Tiêu đề Matrices Engineering Practice Sheet Solution-2025.pdf ✅

g ̈vwUa· I wbY©vqK  Engineering Practice Sheet Solution 1 01 g ̈vwUa· I wbY©vqK Matrices and Determinants WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. A       x y z =       x + y x – y 0 n‡j, A g ̈vwUa·wU wbY©q Ki| [BUET 24-25] mgvavb: GLv‡b, A g ̈vwUa‡·i gvÎv n‡e 3  3 awi, A =       a d g b e h c f i        a d g b e h c f i       x y z =       x + y x – y 0        ax + by + cz dx + ey + fz gx + hy + iz =       x + y x – y 0 ...... (i) (i) n‡Z mnM mgxK...Z K‡i cvB, a = 1, b = 1, c = 0 d = 1, e = – 1, f = 0 g = 0, h = 0, i = 0  wb‡Y©q g ̈vwUa·, A =       1 1 0 1 – 1 0 0 0 0 (Ans.) 2.       1 + x 1 1 1 1 + y 1 1 1 1 + z g ̈vwUa‡·i D = 0 Ges (x, y, z)  0 n‡j, 1 x + 1 y + 1 z =? [BUET 23-24] mgvavb: D = 0        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  xy + xyz + xz + yz = 0  yz + xz + xy = – xyz  1 x + 1 y + 1 z = – 1 (Ans.) [xyz Øviv fvM K‡i] 3. 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     c1 = c1– c2 c2 = 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.) 4. A =       0 1 0 0 0 1 1 0 0 Ges |I – 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         0 – 1 – 1  0 0 – 1  = 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 = A2A =       0 0 1 1 0 0 0 1 0       0 1 0 0 0 1 1 0 0 = I A 12 = (A3 ) 4 = I4 =       1 0 0 0 1 0 0 0 1 (Ans.)