Determine whether each of the following sets is linearly independent. (a){(4, -1

Question

Determine whether each of the following sets is linearly independent. (a){(4, -1, 2), (-4, 1, 2)} (b){(2, 0, 1, 4, 3), (-4, 0, -2, -8, -5), (-2, 1, -1, -4, -3)} (c){(2, 5, 4), (2, 1, 3), (6, 7, 10)} (d){(1, 3, 4, 4, -1), (3, 2, 1, 2, -3), (1, 5, 0, -2, -4)} (e){(2, 3, 5, 7, 11), (1, 1, 2, 3, 5), (1, 3, 5, 7, 9), (0, 2, 4, 6, 8), (1, 3, 6, 10, 15), (1, 4, 9, 16, 25)} Prove or disprove each of the following statements. (a)Let A be an m times n matrix whose column vectors are linearly independent. Suppose that vectors v_1, v_2, ..., V_k in R^n are linearly independent. Then the vectors Av_1, Av_2, ..., Av_k in R^m are linearly independent. (b)Let S = {(a, y, z) R^3: 2x + 3y + 5z = 1}. Then S is a subspace of R^3. (c)Suppose a set Q = {v_1, v_2, v_3, v_4, v_5} of vectors in R^5 is linearly independent. Then R = {v_1- v_2, v_2 - v_3, v_3 - v_4, v_4 - v_5, v_5 - v_1) is also linearly independent. (d)If W_1 and W_2 are subspaces of R^n, then W_+ W_2:= {w_1 + W_2: w_1 W_1, w_2 W_2} is also a subspace of R^n. (e)Let B be an n times n invertible matrix. There exists exactly one vector in R^n which is orthogonal to all row vectors of B.

Explanation / Answer

Solution : 1 a. The 2 vectors are : u =( 4,-1,2 ) and v = ( -4,1,2 ) are LI if au + bv = 0 => both a,b vanish

consider au + bv= a ( 4,-1,2) +b ( -4,1,2 ) = (0,0,0 ) => 4a-4b=0 , -a+b=0 , 2a + 2b =0

first 2 eqns are same - a +b =0 and a+b =0 solving we get a=0 and b=0 . hence the 2 vectors are LI

1,b. considr au +bv +cw = o => a ( 2,0,1,4,3 ) +b ( - 4, 0 -2 , -8 , -5 ) +c ( -2 ,1 ,-1 ,-4 ,-3 ) =( 0,0,0,0,0)

equating the corresponding coords : 2a -4b -2c =0 , c=0 , a - 2b -c= 0, 4a - 8b =0 , 3a - 5b =0.put c=0 in the eqns the eqns reduce to : a -2b =0 abd 3a - 5b =0 solving these 2 eqns we get a=0,b=0 and c=0

.hence the 3 vectors are LI

1 c. Consider au + bv + cw =0 => a (2,5,4 ) +b ( 2,1,3 ) +c ( 6,7,10 ) =(0,0,0 ) =>2a + 2b + 6c =0----(1)

5a+b+ 7c=0 ----(2) , 4a + 3b +10c =0 ----(3) multiply (2) by 2 => 10a + 2b +14c =0 ---(4)

eqn 4 - eqn1 => 8a +8c =0 dividing by 8 weget a+ c=0 ----(5)

multiply eqn 2 by 3 15a + 3b +21 c =0 ----(6) . eqn 6 - eqn 3 => 11a +11c =0 => a+c =0same as eqn 5

substituting c= - a in the eqns 1 ,2 ,3 we get b =2a .hence the system has infinte no of solutins of the form

a = k , b = 2k , c= - k where k is any real no for example a=1 ,b=2 ,c= -1 satisfy the system ie all a ,b, c, are not 0

hence the vectors are not LI

1 ,d. Consider au + bv +cw =0 => a (1,3,4,4,-,1 ) +b (3,2,1,2,-3 ) +c ( 1,5,0,- 2 - 4 ) = ( 0,0,0,0,0 ) =>

a +3b +c=0 ---1 , 3a +2b +5c =0 ------2 , 4a +b =0 ---3 , 4a +2b - 2c =0 ---4 , - a - 3b - 4c =0 -----5

from eqn 3 we get b = - 4a substitute in eqns 1 , 2 we get c =11a and c=a which is possible if a=0 ,c=0,hence b=0 hence the given vectors are LI

1 e . there are 6 vectors given the maximum no of LI vectors in R5 is 5 . hence these 6 vectors cannot be LI

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