Tiêu đề Matrices & Determinants Practice Sheet HSC FRB 24.pdf ✅

g ̈vwUa· I wbY©vqK  Final Revision Batch 1 01 g ̈vwUa· I wbY©vqK Matrices and Determinants Board Questions Analysis m„Rbkxj cÖkœ †evW© mvj XvKv gqgbwmsn ivRkvnx Kzwgjøv h‡kvi PÆMÖvg ewikvj wm‡jU w`bvRcyi 2023 2 2 2 2 1 2 1 2 1 2022 2 1 2 1 2 2 1 2 1 eûwbe©vPwb cÖkœ †evW© mvj XvKv gqgbwmsn ivRkvnx Kzwgjøv h‡kvi PÆMÖvg ewikvj wm‡jU w`bvRcyi 2023 5 5 4 4 7 4 5 4 5 2022 3 5 4 5 4 4 4 5 4 weMZ mv‡j †ev‡W© Avmv m„Rbkxj cÖkœ 1| `„k ̈Kí-1: x + y + z = 1 [XvKv †evW©- Õ23] x + 2y + z = 2 x + y + 2z = 0 `„k ̈Kí-2: D = 8         p – q – r 2 q r p q – r – p 2 r p q r – p – q 2 (K) A = [1 2 3] Ges B =       3 2 1 n‡j, (AB)t wbY©q Ki| (L) `„k ̈Kí-1: G DwjøwLZ mgxKiY †RvU wbY©vq‡Ki mvnv‡h ̈ mgvavb Ki| (M) `„k ̈Kí-2: †_‡K cÖgvY Ki †h, D = S3 , †hLv‡b S = p + q + r DËi: (K) [10] ; (L) (x, y, z)  (1, 1, – 1) 2| P =       4 0 6 – 1 7 – 2 3 5 2 Ges Q =       0 – 3 – 2 4 – 4 1 3 – 5 2 [XvKv †evW©- Õ23] (K) P + Q g ̈vwUa‡·i †Uam wbY©q Ki| (L) cÖgvY Ki †h, (PQ)t = QtP t (M) PR = RP = I n‡j, R g ̈vwUa·wU wbY©q Ki| †hLv‡b I GKwU A‡f`K g ̈vwUa·| DËi: (K) 11 ; (M) 1 30       – 12 30 21 2 5 – 1 13 10 – 14 3| `„k ̈Kí-1: x – 2y + 2z = 1 [ivRkvnx †evW©- Õ23] 2x + 6y – z = 2 x + 3y – 3z = 3 `„k ̈Kí-2:  =       1 1 1 x y z x 2 y 2 z 2 , 1 =       1 yz x 1 zx y 1 xy z (K) †`LvI †h,     2 3 – 1 – 2 GKwU A‡f`NvwZ (involutary) g ̈vwUa·| (L) `„k ̈Kí-1 G ewY©Z mgxKiY †RvUwU wbY©vqK c×wZ‡Z mgvavb Ki| (M) `„k ̈Kí-2 e ̈envi K‡i †`LvI †h,  + 1 = 0. DËi: (L) (x, y, z) =     9 5  – 2 5  – 4 5 4| A =       1 0 3 2 1 – 1 1 – 1 1 ,  =       x – 1 1 3 2 x – 1 2 3 1 x – 1 [ivRkvnx †evW©- Õ23] (K) K Gi †Kvb gv‡bi Rb ̈ A =    K  – 3 –2 – 1 K – 2 e ̈wZμgx g ̈vwUa· n‡e? (L) DÏxcK n‡Z A 3 – 3A2 – A + 9I = 0 Gi mvnv‡h ̈ A –1 wbY©q Ki| (M) DÏxc‡Ki mvnv‡h ̈ | + I| = 0 mgxKi‡Yi mgvavb Ki| †hLv‡b I GKwU A‡f`K g ̈vwUa·| DËi: (K) K = 1, 4 ; (L) 1 9       0 3 3 3 2 – 7 3 – 1 – 1 (M) x = – 4, 1, 3
2  Higher Math 1st Paper Chapter-1 5| A = (1 –2 3) [h‡kvi †evW©- Õ23] X = (x y z), B =       1 1 4 – 2 5 – 2 3 0 1 C =       (m + n) 2 m 2 n 2 l 2 (n + l) 2 n 2 l 2 m 2 (l + m) 2 (K) 3     1 2 – 1 4 + E = I2 n‡j E g ̈vwUa·wU wbY©q Ki| (L) †μgv‡ii wbq‡g BXT = AT mgxKiY †RvU mgvavb Ki| (M) †`LvI †h, |C| = 2lmn(l + m + n)3 . DËi: (K)     – 2 – 6 3 – 11 ; (L) (x, y, z) =     32 59  – 30 59  – 1 59 6| A =       2 1 1 – 1 1 – 1 3 1 2 , B =       x y z , C =       2 5 4 [Kzwgjøv †evW©- Õ23] (K) Kx k‡Z© `yBwU g ̈vwUa‡·i †hvM I ̧Y Kiv m¤¢e? (L) AB = C n‡j †μgv‡ii wbq‡g mgxKiY †RvUwUi mgvavb Ki| (M) A –1 wbY©q Ki| DËi: (L) (x, y, z) = (– 15, 7 13) ; (M)       3 – 1 – 2 – 1 1 1 – 4 1 3 7| M =     a – 5 2 2 a – 2 , N =       – 1 2 4 2 1 – 2 – 3 0 5 P =       – 2 – 2 a + b – c a + b b + c c 2 – c – a ab [Kzwgjøv †evW©- Õ23] (K) a Gi gvb KZ n‡j M GKwU e ̈wZμgx g ̈vwUa· n‡e? (L) N 2 – 5N + 4I wbY©q Ki| (M) †`LvI †h, |P| = (c – a) (a2 + b2 + c2 ) DËi: (K) a = 1, 6 ; (L)       2 – 10 – 8 – 4 4 6 3 – 6 – 8 8| 2x – y – z = 6, x + 3y + 2z = 1 Ges 3x – y – 5z = 1 [PÆMÖvg †evW©- Õ23] (K) we ̄Ívi bv K‡i      a b c 1 1 1 b + c c + a a + b Gi gvb wbY©q Ki| (L) x, y I z Gi mnM ̧‡jv wb‡q MwVZ g ̈vwUa· A n‡j A –1 wbY©q Ki| (M) †μgv‡ii wbq‡g mgxKiY †RvU mgvavb Ki| DËi: (K) 0 ; (L) 1 27       13 – 11 10 4 7 1 – 1 5 – 7 (M) (x, y, z) = (3, – 2, 2) 9| Q =       3 + x 4 2 4 2 + x 3 2 3 4 + x [PÆMÖvg †evW©- Õ23] (K)     1 – 1 2 – k g ̈vwUa·wU e ̈wZμgx g ̈vwUa· n‡j k Gi gvb wbY©q Ki| (L) hw` x = 7 nq, Q 2 – 5Q + 3I3 Gi gvb wbY©q Ki †hLv‡b I3 GKK g ̈vwUa·| (M) |Q| = 0 n‡j, mgvavb †mU wbY©q Ki| DËi: (K) k = 2 ; (L)       73 62 44 62 64 53 44 53 82 ; (M) {– 9  – 3  3} 10| A =       p p + 1 p + 1 p + 1 p p + 1 p + 1 p + 1 p [wm‡jU †evW©- Õ23] (K) we ̄Ívi bv K‡i      1 4 6 2 5 7 3 6 8 Gi gvb wbY©q Ki| (L) DÏxc‡Ki Av‡jv‡K A 2 – 7A – 8I3 wbY©q Ki ; hLb p = 2 (M) AX = B n‡j wbY©vq‡Ki mvnv‡h ̈ ‘X’ wbY©q Ki ; †hLv‡b p = 1, B =       11 10 9 DËi: (K) 0 ; (L)       0 0 0 0 0 0 0 0 0 ; (M) X =       x y z =       1 2 3 11| px + qy + rz = 1 [wm‡jU †evW©- Õ23] p 2 x + q2 y + r2 z = a (p3 – 1)x + (q3 – 1)y + (r3 – 1)z = a2 (K) cÖgvY Ki †h,     4 – 4 3 – 3 GKwU mgNvZx g ̈vwUa·| (L) DÏxc‡Ki mgxKiY ̧‡jv‡K AX = B AvKv‡i cÖKvk K‡i †`LvI †h, pqr = 1, hLb Det(A) = 0 Ges p  q  r (M) l = 1, m = 2, n = – 1 n‡j, A –1 wbY©q Ki| DËi: (M) 1 – 18       15 – 2 – 7 3 2 7 – 6 2 – 2 12| mgxKiY †RvU: tx + uy + vz = 5 [ewikvj †evW©- Õ23] t 2 x + u2 y + v2 z = 5 (t3 – 1)x + (u3 – 1)y + (v3 – 1)z = – 5 (K) M =       2 9 – 3 , N = [– 3 5 6] n‡j, [MN]T wbY©q Ki| (L) t = 1, u = 2, v = 3 n‡j †μgv‡ii wbq‡g mgxKiY †Rv‡Ui mgvavb Ki| (M) x, y, z Gi mnM ̧wj Øviv MwVZ wbY©vqK D n‡j cÖgvY Ki, D = (tuv – 1) (t – u) (u – v) (v – t) DËi: (K)       – 6 10 12 – 27 45 54 9 – 15 – 18 ; (L) (x, y, z) = (2, 3, – 1)
g ̈vwUa· I wbY©vqK  Final Revision Batch 3 13| A =       2 3 5 – 1 1 2 1 – 4 – 3 , B =       p 2 q 2 r 2 qr rp pq 2p 2q 2r [w`bvRcyi †evW©- Õ23] (K)     x – 5 – 1 8 y + 3 =     y – 1 – 1 8 7 n‡j (x, y) Gi gvb wbY©q Ki| (L) A –1 wbY©q Ki| (M) cÖgvY Ki †h, |B| = – 2(p – q) (q – r) (p – r) (pq + qr + rp) DËi: (K) (x, y) = (8, 4) ; (L) 1 22       5 – 11 1 – 1 – 11 –9 3 11 5 T 14| A =     1 0 4 1 , B =     1 0 m n , C =       0 1 2 1 2 0 2 0 4 Ges f(x) = x2 + 5x + 6. [gqgbwmsn †evW©- Õ23] (K) P =       1 2 3 Ges Q = (4 5 6) n‡j (PQ)T wbY©q| (L) AB = I2 n‡j m I n Gi gvb †ei Ki| (M) f(C) wbY©q Ki| DËi: (K)       4 5 6 8 10 12 12 15 18 ; (L) m = – 4 Ges n = 1 (M)       11 7 18 7 21 2 18 2 46 15| A =       (b + c) 2 (c + a) 2 (a + b) 2 a 2 b 2 c 2 bc ca ab , B =       1 0 3 0 2 0 1 0 1 , C =       x y z , D =       1 2 1 [gqgbwmsn †evW©- Õ23] (K) we ̄Ívi bv K‡i cÖgvY Ki †h,       1 1 1 bc ca ab bc(b + c) ca(c + a) ab(a + b) = 0 (L) †`LvI †h, detA = (a2 + b2 + c2 ) (a + b + c) (a – b) (b – c) (c – a). (M) BC = D n‡j †μgv‡ii wbq‡g mgxKiY †RvU mgvavb Ki| DËi: (L) (a 2 + b2 + c2 ) (a + b + c) (a – b) (b – c) (c – a) (M) (x, y,z) = (0, 1, 1) 16| A =       x – 1 – 2 2 1 x + 1 0 2 3 x Ges B =     5 2 –10 –4 [XvKv †evW©- Õ22] (K) †`LvI †h, B GKwU mgNvwZ g ̈vwUa·| (L) |A| = 0 n‡j, x Gi gvb wbY©q Ki| (M) (A T ) –1 wbY©q Ki hLb, x = 0 nq| DËi: (L) x = – 2, 1 ; (M) 1 2       0 0 1 6 – 4 – 1 – 2 2 1 17| A =       3 + x 4 1 4 1 + x 3 1 3 4 + x , B =       2 + x 2 + y 4 b + x b + y b 2 c + x c + y c 2 [ivRkvnx †evW©- Õ22] (K) B =     2 0 1 2 n‡j, B.Bt wbY©q Ki| (L) †`LvI †h, det (B) = (2 – b) (b – c) (c – 2) (x – y) (M) det(A) = 0 mgxKi‡Yi ev ̄Íe g~j wb‡q A Gi †Uam wbY©q Ki| DËi: (K)     5 2 2 4 ; (M) –16 ; 8 + 3 7 ; 8 – 3 7 18| A =       3 4 2 2 2 1 3 1 – 2 , B =       2 3 1 10 8 8 3 2 1 , C =       a b 3 12 10 9 6 3 – 1 , X =       x y z , D =       5 10 15 [Kzwgjøv †evW©- Õ22] (K) A + B = C n‡j a, b Gi gvb wbY©q Ki| (L) B –1 wbY©q Ki| (M) wbY©vq‡Ki mvnv‡h ̈ AX = D Gi mgvavb Ki| DËi: (K) a = 5 I b = 7 ; (L) 1 22       – 8 – 1 16 14 – 1 – 6 – 4 5 – 14 (M) (x, y, z) = (– 3, 13, – 4) 19| P =       1 1 1 5 – 1 2 3 6 – 5 , Q =       6 9 0 , R =       x y z [h‡kvi †evW©- Õ22] (K) `yBwU g ̈vwUa‡·i ̧Yb †hvM ̈Zv e ̈vL ̈v Ki| (L) f(x) = x2 – 3x n‡j, f(P) wbY©q Ki| (M) P TR = Q †_‡K cÖvß mgxKiY †RvU‡K wbY©vq‡Ki mvnv‡h ̈ mgvavb Ki| DËi: (L)       6 3 – 5 – 9 21 – 13 9 – 51 55 ; (M) (x, y, z) = (1, 2, 3) 20| S =     – 1 2 1 – 3 , T =     3 – 1 – 5 2 , U =       a 2a3 + 1 a 2 b 2b3 + 1 b 2 c 2c3 + 1 c 2 [h‡kvi †evW©- Õ22] (K) we ̄Ívi bv K‡i cÖgvY Ki      a b c – x – y – z a + x b + y c + z = 0. (L) †`LvI †h, (ST)–1 – T –1 S –1 GKwU k~b ̈ g ̈vwUa·| (M) cÖgvY Ki †h, |U| = – (2abc + 1) (a – b) (b – c) (c – a).