Title Matrices CQ & MCQ Practice Sheet Solution (HSC 27).pdf ✅

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