Title Matrices Engineering Practice Sheet Solution-2025.pdf ✅

2  Higher Math 1st Paper Chapter-1 5. 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 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.) 6. `ywU g ̈vwUa· A Ges B †`qv Av‡Q| AB I BA Gi g‡a ̈ †Kv‡bv 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 – 2 – 1 – 4 1 – 1 2 0 1 . x       1 2 3 2 5 7 – 2 – 4 – 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.) 7. 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 – 3 7 – 1 7 5 7 =     2 – 1 – 3 5  A 2 + 2A =     2 – 3 – 1 5     2 – 3 – 1 5 + 2     2 – 3 – 1 5 =     11 – 27 – 9 38 (Ans.) 8. 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.) 9. 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  A –1 AB =      –  1 2 3 2 1 – 2     10 17 4 7 =     1 2 2 3  B =     1 2 2 3 (Ans.) 10. 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.) 11. 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 = 0 [c1 = c1 – c2]  (x + 1) 1 3 3 – 1 x + 4 5 0 5 x + 1 = 0