Determine whether or not the vector (6, 10, 11, -4, -3, 12) is a linear combinat

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(6,10,11,-4,-3,12) writing it in linear combination of u,v and w as (6,10,11,-4,-3,12) = a * (5,3,-1,7,1,-2) + b * (-2,4,-3,1,2,-1) + c * (-1,5,2,-3,0,4) Comparing on both the sides will give 6 = 5a-2b-c -------------------------> 1 10=3a+4b+5c -------------------------> 2 11=-a-3b+2c -------------------------> 3 -4=7a+b-3c -------------------------> 4 -3=a+2b -------------------------> 5 12=-2a-b+4c -------------------------> 6 i will use equations 1,2 and 5 from equation 5 ==> 2b = -3-a substitute '2b' in equation 1 5a-(-3-a)-c=6 ==> 6a-c=3 --------------->7 from equation 5 ==> 2b = -3-a substitute '2b' in equation 2 3a+2(-3-a)+5c=10 ==> a+5c=16 ----------->8 multiplying Equation 7 by '5' on both sides ==> 5*(6a-c=3) ==> 30a-5c=15 --------->9 Adding Equation 8 and Equation 9 ==> a+5c=16 30a-5c=15 will give 31a=31 ==> a=1 substitute 'a' in Equation 8 ==> 1+5c=16 ==> c=3 substitute 'a' in Equation 5 ==> 1+2b=-3 ==>b=-2 Lets check whether the system is consistent or not by substituting 'a', 'b', 'c' in other three Equations Equation 3 ==> 11=-a-3b+2c ==> 11=-1+6+6 ==> 11=11 (satisfied) Equation 4 ==> -4=7a+b-3c ==> -4=7-2-9 ==> -4=-4 (satisfied) Equation 6 ==> 12=-2a-b+4c ==> 12=-2+2+12 ==> 12=12 (satisfied) Hence (6,10,11,-4,-3,12) can be written in linear combination of u,v and w

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