Title Matrices Practice Sheet Solution (HSC FRB 25).pdf ✅

g ̈vwUa· I wbY©vqK  Final Revision Batch '25 1 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 mgvavb: (K) †`Iqv Av‡Q, A = [1 2 3] Ges B =       3 2 1  AB = [1 2 3]       3 2 1 = [3 + 4 + 3] = [10]  (AB)t = [10]t = [10] (Ans.) (L) †`Iqv Av‡Q, x + y + z = 1 x + 2y + z = 2 x + y + 2z = 0 x, y I z Gi mnM ̧‡jv wb‡q MwVZ wbY©vqK, D =       1 1 1 1 2 1 1 1 2 = 1(4 – 1) – 1(2 – 1) + 1(1 – 2) = 1  0 GLb, Dx =       1 2 0 1 2 1 1 1 2 = 1(4 – 1) – 1(4 – 0) + 1(2 – 0) = 1 Dy =       1 1 1 1 2 0 1 1 2 = 1(4 – 0) – 1(2 – 1) + 1(0 – 2) = 1 Dz =       1 1 1 1 2 1 1 2 0 = 1(0 – 2) – 1(0 – 2) + 1(1 – 2) = – 1  x = Dx D = 1 1 = 1 y = Dy D = 1 1 = 1 z = Dz D = – 1 1 = – 1  wb‡Y©q mgvavb: (x, y, z)  (1, 1, – 1) (Ans.) (M) D = 8         p – q – r 2 q r p q – r – p 2 r p q r – p – q 2 = 8  1 2  1 2  1 2       p – q – r 2q 2r 2p q – r – p 2r 2p 2q r – p – q [cÖ‡Z ̈KUv mvwi‡K 2 Øviv ̧Y K‡i|] = 8  1 8       p + q + r 2q 2r p + q + r q – r – p 2r p + q + r 2q r – p – q [r1 = r1 + r2 + r3]
2  Higher Math 1st Paper Chapter-1 = (p + q + r)      1 2q 2r 1 q – r – p 2r 1 2q r – p – q = (p + q + r)       0 p + q + r 0 0 – (p + q + r) p + q + r 1 2q r – p – q     c1 = c1 – c2 c2 = c2 – c3 = (p + q + c) {(p + q + r)2 + 0} [1g mvwi mv‡c‡ÿ we ̄Ívi K‡i] = (p + q + r)3 = S3 [S = p + q + r]  D = S3 (Proved) 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·| mgvavb: (K) †`Iqv Av‡Q, P =       4 0 6 – 1 7 – 2 3 5 2 Ges Q =       0 – 3 – 2 4 – 4 1 3 – 5 2  P + Q =       4 0 6 – 1 7 – 2 3 5 2 +       0 – 3 – 2 4 – 4 1 3 – 5 2 =       4 + 0 0 – 3 6 – 2 – 1 + 4 7 – 4 – 2 + 1 3 + 3 5 – 5 2 + 2 =       4 – 3 4 3 3 – 1 6 0 4  (P + Q) g ̈vwUa‡·i †Uam = 4 + 3 + 4 = 11 (Ans.) (L) †`Iqv Av‡Q, P =       4 0 6 – 1 7 – 2 3 5 2  P t =       4 0 6 – 1 7 – 2 3 5 2 t =       4 – 1 3 0 7 5 6 – 2 2 Ges Q =       0 – 3 – 2 4 – 4 1 3 – 5 2 Q t =       0 – 3 – 2 4 – 4 1 3 – 5 2 t =       0 4 3 – 3 – 4 – 5 – 2 1 2 PQ =       4 0 6 – 1 7 – 2 3 5 2       0 – 3 – 2 4 – 4 1 3 – 5 2 =       0 + 3 – 6 0 – 21 – 10 0 + 6 – 4 16 + 4 + 3 0 – 28 + 5 24 + 8 + 2 12 + 5 + 6 0 – 35 + 10 18 + 10 + 4 =       –3 – 31 2 23 – 23 34 23 – 25 32 L.H.S. = (PQ)t =       –3 – 31 2 23 – 23 34 23 – 25 32 t =       – 3 23 23 – 31 – 23 – 25 2 34 32 R.H.S. = Qt P t =       0 4 3 – 3 – 4 – 5 – 2 1 2       4 – 1 3 0 7 5 6 – 2 2 =       0 + 3 – 6 16 + 4 + 3 12 + 5 + 6 0 – 21 – 10 0 – 28 + 5 0 – 35 + 10 0 + 6 – 4 24 + 8 + 2 18 + 10 + 4 =       – 3 23 23 – 31 – 23 – 25 2 34 32 = (PQ)t L.H.S. = R.H.S. (Proved) (M) †`Iqv Av‡Q, P =       4 0 6 – 1 7 – 2 3 5 2 PR = RP = I n‡j, R = P–1 GLb, |P| =       4 0 6 – 1 7 – 2 3 5 2 = 4(14 + 10) + 1(0 – 30) + 3(0 – 42) = – 60  0  P –1 we` ̈vgvb| P Gi mn ̧YK mg~n: P11 =     7 – 2 5 2 = 14 + 10 = 24 P12 = –     0 6 5 2 = – (0 – 30) = 30 P13 =     0 6 7 2 = 0 – 42 = – 42 P21 =     – 1 – 2 3 2 = – (– 2 + 6) = – 4 P22 =     4 6 3 2 = 8 – 18 = – 10 P23 = –     4 6 – 1 – 2 = – (– 8 + 6) = 2 P31 =     – 1 7 3 5 = – 5 – 21 = – 26 P32 = –     4 0 3 5 = – (20 – 0) = – 20 P33 =     4 0 – 1 7 = 28 – 0 = 28 Adj P =       P11 P21 P31 P12 P22 P12 P13 P23 P33 T =       24 – 4 – 26 30 – 10 – 20 – 42 2 28 T =       24 30 – 42 – 4 – 10 2 – 26 – 20 28
g ̈vwUa· I wbY©vqK  Final Revision Batch '25 3 GLb, R = P–1 = 1 |P| AdjP = 1 – 60       24 30 – 42 – 4 – 10 2 – 26 – 20 28 = 1 30       – 12 – 15 21 2 5 – 1 13 10 – 14 (Ans.) 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. mgvavb: (K) awi, P =     2 3 – 1 – 2 GLb, P 2 = P.P =     2 3 – 1 – 2     2 3 – 1 – 2 =     4 – 3 6 – 6 – 2 + 2 – 3 + 4 =     1 0 0 1 = I †h‡nZz P 2 = I myZivs, P =     2 3 – 1 – 2 GKwU A‡f`NvwZ g ̈vwUa·| (Showed) (L) †`Iqv Av‡Q, x – 2y + 2z = 1 2x + 6y – z = 2 x + 3y – 3z = 3 GLb, D =       1 2 1 – 2 6 3 2 – 1 – 3 = 1(– 18 + 3) + 2(– 6 + 1) + 2(6 – 6) = – 25  0 GLb, Dx =       1 2 3 – 2 6 3 2 – 1 – 3 = 1(– 18 + 3) + 2(– 6 + 3) + 2(6 – 18) = – 45 Dy =       1 2 1 1 2 3 2 – 1 – 3 = 1(– 6 + 3) – 1(– 6 + 1) + 2(6 – 2) = 10 Dz =       1 2 1 – 2 6 3 1 2 3 = 1(18 – 6) + 2(6 – 2) + 1(6 – 6) = 20  x = Dx D = – 45 – 25 = 9 5 y = Dy D = 10 – 25 = – 2 5 z = Dz D = 20 – 25 = – 4 5  wb‡Y©q mgvavb: (x, y, z) =     9 5  – 2 5  – 4 5 (Ans.) (M) L.H.S. =  + 1 =       1 1 1 x y z x 2 y 2 z 2 +       1 yz x 1 zx y 1 xy z =       1 1 1 x y z x 2 y 2 z 2 +       1 1 1 yz zx xy x y z =       1 1 1 x y z x 2 y 2 z 2 + 1 xyz       x y z xyz xyz xyz x 2 y 2 z 2 =       1 1 1 x y z x 2 y 2 z 2 + 1 xyz . xyz       x y z 1 1 1 x 2 y 2 z 2 =       1 1 1 x y z x 2 y 2 z 2 –       1 1 1 x y z x 2 y 2 z 2 = 0 = R.H.S. (Showed) 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·| mgvavb: (K) †`Iqv Av‡Q, A =     K – 3 –2 – 1 K – 2 g ̈vwUa·wU e ̈wZμgx n‡j,     K – 3 –2 – 1 K – 2 = 0  (K – 3) (K – 2) – 2 = 0  K 2 – 2K – 3K + 6 – 2 = 0  K 2 – 5K + 4 = 0  K 2 – K – 4K + 4 = 0  K(K – 1) – 4(K – 1) = 0  (K – 1) (K – 4) = 0 nq, K – 1 = 0  K = 1 A_ev, K – 4 = 0  K = 4 wb‡Y©q K = 1, 4 (Ans.)
4  Higher Math 1st Paper Chapter-1 (L) †`Iqv Av‡Q, A =       1 0 3 2 1 – 1 1 – 1 1 Ges A 3 – 3A2 – A + 9I = 0  A –1 (A3 – 3A2 – A + 9I) = 0  A –1 .A3 – 3A–1 .A2 – A –1A + 9A–1 .I = 0 [⸪ A–1 .A = I ; A–1 I = A–1 ]  I.A2 – 3I.A – I + 9A–1 = 0  9A–1 = 3A – A.A + I = 3      1 0 3 2 1 – 1 1 – 1 1 –       1 0 3 2 1 – 1 1 – 1 1       1 0 3 2 1 – 1 1 – 1 1 +       1 0 0 0 1 0 0 0 1 =       3 0 9 6 3 – 3 3 – 3 3 –       4 – 3 6 3 2 4 0 – 2 5 +       1 0 0 0 1 0 0 0 1 =       3 – 4 + 1 0 + 3 + 0 9 – 6 + 0 6 – 3 + 0 3 – 2 + 1 – 3 – 4 + 0 3 – 0 + 0 – 3 + 2 + 0 3 – 5 + 1  9A–1 =       0 3 3 3 2 – 7 3 – 1 – 1  A –1 = 1 9       0 3 3 3 2 – 7 3 – 1 – 1 (Ans.) (M) †`Iqv Av‡Q,  =       x – 1 1 3 2 x – 1 2 3 1 x – 1   + I =       x – 1 1 3 2 x – 1 2 3 1 x – 1 +       1 0 0 0 1 0 0 0 1 =       x 1 3 2 x 2 3 1 x GLv‡b,| + I| = 0        x 1 3 2 x 2 3 1 x = 0  x(x2 – 2) – 2(x – 3) + 3(2 – 3x) = 0  x 3 – 2x – 2x + 6 + 6 – 9x = 0  x 3 – 13x + 12 = 0  x 3 – x 2 + x2 – x – 12x + 12 = 0  x 2 (x – 1) + x(x – 1) – 12(x – 1) = 0  (x – 1) (x2 + x – 12) = 0  (x – 1) (x2 – 3x + 4x – 12) = 0  (x – 1) {x(x – 3) + 4(x – 3)} = 0  (x – 1) (x – 3) (x + 4) = 0 nq, x – 1 = 0  x = 1 A_ev, x – 3 = 0  x = 3 A_ev, x + 4 = 0  x = – 4  wb‡Y©q mgvavb: x = – 4, 1, 3 (Ans.) 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 . mgvavb: (K) †`Iqv Av‡Q, 3     1 2 – 1 4 + E = I2      3 6 – 3 12 + E =     1 0 0 1  E =     1 0 0 1 –     3 6 – 3 12 =     – 2 – 6 3 – 11 (Ans.) (L) A T = (1 –2 3)T =       1 – 2 3 B =       1 1 4 – 2 5 – 2 3 0 1 Ges X = (x y z) X T = (x y z)T =       x y z GLb, BXT = AT        1 1 4 – 2 5 – 2 3 0 1       x y z =       1 – 2 3        x – 2y + 3z x + 5y + 0z 4x – 2y + z =       1 – 2 3 g ̈vwUa‡·i mgZv Abymv‡i cvB, x – 2y + 3z = 1 x + 5y + 0z = – 2 4x – 2y + z = 3 x, y I z Gi mnM ̧‡jv wb‡q MwVZ wbY©vqK, D =       1 1 4 – 2 5 – 2 3 0 1 = 1(5 + 0) + 2(1 – 0) + 3(– 2 – 20) = – 59 GLb, Dx =       1 – 2 3 – 2 5 – 2 3 0 1 = 1(5 + 0) + 2(– 2 – 0) + 3(4 – 15) = – 32 Dy =       1 1 4 1 – 2 3 3 0 1 = 1(– 2 – 0) – 1(1 – 0) + 3(3 + 8) = 30