Title Matrices Varsity Question Bank Solution (HSC 27).pdf ✅

g ̈vwUa· I wbY©vqK  Varsity Question Bank Solution (HSC 27) 1 Rhombus Publications weMZ mv‡j DU-G Avmv cÖkœvejx 1.         bc 1 a a ca 1 b b ab 1 c c =? [DU 24-25] 1 0 1 2 abc DËi: 0 e ̈vL ̈v:         bc 1 a a ca 1 b b ab 1 c c = 1 abc      abc 1 a 2 abc 1 b 2 abc 1 c 2 =       1 1 a 2 1 1 b 2 1 1 c 2 = 0 [⸪ `ywU mvwi GKB] 2.       1 2 1 3 0 – 1 2 3 p g ̈vwUa·wU e ̈wZμgx n‡j, p Gi gvb KZ? [DU 23-24] 4 3 3 4 5 3 3 5 DËi: 4 3 e ̈vL ̈v: e ̈wZμgx n‡j,       1 2 1 3 0 – 1 2 3 p = 0  1(0 + 3) – 3(2p – 3) + 2(– 2 – 0) = 0  3 – 6p + 9 – 4 = 0  8 – 6p = 0  p = 8 6 = 4 3 3. A =     1 2 2 5 n‡j, det(AA–1 ) Gi gvb KZ? [DU 21-22] 1 – 1 0 – 1 2 DËi: 1 e ̈vL ̈v: AA–1 = I [I n‡jv A‡f`K g ̈vwUa·]  det(AA–1 ) = det(I) = 1 Note: A‡f`K g ̈vwUa‡·i wbY©vq‡Ki gvb 1 4. hw` A, B, C g ̈vwUa· wZbwUi AvKvi h_vμ‡g 4  5, 5  4 Ges 4  2 nq, Z‡e (AT + B)C g ̈vwUa·wUi AvKvi wK? [DU 20-21] 4  2 5  4 2  5 5  2 DËi: 5  2 e ̈vL ̈v: B g ̈vwUa‡·i μg 5  4  A T + B g ̈vwUa‡·i μg 5  4 C g ̈vwUa‡·i μg 4  2  (AT + B)C Gi †ÿ‡Î, (5  4)  (4  2)  (AT + B)C Gi μg: 5  2 5. A =     3 2 – 4 – 3 n‡j, det(2A–1 ) Gi gvb n‡jvÑ [DU 19-20] 1 4 – 4 4 – 1 4 DËi: – 4 e ̈vL ̈v: |A| =     3 2 – 4 – 3 = – 9 + 8 = – 1  det(2A–1 ) = 2 n |A| [†hLv‡b, n gvÎv] = 2 2 – 1 = – 4 6. A =       a – 2 – 5 2 b 3 5 – 3 c GKwU eμ cÖwZmg g ̈vwUa· n‡j, a, b, c Gi gvb ̧‡jvÑ [DU 17-18] – 2, – 5, 3 0, 0, 0 1, 1, 1 2, 5, 3 DËi: 0, 0, 0 e ̈vL ̈v: AcÖwZmg/wecÖwZmg/eμ cÖwZmg g ̈vwUa‡·i †ÿ‡Î gyL ̈ K‡Y©i mKj fzw3 0 nq| 7. k Gi †Kvb gv‡bi Rb ̈      1 1 1 1 k k 2 1 k 2 k 4 wbY©vqKwUi gvb k~b ̈ n‡e bv? [DU 17-18] k = 1 k = – 1 k = 3 k = 0 DËi: k = 3
2  Higher Math 1st Paper Chapter-1 Rhombus Publications e ̈vL ̈v: wbY©vq‡Ki ag© ̧‡jv g‡b ivL‡Z n‡e| k = 1 n‡j wbY©vq‡Ki wZbwU Kjvg B mgvb nq| gvb 0 k = – 1 n‡j wbY©vq‡Ki 1g I 3q Kjvg mgvb nq|  gvb 0 k = 0 n‡j wbY©vq‡Ki 2q I 3q Kjvg mgvb nq  gvb 0 wKš‘, k = 3 n‡j wbY©vq‡Ki †Kv‡bv Kjvg ev mvwi ci ̄úi mgvb n‡e bv|  gvb k~b ̈ n‡e bv| 8.            x x   = 0 n‡j, x = ? [DU 14-15] , ,  ,  ,  ,  DËi: ,  e ̈vL ̈v: x =  n‡j 1g I 3q Kjvg mgvb nq| d‡j wbY©vq‡Ki gvb k~b ̈| x =  n‡j 1g I 2q Kjvg mgvb nq| d‡j wbY©vq‡Ki gvb k~b ̈| wKš‘ x =  n‡j †Kv‡bv mvwi ev Kjvg ci ̄úi mgvb nqbv| ZvB x = , n‡j wbY©vq‡Ki gvb k~b ̈ n‡e| 9.     cos – sin sin cos Gi wecixZ g ̈vwUa·Ñ [DU 13-14]     cos – sin – sin cos     cos sin – sin – cos     cos sin – sin cos     cos sin sin cos DËi:     cos sin – sin cos e ̈vL ̈v:     cos – sin sin cos –1 = 1 cos2  + sin2      cos sin – sin cos =     cos sin – sin cos [⸪ cos2  + sin2  = 1] 10.       0 2 0 3 7x 0 2x + 7 9 + 5x 2x + 5 = 0 n‡j, x Gi gvb KZ? [DU 13-14] – 9 5 – 7 2 – 5 2 0 DËi: – 5 2 e ̈vL ̈v: (2x + 5)(0 – 3  2) = 0  2x + 5 = 0  x = – 5 2 11. BA Gi gvb wbY©q Ki, hw` A =     1 – i i 1 Ges B =     i – 1 – 1 – i I i = – 1 nq| [DU 12-13; CU 18-19; RU 17-18]     – 1 – i i – 1     1 7 1 3     1 0 0 1     2i – 2 – 2 – 2i DËi:     2i – 2 – 2 – 2i e ̈vL ̈v: BA =     i – 1 – 1 – i     1 – i i 1 =     i + i – 1 + i2 i 2 – 1 – i – i =     2i – 2 – 2 – 2i [⸪ i 2 = – 1] 12. hw` A =       – 2 3 2 1 – 1 2 nq, Z‡e A –1 mgvbÑ [DU 10-11; JU 14-15]     1 2 3 4     1 0 0 1     3 1 4 2     1 3 2 4 DËi:     1 3 2 4 e ̈vL ̈v: A –1 = 1 (– 2)    –  1 2 – 3 2        –  1 2 – 3 2 – 1 – 2 = 1 1 – 3 2        –  1 2 – 3 2 – 1 – 2 =     1 3 2 4 13. hw` A =     1 0 0 5 , B =     5 0 0 1 nq, Z‡e AB n‡jvÑ [DU 05-06, 03-04; JU 18-19, 16-17, 14-15; RU 08-09, 06-07]     5 0 0 5     5 10 0 5     10 0 0 5     0 5 5 10 DËi:     5 0 0 5 e ̈vL ̈v: AB =     1 0 0 5     5 0 0 1 =     5 + 0 0 + 0 0 + 0 0 + 5 =     5 0 0 5 14. hw` A =     2 3 – 3 2 nq, Z‡e A 2 gvbÑ [DU 04-05; JU 14-15; RU 08-09]     – 5 – 12 12 5     5 – 12 – 12 5     – 5 12 12 – 5     – 5 12 – 12 – 5 DËi:     – 5 12 – 12 – 5 e ̈vL ̈v: A 2 =     2 3 – 3 2     2 3 – 3 2 =     4 – 9 6 + 6 – 6 – 6 – 9 + 4 =     – 5 12 – 12 – 5