i j gives us the ( i, j )th cofctor of A. Exmple The determinnt of the mtrix 0 2 1 A 1 1 3 3 3 2 is 1 3 1 3 1 1 12 (0)( 1) (2)( 1) ( 1)( 1) 28 3 2 3 2 3 3 ; The cofctor expnsion for computing determinnts is not unique. For instnce, we could hve written the originl formul s 22 12 23 31 21 32 22 31 23 32 12 21 12 21 21 32 22 22 31 23 32 12 23 31 ( ) ( ) ( ) 21 12 32 22 31 23 32 12 31 which cn be written s Mtrix Algebr Notes Anthony Ty 9-2
12 12 12 12 21 ( 1) 21 21 22 23 22 ( 1) 22 21 22 23 23 ( 1) 23 21 22 23 31 32 31 32 (2,1)th Minor of A (2,2)th Minor of A (2,3)th Minor of A (2,1)th Cofctor of A (2,2)th Cofctor of A (2,3)th Cofctor of A Alterntively, we could hve expnded long column: in which cse, we hve 12 22 12 23 31 21 32 22 31 23 32 12 21 12 23 31 12 21 22 22 31 21 32 23 32 ( ) ( ) ( ) 12 21 23 31 22 31 32 23 21 12 ( 1) 12 12 21 22 23 31 32 22 ( 1) 22 12 21 22 23 31 32 32 ( 1) 32 12 21 22 23 31 32 (1,2)th Minor of A (2,2)th Minor of A (3,2)th Minor of A (1,2)th Cofctor of A (2,2)th Cofctor of A (3,2)th Cofctor of A Note tht we chieved the correct signs by multiplying the ( i, j )th minor with ( 1) i j. Expnding long ny row or column would in fct give us the sme expression; we cn write or 3 ( 1) i A j M for ny row i j1 3 i1 ( 1) i A j M for ny column j. where M is the ( i, j )th minor of A. This formul is known s the Lplce Expnsion. Mtrix Algebr Notes Anthony Ty 9-3
The fct tht you cn expnd long ny row or column cn simplify computtions substntilly if there is row or column with mny zeros. Exmple Compute the determinnt of the mtrix 1 0 A 2 0 3 5 4 6 using the Lplce Expnsion (i) expnding long the first row, (ii) expnding down the second column. 1 0 0 3 2 3 2 0 2 0 3 (1)( 1) (0)( 1) ()( 1) 12 0 88 76 4 6 5 6 5 4 5 4 6 (i) 12 1 0 (ii) 12 2 3 2 0 3 (0)( 1) 5 6 5 4 6 22 (0)( 1) 1 5 6 32 (4)( 1) 1 76 2 3 (nn) Determinnts Crmer s Rule nd the Lplce Expnsion extend to lrger systems. The determinnt for generl ( n n) mtrix is or 12 1n 21 22 2n A n1 n2 nn n ( 1) i j j1 A M for ny row i n ( 1) i j i1 A M for ny column j where the (, )th i j minor M of n nn mtrix A is the determinnt of the (n1)(n1) mtrix tht remins fter removing the ith row nd jth column of A. Mtrix Algebr Notes Anthony Ty 9-4
Exmple The determinnt of 2 2 3 4 0 1 0 D is 1 1 0 3 2 0 1 3 0 1 2 2 4 23 D 3( 1) 1 1 3 0( 1) 1 1 3... 2 0 3 2 0 3 39 0 2 2 4 2 2 4 43 0( 1) 0 1 ( 1)( 1) 0 1 3 2 0 3 1 1 3 0 42 where we hve expnded down the third column. The Lplce expnsion includes the (2 2) cse. Define the determinnt of single number s. Note tht in this context the symbol does not refer to bsolute vlues, e.g. 2 = 2, not 2. Then tking, sy, first row expnsion, we hve A 2 1 j 12 1 1 j( 1) M1 j ( 1) M 12( 1) M j 12 22 12 21. Crmer s Rule lso extends to generl ( n n) systems of equtions. The solution to x x... x b 1 12 2 1n n 1 x x... x b... 21 1 22 2 2n n 2 x x... x b n1 1 n2 2 nn n n is i ( ) xi A b, i 1,..., n, where 12 1n b1 21 22 2n b 2 A, b n1 n2 nn bn nd Ai ( b) is the mtrix A with its i th column replced by b. Mtrix Algebr Notes Anthony Ty 9-5
Exercises 1. Find the determinnts of the following mtrices using the Lplce expnsion. (i) 4 0 1 19 1 3 7 1 0 (ii) 0 2 0 3 0 4 2 3 0 (iii) 41 42 43 (vi) 0 0 21 22 0 (v) 0 0 0 12 22 23 (vi) 0 0 0 22 0 0 0 (vii) 0 0 23 0 32 0 41 0 0 (viii) 4 3 2 6 4.5 3 7 1 0 (ix) 4 3 0 9 1 3 7 1 0 2. Use Crmer s Rule to solve 4x z 4 19x y 3z 3 7x y 1 3. Solve the following system of equtions 2x 2x 3x 4x 2 1 2 3 4 x x 3 2 4 x x 3x 1 1 2 4 2x x 3x 2 1 3 4 4. Find the determinnts using the Lplce expnsion (i) 0 0 41 21 22 0 0 31 0 0 41 42 43 44 (ii) 0 0 0 21 22 0 0 0 41 42 43 44 (iii) 0 0 0 0 0 0 21 31 41 22 32 42 43 44 (iv) 0 0 0 0 22 0 0 0 0 0 0 0 0 44 (v) 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 (vi) 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 Mtrix Algebr Notes Anthony Ty 9-6
5. Let A be n rbitrry ( n n) mtrix. Show tht if we multiply every element of single row or column by c, then the determinnt of the new mtrix is c. Wht is the determinnt of ca? 6. Let E be the mtrix obtined by switching the 1st nd lst rows of n ( n n) identity mtrix, i.e. 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 E 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 () Show using the Lplce expnsion tht det E 1; (b) Does your nswer depend on whether n is even or odd? (c) Use your result to prove tht the determinnt of the mtrix formed by switching ny two rows of the identity mtrix is 1. 7. () Let 1 0 0 E 0 1 0 0 1 nd suppose A is some (3 n) mtrix. Describe the rows of the product EA in terms of the rows of A ; Wht is the determinnt of the mtrix E? (b) Let E be mtrix tht crries out the elementry row opertion of subtrcting multiple of one row from nother row. Wht is the determinnt of E? Mtrix Algebr Notes Anthony Ty 9-7