Title 03. MATRICES and DETERMINANTS.pdf ✅

#QID# 85857 (1.) The value of 10 10 11 4 5 m 11 11 12 6 7 m 2 12 12 13 8 9 m 4 C C C C C C C C C + +  = is equal to zero, where m is (a) 6 (b) 4 (c) 5 (d) None of these c (c) 4 13 9 12 8 12 2 12 7 11 6 11 11 5 10 4 10 +  = + m m m C C C C C C C C C = 0 Applying C2 → C1 + C2 0 4 13 9 12 8 12 8 12 2 12 7 11 6 11 6 11 11 5 10 4 10 4 10 = + + +  = + + m m m C C C C C C C C C C C C  4 13 9 13 8 12 2 12 7 12 6 11 11 5 11 4 10 +  = + m m m C C C C C C C C C = 0 Clearly m = 5 satisfies the above result [ , will be identical]  C2 C3 #QID# 85858 (2.) If 1 2 3 n a , a , a ,......., a , ...... are in G.P. then the value of the determinant n n 1 n 2 n 3 n 4 n 5 n 6 n 7 n 8 log a log a log a log a log a log a log a log a log a + + + + + + + + is (a) –2 (b) 1 (c) 2 (d) 0 d (d)  , , ,......., , ...... 1 2 3 n a a a a are in G.P. 2 2 1 .  an+ = an an+ 1 2 2log an+ =log an + log an+ 3 5 2 4 . an+ = an+ an+ 4 3 5 2log log log  n+ = n+ + n+ a a a 6 8 2 7 . an+ = an+ an+ 7 6 8 2log an+ =log an+ + log an+ Putting these values in the second column of the given determinant, we get 6 6 8 8 3 3 5 5 2 2 log log log log log log log log log log log log 2 1 + + + + + + + + + + + + +  = n n n n n n n n n n n n a a a a a a a a a a a a (0) 0 2 1 = = a [  c2 is the sum of the elements, first identical with c1 and second with c3 ]
#QID# 85859 (3.) The value of x x 2 x x 2 x x 2 x x 2 x x 2 x x 2 1 1 1 (2 2 ) (3 3 ) (5 5 ) (2 2 ) (3 3 ) (5 5 ) − − − − − − + + + − − − (a) 0 (b) x 30 (c) x 30− (d) None of these a (a) Applying R2 → R2 − R3 2 2 2 (2 2 ) (3 3 ) (5 5 ) 2.2 .2.2 2.3 .2.3 2.5 .2.5 1 1 1 x x x x x x x x x x x x − − − − − − − − −  = 0 (2 2 ) (3 3 ) (5 5 ) 1 1 1 1 1 1 4 2 2 2 = − − − = x −x x −x x −x [ and are identical] R1 R2 Trick : Putting x = 0 , we get option (a) is correct #QID# 85860 (4.) If x, y, z are integers in A.P. lying between 1 and 9 and x51, y41 and z31 are three digit numbers then the value of 5 4 3 x 51 y 41 z31 x y z is (a) x y z + + (b) x y z − + (c) 0 (d) None of these c (c)  x 51 = 100x + 50 +1, y 41 = 100y + 40 + 1 z 31 = 100z + 30 + 1 x y z 100x 50 1 100y 40 1 100z 30 1 5 4 3   = + + + + + + Applying R2 → R2 −100R3 −10R1 x y z x y z  = 1 1 1 = − 2 + 5 4 3  x, y, z are in A.P. ,  x − 2y + z = 0 ,   = 0 #QID# 85861 (5.) If a b c   , the value of x which satisfies the equation 0 x a x b x a 0 x c 0 x b x c 0 − − + − = + + is (a) x 0 = (b) x a =
(c) x b = (d) x c = a (a) Expanding determinant, we get,  = −(x − a)[−(x + b)(x − c)] + (x + b)[(x + a)(x + c)] = 0  2 (2 ) 0 3 x − ab x =  Either x = 0 or x = ab 2 . Since x = 0 satisfies the given given equation. Trick : On putting x = 0 , we observe that the determinant becomes 0 0 0 0 0 − = − −  = = b c a c a b x  x = 0 is a root of the given equation. #QID# 85862 (6.) The number of distinct real roots of sin x cos x cos x cos x sin x cos x 0 cos x cos x sin x = in the interval x 4 4   −   is (a) 0 (b) 2 (c) 1 (d) 3 c (c) 0 1 cos sin 1 sin cos 1 cos cos (2cos + sin ) = x x x x x x x x Applying, R2 → R2 − R1 and R3 → R3 − R1 0 0 0 sin cos 0 sin cos 0 1 cos cos (2cos sin ) = − + − x x x x x x x x  (2cos sin )(sin cos ) 0 2 x + x x − x =  tanx = −2,1 But tanx  −2 in       − 4 , 4   . Hence 4 tan 1  x =  x = #QID# 85863 (7.) If 2 4 3 2 3 1 3 p q r s t 1 2 4 3 4 3               + − + + + + + = + − − − + , then value of t is (a) 16 (b) 18 (c) 17 (d) 19 b (b)
Since it is an identity in  so satisfied by every value of  . Now put  = 0 in the given equation, we have 12 30 18 3 4 0 1 2 4 0 1 3 = − + = − − − t = #QID# 85864 (8.) If 2 p 3 p 15 8 D p 35 9 p 25 10 = , then D D D D D 1 2 3 4 5 + + + + = (a) 0 (b) 25 (c) 625 (d) None of these d (d)  1 25 10 1 35 9 1 15 8 D1 = , 8 25 10 4 35 9 2 15 8 D2 = , 27 25 10 9 35 9 3 15 8 D3 = , 64 25 10 16 35 9 4 15 8 D4 = , 125 25 10 25 35 9 5 15 8 D5 =  D1 + D2 + D3 + D4 + D5 = 225 125 50 55 175 45 15 75 40 = 15(3125) − 75(−7375) + 40(−32500) = 46875 + 553125 − 1300000 = −700000 #QID# 85865 (9.) The value of N n n 1 U =  , if 2 n 3 2 n 1 5 U n 2N 1 2N 1 n 3N 3N = + + is (a) 0 (b) 1 (c) –1 (d) None of these a (a) =       + + + + + + = N n n N N N N N N N N N N N U 1 2 2 3 3 2 ( 1) 2 1 2 1 6 ( 1)(2 1) 1 5 2 ( 1) N N N N N N N N N 3 ( 1) 3 3 4 2 2 1 2 1 6 1 5 12 ( 1) 2 + + + + + = Applying C3 → C3 + C2 3 ( 1) 3 3 ( 1) 4 2 2 1 4 2 6 1 6 12 ( 1) 2 + + + + + + = N N N N N N N N N N = 0 [  C1 and C3 are identical] #QID# 85866