Question #313549
  1. For what values of h the vectors


⟶ [1 ⟶ [-1 ⟶ [5

u1 = -3 u2 = 9 u3 = -7

-2] -6] h]


are linearly independent? (Show all working)


1
Expert's answer
2022-03-19T02:39:43-0400

Solution


Given the three vectors are


u1=[1,3,2]\vec{u_{1}}=[1, -3, -2]


u2=[5,9,6]\vec{u_{2}}=[5, 9, -6]


u3=[5,7,h]\vec{u_{3}}=[5, -7, h]


Now the three vectors will be linearly independent when for three scalars, a,ba, b and cc , we can write,


au1+bu2+cu3=0a\vec{u_{1}}+b\vec{u_{2}}+c\vec{u_{3}}=0 if and only iff when a=b=c=0a=b=c=0


Therefore,


au1+bu2+cu3=a[1,3,2]+b[5,9,6]+c[5,7,h]=0a\vec{u_{1}}+b\vec{u_{2}}+c\vec{u_{3}}=a[1, -3, -2]+b[5, 9, -6]+c[5, -7, h]=0


a[1,3,2]+b[5,9,6]+c[5,7,h]=[0,0,0]a[1, -3, -2]+b[5, 9, -6]+c[5, -7, h]=[0, 0, 0]


[a,3a,2a]+[5b,9b,6b]+[5c,7c,hc]=[0,0,0][a, -3a, -2a]+[5b, 9b, -6b]+[5c, -7c, hc]=[0, 0, 0]


a+5b+5c=03a+9b7c=02a6b+hc=0a+5b+5c=0\\ -3a+9b-7c=0\\ -2a-6b+hc=0\\


Solving these equations,





since for the linearly independent, a=b=c=0a=b=c=0 , which requires,


h+2630h+\frac{26}{3}\neq0


h263h\neq\frac{26}{3}


Hence other than this value, the three vectors will be linearly independent.





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