Tiêu đề Matrices Varsity Practice Sheet Solution.pdf ✅

g ̈vwUa· I wbY©vqK  Varsity Practice Sheet Solution 1 01 g ̈vwUa· I wbY©vqK Matrices and Determinants weMZ mv‡j DU-G Avmv cÖkœvejx 1.       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 2. 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 3. 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 4. 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 5. 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| 6. 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 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|
2  Higher Math 1st Paper Chapter-1 7.            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| 8.     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] 9.       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 10. 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] 11. 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 12. 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 13. 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
g ̈vwUa· I wbY©vqK  Varsity Practice Sheet Solution 3 14. hw`       1 x x 2 1 a a 2 1 b b 2 = 0 nq, Z‡e x = ? [DU 03-04; CU 02-03] – a ev b a ev – b – a ev – b a ev b DËi: a ev b e ̈vL ̈v: Option Test: x = a n‡j 1g I 2q Kjvg mgvb|  gvb k~b ̈ x = – a n‡j †Kv‡bv mvwi ev Kjvg mgvb nq bv| x = b n‡j 1g I 3q Kjvg mgvb|  gvb k~b ̈ x = – b n‡j †Kv‡bv mvwi ev Kjvg mgvb nq bv|  x = a ev b 15.       10 11 12 13 14 15 16 17 18 wbY©vq‡Ki gvbÑ [DU 02-03; RU 11-12, 10-11, 06-07; JU 11-12, 10-11; JnU 10-11, 09-10] 0 1 10 5 DËi: 0 e ̈vL ̈v: Kjvg ̧‡jv mgvšÍi avivq Av‡Q|  mwVK DËi 0| weMZ mv‡j GST-G Avmv cÖkœvejx 1.       2 0 1 3 3 3 x x 1 – x wbY©qv‡Ki (2, 1) Zg fzw3i mn ̧YK 9 n‡j, x Gi gvb KZ? [GST 23-24] 2 1.5 0.5 0 DËi: 2 e ̈vL ̈v:       2 0 1 3 3 3 x x 1 – x (2, 1) Zg fzw3i mn ̧YK, (– 1)2+1     3 3 x 1 – x = 9  – (3 – 3x – 3x) = 9  3 – 6x = – 9  – 6x = – 12  x = 2 Note: mn ̧YK = (– 1)r + c  Abyivwk 2. hw` A =     4 3 2 5 ; B =     6 1 ; X =     x y Ges AX = B nq, Zvn‡j †μgv‡ii m~Î g‡Z x I y Gi gvb KZ? [GST 23-24] 2, – 1 3, – 1 – 1, 3 – 1, 1 DËi: 2, – 1 e ̈vL ̈v: D =     4 3 2 5 = 14 Dx =     6 1 2 5 = 28 Dy =     4 3 6 1 = – 14  x = Dx D = 2  y = Dy D = – 1 3. A Ges (AT + B)C g ̈vwUa· `yBwUi μg h_vμ‡g 4  5 Ges 5  2 n‡j, C g ̈vwUa· Gi μg wK n‡e? [GST 22-23] 4  2 4  3 4  4 4  5 DËi: 4  2 e ̈vL ̈v: A g ̈vwUa‡·i μg 4  5  A T g ̈vwUa‡·i μg 5  4  A T + B g ̈vwUa‡·i μg 5  4 GLv‡b, (AT + B)C g ̈vwUa‡·i μg 5  2 (AT + B)C wbY©q‡hvM ̈ n‡e hw`,  C g ̈vwUa· Gi μg: 4  2 (5  4)  (4  2) nq| 4. 3  2 Ges 2  3 μg wewkó `ywU g ̈vwUa· h_vμ‡g A I B Gi fzw3 0 ev 1 n‡j Tr(BA) Gi m‡ev©”P gvb n‡eÑ [GST 21-22] 0 1 6 9 DËi: 6 e ̈vL ̈v: BA g ̈vwUa‡·i μg n‡e 2  2| fzw3 0 n‡j BA =     0 0 0 0 ZLb me©wb¤œ Trace = 0 Avevi, A I B Gi fzw3 1 n‡j BA =     1 1 1 1 1 1       1 1 1 1 1 1 =     3 3 3 3 ZLb m‡e©v”P Trace = 3 + 3 = 6 5. i 2 = – 1 n‡j,      i i 3 i 5 i 3 i 5 i 7 i + i3 i 3 + i5 i 5 + i7 = ? [GST 20-21] – 1 0 1 i DËi: 0 e ̈vL ̈v:      i i 3 i 5 i 3 i 5 i 7 i + i3 i 3 + i5 i 5 + i7 =       i i 3 i 5 i 3 i 5 i 7 0 0 0 = 0 Note: i n + in + 2 = 0
4  Higher Math 1st Paper Chapter-1 weMZ mv‡j Agri-G Avmv cÖkœvejx 1. (x + 5, 2y + 1) = (2y + 4, 3y) n‡j, x Gi gvb KZ? [Agri. Guccho 20 -21] – 1 0 1 2 DËi: 1 e ̈vL ̈v: x + 5 = 2y + 4  x = 2y – 1 .... (i) 2y + 1 = 3y  y = 1 (i) G y Gi gvb ewm‡q, x = 2  1 – 1  x = 1 weMZ mv‡j JU-G Avmv cÖkœvejx 1. wb‡Pi †KvbwU cÖwZmg g ̈vwUa·? [JU 22-23; RU 17-18]     0 – b b 0     b 0 0 – b     b – b 0 0     0 0 – b b DËi:     b 0 0 – b e ̈vL ̈v: cÖwZmg g ̈vwUa‡·i †ÿ‡Î A T = A nq| GLv‡b, bs Ack‡b A T = A kZ©wU cÖ‡hvR ̈ n‡q‡Q| Note: cÖwZmg g ̈vwUa· n‡j, aij = aji eμ cÖwZmg g ̈vwUa· n‡j, aij = – aji 2. A =       3 4 1 5 0 6 1 2 4 n‡j, A – 2I Gi gvb †KvbwU? [JU 22-23]       5 4 1 5 2 6 1 2 6       1 4 1 5 – 2 6 1 2 2       1 4 1 3 0 6 – 1 2 4       1 2 – 1 3 – 2 4 1 2 6 DËi:       1 4 1 5 – 2 6 1 2 2 e ̈vL ̈v: A – 2I =       3 4 1 5 0 6 1 2 4 – 2       1 0 0 0 1 0 0 0 1 =       3 4 1 5 0 6 1 2 4 –       2 0 0 0 2 0 0 0 2 =       1 4 1 5 – 2 6 1 2 2 3. A g ̈vwUa‡·i μg 2  3 Ges B g ̈vwUa‡·i μg 3  2 n‡j, AB Gi μg †KvbwU? [JU 22-23] 2  2 2  3 3  2 3  3 DËi: 2  2 e ̈vL ̈v: AB g ̈vwUa‡·i †ÿ‡Î,  μg: 2  2 (2  3)  (3  2) 4. A =       1 4 7 2 5 8 3 6 9 Ges B =       0 2 4 1 3 5 n‡j, A + B = ? [JU 21-22]       1 6 11 3 8 13 3 6 9       1 4 7 2 7 12 3 9 14       1 6 7 3 8 8 Am¤¢e DËi: Am¤¢e e ̈vL ̈v: A + B wbY©q Kiv hv‡e hw`, A I B g ̈vwUa‡·i mvwi msL ̈v I Kjvg msL ̈v ci ̄úi mgvb nq| GLv‡b, A I B g ̈vwUa·Ø‡qi mvwi msL ̈v I Kjvg msL ̈v mgvb bq|  A + B wbY©q Am¤¢e| 5. A =       1 2 3 Ges B = (4 5 6) n‡j, AB = ? [JU 21-22] (4 10 18)       4 10 18       4 8 12 5 10 15 6 12 18 Am¤¢e DËi:       4 8 12 5 10 15 6 12 18 e ̈vL ̈v: A Gi μg 3  1 B Gi μg 1  3  AB Gi μg 3  3 GLv‡b ïaygvÎ bs Ack‡b 3  3 μ‡gi g ̈vwUa· we` ̈gvb| 6. k Gi †Kvb gv‡bi Rb ̈     k – 2 3 4 9 g ̈vwUa·wU Ae ̈wZμgx bq? [JU 21-22] 10 3 30 3 9 4 DËi: 10 3 e ̈vL ̈v: kZ©vbyhvqx,     k – 2 3 4 9 = 0  9k – 18 – 12 = 0  9k = 30  k = 10 3