Title Matrices Engineering Practice Sheet Solution 25.pdf ✅

4  Higher Math 1st Paper Chapter-1 Rhombus Publications 16. 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) 17. (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.) 18. 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.) weMZ mv‡j KUET-G Avmv cÖkœvejx 19. 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 = cos  3  2 = 2n   3   = n   6 ; n  Z (Ans.) 20. 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.) 21. x-Gi mgvavb Ki: x + 4 3 3 3 x + 4 5 5 5 x + 1 = 0 [KUET 04-05; BUET 13-14, 01-02; RUET 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 = 0 [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.)