Page 1 : log.x logey loge 2!, log.x loge y loge 2’, logex logey loge 2!, 111, =logex logey logez|1 1 1, , _ loge x loge y loge z, ~ log. x log. y log. z, , 11, =logex logey logez {0}=0, 1, , 5 @, , 14.1f A=|? _, |, Prove that Y7_, det(A*), Oo , 2, , a 1, 1 then || = 7-0, 2, , a, a, 2., a, i, , |, Her det(Ak yai+t+, , 6, Above series is like a GP, , wy) 4, 5, = 4G) _ 059) _1(1 4) Hence Proved, 4 ;, 15.With out expanding , Evaluate the following determinants, 2 3 4, @|5 6 8, 6x 9x 12x, 2 3 4 2 3 4, 5 6 =3x15 6 al =2100 =0 [= Ri = Rs, 6x 9x 12x 2 3 4, x+y ytz ztx, , (ii)| z a y, , 1 1 1, xtytz xt+yt+z x+y+z|, x 7, 1 1, ah oe, , =(xty+2)|z x yl=(xt+y+z){0}=0 [= Ry = Ry, 111, 16.1f A is a square matrix and |A| = 2,find the value of |AA"|, |aa?| = |A| |A"| = [A] [A] = 2x 2 = 4 [- 14" = lal, 17.1f A and B are square matrices of order 3 such that, |A| = -1 and |B| = 3, find the value of |3AB| ., [3AB| = 3°|AB] [+ [kal =, 33|Al|B|, = 27(-1)(3) = -81, , Ry > Ry +Rz, , k"|Al, , 0 2a 1, 18. If A = —2, Determine the value of |27, 034° +1, 1 6A-1 0, 1 Oo -4 1, 3¥41)=|4 0 13, i—1 64-1 0 -1 -13 0, = 4(0 — 13) + 1(—52 — 0) = 52-52 =0, _ ft 4 20, 19.Determine the roots of the equation |i -2 5 | =0, , 1 2x 5x?, 4 20, , -2 5|=0, 2x 5x", 4 20, , -6 -15, 2x-4 5x?-20l, 1(-—30x? + 120 + 30x - 60) = 0, —30x? + 30x + 6, x?-x-2=0, , (e-2)@+D)=0, x=2 andx=-1, , RoR , RyRy -R, 7", , , , , , , , , , 43 -2, 20.Verify det(AB) = det(A) det(B) for a 0 7 jand, 23 , To Prove: |AB| = |A| |B|, 4 3 -2 3-3), ani 0 7{/-2 4 0, 23 -sIlo 755., 4-6-18 12+12-14 12+0-10, 1+04+63 340449 340435, , 2-6-45 6+12-35 6+0-25, —20 10 |, , =|64 52 38, -49 -17 -19, = —20(—988 + 646) — 10(—1216 + 1862) + 2(—1088 + 2548), =-20(-342)-10(646)+2(1460)= 6840-6460+2920 = 3300 --------(1), |A| (0 — 21) — 3(-5 — 14) - 2(3 - 0), = 4(—21) — 3(-19) — 2(3) = -84 + 57 —- 6 = -33, |B|= 1(20 — 0) — 3(-10 - 0) + 3(—-14 - 36), = 1(20) — 3(—10) + 3(—50) = 20 + 30 —- 150 = -100, JA| |B| =-33 (-100) = 3300 —., From (1) and (2) |AB | = |A| |B|, 21.Using co factors element second row, Evaluate | A|, 5 3 8, where af 0 i, 12 3, |Al= ee ee ae, = lett 3l-2h, = -2(9 — 16) — 1(10 — 3), , 01.Show that |a = (x-a)*(x + 2a)., , Let |A] =, , Putting x=a we get |A| =, aaa, Since all the three rows are identical (x — a)? is, a factor of |A]., -2a a a, Putting x=-2a we get |A|=| a -2a a, a a -—2a, 0 0 0, ja -2a a, la a ~-—2a, |A| = 0,(%+ 2a) _ isa factor of [A]., The product of (x — a)?(x + 2a) is a factor of |]., Now determinant is a cubic polynomial in x., Degree of the product of known factors are 3., Degree of the leading diagonal elements are 3., Difference between them m=3-3=0, , R, > Ry +R, + Ry, , so that remaining factor must be a constant ‘k’., x aa, , a x al|=k(x—a)*(x+2a)