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

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