Tiêu đề Matrices & Determinants Engineering Practice Sheet Solution.pdf ✅

g ̈vwUa· I wbY©vqK  Engineering Question Bank Solution 1 01 g ̈vwUa· I wbY©vqK Matrices and Determinants WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1|       1 + x 1 1 1 1 + y 1 1 1 1 + z = 0 n‡j Ges (x, y, z)  0 n‡j 1 x + 1 y + 1 z = ? [BUET 23-24] mgvavb:       1 + x 1 1 1 1 + y 1 1 1 1 + z = 0        x – y 0 0 y – z 1 1 1 + z = 0     c1 = c1 – c2 c2 = c2 – c3  x {y + yz + z} + yz = 0 [1g mvwi eivei we ̄Ívi K‡i]  xy + xyz + xz + yz = 0  1 z + 1 + 1 y + 1 x = 0 [xyz Øviv fvM K‡i]  1 x + 1 y + 1 z = – 1 2| hw`      x – 4 2x 2x 2x x – 4 2x 2x 2x x – 4 = (A + Bx) (x – A)2 nq, Zvn‡j A I B Gi gvb wbY©q Ki| [BUET 22-23] mgvavb:       x – 4 2x 2x 2x x – 4 2x 2x 2x x – 4 =       – (x + 4) x + 4 0 0 – (x + 4) x + 4 2x 2x x – 4 [ ] c1 = c1– c2 c2 = c2 – c3 = (x + 4)2       –1 1 0 0 –1 1 2x 2x x – 4 = (x + 4)2 {– (– x + 4 – 2x) + 2x} = (x + 4)2 (5x – 4) = (x – A)2 (Bx + A) A = – 4 Ges B = 5 (Ans.) 3| A =       0 1 0 0 0 1 1 0 0 Ges |l – A| = 0 n‡j  Gi gvb wbY©q Ki Ges A 12 Gi gvb †ei Ki| [BUET 21-22] mgvavb: I – A =        1 0 0 0 1 0 0 0 1 –       0 1 0 0 0 1 1 0 0 = 0         0 0 0  0 0 0  –       0 1 0 0 0 1 1 0 0 = 0         –1 0 0  –1 –1 0  = 0  ( 2 – 0) + 0 – 1 (1 – 0) = 0   3 – 1 = 0   = 3 1 = 1, ,  2 (Ans.) Avevi, A 2 =       0 1 0 0 0 1 1 0 0 .       0 1 0 0 0 1 1 0 0 =       0 0 1 1 0 0 0 1 0 A 3 = A2 A =       0 0 1 1 0 0 0 1 0       0 1 0 0 0 1 1 0 0 =       1 0 0 0 1 0 0 0 1 = I A 12 = (A3 ) 4 = I 4 =       1 0 0 0 1 0 0 0 1 (Ans.) 4| A =     1 2 – 1 4 , C =     2 4 5 – 2 Ges ABC =     – 1 2 0 3 n‡j g ̈vwUa· B = ? [BUET 20-21] mgvavb: A –1 =         2 3 – 1 3 1 6 1 6 ; C–1 =         1 12 1 6 5 24 – 1 12 B = A –1 ABC C –1 =         2 3 – 1 3 1 6 1 6     – 1 2 0 3         1 12 1 6 5 24 – 1 12  B =         1 72 – 5 36 23 144 – 7 72 (Ans.)
2  Higher Math 1st Paper Chapter-1 5| `ywU g ̈vwUa· A Ges B †`qv Av‡Q| AB I BA Gi g‡a ̈ †Kvb m¤úK© _vK‡j Zv wbY©q Ki| B –1 †K x I A Gi gva ̈‡g cÖKvk Ki| A =       3x – 4x 2x – 2x x 0 – x – x x Ges B =       x 2x – 2x 2x 5x – 4x 3x 7x – 5x [BUET 19-20; MIST 19-20] mgvavb: AB =       3x – 4x 2x – 2x x 0 – x – x x       x 2x – 2x 2x 5x – 4x 3x 7x – 5x = x       3 – 4 2 –2 1 0 –1 –1 1 . x       1 2 –2 2 5 –4 3 7 –5 = x 2       1 0 0 0 1 0 0 0 1 = x 2 I Abyiƒcfv‡e, BA =       x 2 0 0 0 x2 0 0 0 x2 = x 2       1 0 0 0 1 0 0 0 1 = x 2 I  AB = BA (Ans.) Ges BA = x2 I A = x2B –1 B –1 = A x 2 (Ans.) 6| hw` A –1 =         5 7 1 7 3 7 2 7 nq, Zvn‡j A 2 + 2A Gi gvb wbY©q Ki| [BUET 18-19; MIST 18-19] mgvavb: A = (A–1 ) –1 = 1 1 7         2 7 – 1 7 – 3 7 5 7 =     2 –1 –3 5  A 2 + 2A =     2 –1 –3 5     2 –1 –3 5 + 2     2 –1 –3 5 =     11 –9 –27 38 (Ans.) 7| hw`       4 1 3 A =       – 4 8 4 – 1 2 1 – 3 6 3 nq Zvn‡j A g ̈vwUa·wU wbY©q Ki| [BUET 17-18] mgvavb: GLv‡b A g ̈vwUa‡·i μg n‡e = 1 × 3 awi, A = [x y z]       4 1 3 × [x y z] =       – 4 8 4 – 1 2 1 – 3 6 3        4x 4y 4z x y z 3x 3y 3z =       – 4 8 4 – 1 2 1 – 3 6 3  x = – 1; y = 2; z = 1  A = [– 1 2 1] (Ans.) 8| hw` A =     4 3 2 1 Ges AB =     10 17 4 7 nq Z‡e B g ̈vwUa· Gi Dcv`vbmg~n †ei Ki| [BUET 16-17; MIST 21-22] mgvavb: A =     4 3 2 1  A –1 =      –  1 2 3 2 1 – 2 B = A –1AB =      –  1 2 3 2 1 – 2     10 17 4 7 =     1 2 2 3  B =     1 2 2 3 (Ans.) 9| g ̈vwUa‡·i mvnv‡h ̈ mgvavb Ki:     2 3 1 – 1     x y =     4 7 [BUET 15-16] mgvavb:     2 3 1 – 1     x y =     4 7      2x + 3y x – y =     4 7  2x + 3y = 4 .............(i) x – y = 7 ............. (ii) (i) I (ii) bs mgvavb K‡i cvB, (x, y)  (5, – 2) (Ans.) 10| x-Gi mgvavb Ki: x + 4 3 3 3 x + 4 5 5 5 x + 1 = 0 [BUET 13-14, 01-02; RUET 04-05; KUET 04-05; CUET 13-14] mgvavb: x + 4 3 3 3 x + 4 5 5 5 x + 1 = 0  x + 1 3 3 – x – 1 x + 4 5 0 5 x + 1 [c1 = c1 – c2]  (x + 1) 1 3 3 – 1 x + 4 5 0 5 x + 1 = 0  (x + 1) 1 3 3 0 x + 7 8 0 5 x + 1 = 0 [r2 = r1 + r2]  (x + 1) [(x + 7) (x + 1) – 40] = 0  (x + 1) (x + 11) (x – 3) = 0  x = –1, –11, 3 (Ans.)
g ̈vwUa· I wbY©vqK  Engineering Question Bank Solution 3 11| ln x ln y ln z ln 2x ln 2y ln 2z ln 3x ln 3y ln 3z = KZ? [BUET 09-10] mgvavb: ln x ln y ln z ln 2x ln 2y ln 2z ln 3x ln 3y ln 3z = ln x – ln y ln y – ln z ln z ln 2x – ln 2y ln 2y – ln 2z ln 2z ln 3x – ln 3y ln 3y – ln 3z ln 3z [c1 = c1 – c2, c2 = c2 – c3] = ln x y ln y z ln z ln x y ln y z ln 2z ln x y ln y z ln 3z = ln x y ln y z 1 1 ln z 1 1 ln 2z 1 1 ln 3z = 0 (Ans.) 12| KviY cÖ`k©b K‡i Ges we ̄Ívi bv K‡i mZ ̈ A_ev wg_ ̈v DËi Ki| [BUET 02-03] (i) 1 2 3 5 6 7 8 7 6 = 2 1 3 6 5 7 7 8 6 (ii) 1 3 – 4 2 8 3 0 – 2 5 = 1 2 0 3 8 – 2 – 4 3 5 (iii) Cofactor (mn ̧YK) of 2 in 2 4 3 4 is (– 3) (iv) x 2 – y 2 x + y x x – y 1 1 x – y 1 y = 0 mgvavb: (i) wg_ ̈v| KviY, 1 2 3 5 6 7 8 7 6 = – 2 1 3 6 5 7 7 8 6 (ii) mZ ̈| KviY, 1 3 – 4 2 8 3 0 – 2 5 = 1 2 0 3 8 – 2 – 4 3 5 (iii) wg_ ̈v| wbY©vqKwU‡Z 2 Gi mn ̧YK n‡e 4| (iv) mZ ̈| x 2 – y 2 x + y x x – y 1 1 x – y 1 y = (x – y) x + y x + y x 1 1 1 1 1 y = 0 13| a1 b1 c1 a2 b2 c2 a3 b3 c3 G cÖgvY Ki †h, a2A1 + b2B1 + c2C1 = 0 †hLv‡b A1, B1, C1 h_vμ‡g a1, b1, c1 Gi mn ̧YK| [BUET 99-00] mgvavb: A1 = (– 1)1+1 b2 c2 b3 c3 = b2c3 – b3c2 B1 = (– 1)1+2 a2 c2 a3 c3 = – (a2c3 – a3c2) = a3c2 – a2c3 C1 = (– 1)1+3 a2 b2 a3 b3 = a2b3 – a3b2 L.H.S. = a2A1 + b2B1 + c2C1 = a2(b2c3 – b3c2) + b2(a3c2 – a2c3) + c2(a2b3 – a3b2) = a2b2c3 – a2b3c2 + a3b2c2 – a2b2c3 + a2b3c2 – a3b2c2 = 0 = R.H.S (Proved) 14| mgvavb Ki: 3 – 2x 6 6 4 – x 12 12 1 – x 13 14 = 0 [BUET 98-99] mgvavb: 3 – 2x 6 6 4 – x 12 12 1 – x 13 14 = 0  3 – 2x 0 6 4 – x 0 12 1 – x –1 14 = 0 [c2 = c2 – c3]  (3 – 2x)12 + 6(– 4 + x) = 0  36 – 24x – 24 + 6x = 0  12 – 18x = 0  x = 2 3 (Ans.) 15| cÖgvY Ki †h, (b + c) 2 a 2 1 (c + a) 2 b 2 1 (a + b) 2 c 2 1 = – 2(a + b + c) (b – c) (c – a) (a – b) [BUET 97-98] mgvavb: L.H.S. = (b + c) 2 a 2 1 (c + a) 2 b 2 1 (a + b) 2 c 2 1 = (b + c) 2 – a 2 a 2 1 (c + a) 2 – b 2 b 2 1 (a + b) 2 – c 2 c 2 1 [c1 = c1 – c2] = (a + b + c) b + c – a a2 1 c + a – b b2 1 a + b – c c2 1
4  Higher Math 1st Paper Chapter-1 = (a + b + c) 2b – 2a a2 – b 2 0 2c – 2b b2 – c 2 0 a + b – c c2 1     r1 = r1 – r2 r2 = r2 – r3 = (a + b + c) –2(a – b) a 2 – b 2 –2(b – c) b 2 – c 2 = – 2 (a + b + c) (a – b) (b – c) 1 a + b 1 b + c = – 2 (a + b + c) (a – b) (b – c) (b + c – a – b) = – 2 (a + b + c) (b – c) (c – a) (a – b) = R.H.S. (Proved) 16| (we ̄Ívi bv K‡i) wbY©vqKwUi gvb wbY©q Ki: bc ca ab 1 a 1 b 1 c 1 a + b 1 b + c 1 c + a [BUET 95-96] mgvavb: bc ca ab 1 a 1 b 1 c 1 a + b 1 b + c 1 c + a = 1 abc abc abc abc 1 1 1 1 + ab 1 + bc 1 + ca [1g Kjvg‡K a, 2q Kjvg‡K b, 3q Kjvg‡K c Øviv ̧Y K‡i] = 1 abc × abc 1 1 1 1 1 1 1 + ab 1 + bc 1 + ca = 0 (Ans.) 17| hw` F (x) = f(x) (x) g(x) (x) nq, cÖgvY Ki †h, F(x + h) – F(x) = f(x + h) – f(x) (x + h) g(x + h) – g(x) (x + h) + f(x) (x + h) – (x) g(x) (x + h) – (x) [BUET 95-96] mgvavb: R.H.S. = f(x + h) – f(x) (x + h) g(x + h) – g(x) (x + h) + f(x) (x + h) – (x) g(x) (x + h) – (x) = f(x + h) (x + h) g(x + h) (x + h) – f(x) (x + h) g(x) (x + h) + f(x) (x + h) g(x) (x + h) – f(x) (x) g(x) (x) = F(x + h) – F(x) = L.H.S.  L.H.S = R.H.S (Proved) weMZ mv‡j KUET-G Avmv cÖkœvejx 18| hw` A =     cos sin –sin cos Ges A 2 = 1 2     1 3 – 3 1 ;  Gi gvb wbY©q Ki| [KUET 09-10] mgvavb: A 2 = A.A =     cos sin –sin cos     cos sin –sin cos =     cos2  – sin2  2cossin –2sincos cos2  – sin2  Avevi, A 2 =       1 2 3 2 – 3 2 1 2  cos2  – sin2  = 1 2  cos2 = cos60 = cos  3  2 = 2n   3   = n   6 (Ans.) 19| hw` A =       1 2 2 2 1 2 2 2 1 nq, gvb wbY©q Ki: A 2 – 4A – 5I| [KUET 05-06] mgvavb: A 2 =       1 2 2 2 1 2 2 2 1       1 2 2 2 1 2 2 2 1 =       9 8 8 8 9 8 8 8 9  A 2 – 4A – 5I =       9 8 8 8 9 8 8 8 9 –       4 8 8 8 4 8 8 8 4 –       5 0 0 0 5 0 0 0 5 =       0 0 0 0 0 0 0 0 0 (Ans.) 20| hw` A =     3 1 –2 –2 , B =     1 0 2 1 3 5 Ges C =       2 –3 0 , nq Z‡e †`LvI †h, (AB) C = A (BC). [KUET 04-05] mgvavb: AB =     3 1 –2 –2     1 0 2 1 3 5 =     3 1 4 0 –1 –7 BC =     1 0 2 1 3 5       2 –3 0 =     – 4 – 3  L.H.S  (AB)C =     3 1 4 0 – 1 – 7       2 – 3 0 =     – 6 2  R.H.S  A(BC) =     3 1 – 2 – 2     – 4 – 3 =     – 6 2  L.H.S = R.H.S (Showed)