Question

1. Algebra: Why are all these determinants zero? 3 -1 6 21 1 0 0 0 1 -2 3 -4 1 3 1 2 7 -1 2 2 1 3 5 2 3 4 5 (a) 1 1 1 1 1; (b) 2 5 2; (c) 3 0 1 -73 (d) 3 1 7 11 3 7 3 3 -1 6 21 5 2 6 10 3 3 3 3 2. Algebra: - la b c Given that d e f = 6, find 9 h i de fa b c 3b ac та с ь 1 (a) g h il; (b) d+g eth ftil; (c) 3e d f ; (d) d - 3a f-3c e- 36. a b c 29 2h 21 3h 9 il 29 21 26 3. Algebra: 2 1 11 Use elementary row operations on the matrix A = 4 2 3 to obtain an upper triangular (1 3 0] matrix. Hence, evaluate det(A). 4. Algebra: [2 1 11 Consider the matrix A= 1 2 3 [1 -1 1 (a) Find the matrix [Mij] of minors of A. (b) Find the matrix (A j] of cofactors of A. (c) Find the adjugate, adj(A) of A. (d) Find det(A). (e) Hence, write down the inverse of A.

Answer 1

1. (a) 1 3 12 -1 6 7 -1 2 2 | 3 -1 6 2 ist and 4th Thise determinat is zero as rows are reme. Zero as (6) II 3 il 1 252 1 3 7 31 This determinant is zero as ist Columns are same. 1st and 3rd at is zero as Ru+(-2) R₂ Troo 01 2 135 | 21 3 5 30 1.-7 301-7 | 5 2 6 10 lioool After row operation, there are tero rows Row - and row-u are some so the determinant value is zero. (d) 1 1 2 3 1 3 - 2 3 i 3 3 -41 4 5 1 R₂+R, 7 1 3 1 | 1-2 3-41 3 1 7 1 131 7.1 1 3 3 3 3 3 After row operation two rows Row-2 and Row- are same. So the determinat value is zero.

2. Given 1 a b c | о ь с Cal deti g hilo a bel > -1 g h i I de fl la bal det R 23 (-126-1) I g h i - la bici I det = 6 ... Tahil (b) la b i deg eth fti = 29 zh zil la bclla b c] def + g h i [29 zh zi | 129 ر ا = 2 la bch la b c 1. Id efl + 21 g h il و ر ر = (2x6) + 2x0 = 12. tworows of the and determinant are some] . 136 ac 16 a el se df = 3e.df. 13hg il 236-0 lab crimes ch Chann re-column 1 and celum 2 (316. 2-18

2.1d) ja e b Id-3a f-3c e-36 129 zi zkl - a chl | а с Е d felt 1-3a -30 - 3b | 29 zi za 29 zi ahl = 26. a b ci I d e f +(-3)(2) I g h il a c | a co I g i hl - (-2)(6) + (-3)(21.0 [ and determinat I is zero as two rows, rowland row 2 are sene] = -12 Given matrix A= ů 2 3 row operation on A 1 3 0 How apply elementary А R2+(-29R4 21 17 R3+ (£) , 10 ö | LO - 52111 - 223, =B LO olj A. B is the reduced upper triangular matrix. How def(A) = (-1212213212 = -5 swe ndultiply (-19 as we enterch I two rows rows and row 3 77

4. Given matrix A = 12!! 1 2 3 (a) [Mij] of minor of A . The matrix i's [5 -2 | 2 1 L' 5 -37 -3 | 3] (6) The matrix cofactor [Aij ] of A. is . 55 +2 -37 1- 23 Li -5 3] T 5 -2 17 adj (A) is 2 1 -5 1-3 3 3 (4) det A = 1 2 1 1 11-111 = 12)(5) - (1+1) +(3-2) = 10 – 2 +1 = 9 (5) We have Aiad; A = adjA: A = (deta).In. ..A = deta (adj A) .

5 -2 |] 1-3 3 3 , 이 3 43 아 L -