Answer to Question #313549 in Linear Algebra for Yassuo_thewindLord

Question

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)

Answer 1

Solution

Given the three vectors are

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

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

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

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

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

Therefore,

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

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

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

$a+5b+5c=0\\ -3a+9b-7c=0\\ -2a-6b+hc=0\\$

Solving these equations,

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

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

$h\neq\frac{26}{3}$

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