By hand determine if each set of vectors below are linearly

By hand, determine if each set of vectors below are linearly independent or linearly dependent.

Determine if each set of vectors below are linearly independent or linearly dependent. To do this, represent each set as a linear combination of column vectors c_1v_1 + c_2v_2 +...+ c_nv_n = 0 (where v_1 is the 1^st vector, v_2 is the 2^nd vector, etc.) and solve for c_1, c_2 ... c_n. Do your work by hand and then check it in MATLAB. Show your MATLAB work. a. S_a = {(-2, 1, 1), (3, -4, -2), (5, -10, -8)} b. S_b = {(1, 2, 1), (1, 0, -1), (1, 1, 1)} c. S_c = {(1, 1, 2, 1), (0, 2, 1, 1), (3, 1, 2, 0)}

Solution

1. Let a,b,c be scalars such that a(-2,1,1)+b(3,-4,-2)+c(5,-10,-8) = 0. Then ( -2a+3b+5c, a-4b-10c, a-2b-8c)= 0 so that -2a+3b+5c =0 ...(1),  a-4b-10c= 0...(2) and a-2b-8c = 0...(3). From equations 2 and 3, we get a = 4b+10c = 2b+8c. Hence 2b+8c = 4b+10c or, b+4c = 2b+5c or, 5c -4c = b-2b or, c = -b. On substituting c = -b in the 1st and the 2nd equations, we get -2a+3b-5b= 0 or, -2a-2b= 0 or, a +b = 0...(4) and a-4b+10b = 0 or, a+6b = 0...(5). Then 6b = - a = b so that b= 0. Then a = -b = 0 and c = -b = 0. Hence, if a linear combination of the vectors in S_a equals 0, then the coefficients of all the vetors in S_a are 0. Hence S_a is a linearly independent set.

2. Let a,b,c be scalars such that a(1,2,1)+b(1,0,-1)+c(1,1,1) = 0. Then (a+b+c, 2a+c, a-b+c)=0 so that a+b+c= 0...(1), 2a+c=0...(2) and a-b+c = 0...(3). From equation2, we get c = -2a . On substituting this value of c in equations (1) and (2), we get a+b-2a = 0 or, b-a= 0 or b = a..(4) and a-b-2a = 0 or, -b-a = 0 or, b = -a...(5). From equations 4 and 5, we get -a = a (=b). Hence a = 0 which implies that b = 0 and c = 0. Thus if a linear combination of the vectors in Sb equals 0, then the coefficients of all the vetors in Sbare 0. Hence Sb is a linearly independent set.

3. Let a,b,c be scalars such that a(1,1,2,1)+b( 0,2,1,1)+c(3,1,2,0) = 0. Then (a+3c,a+2b+c, 2a+b+2c, a+b)=0 so that a+3c = 0...(1), a+2b +c= 0...(2), 2a+b +2c = 0 ...(3) and a+b = 0...(4). From equation numbers (1) and (4), we get c = -a/3 and b = -a. On substituting these values of b and c in equation no. (2) and (3), we get a -2a-a/3= 0 or, -4a/3 = 0 so that a = 0 and 2a -a -2a/3 = 0 or, a/3 = 0 so that a = 0. Then b = -a = 0 and c = -a/3 = 0.Hence, if a linear combination of the vectors in Sc equals 0, then the coefficients of all the vetors in Scare 0. Hence Sc is a linearly independent set.