Given the three vectors are
u 1 ⃗ = [ 1 , − 3 , − 2 ] \vec{u_{1}}=[1, -3, -2] u 1 = [ 1 , − 3 , − 2 ]
u 2 ⃗ = [ 5 , 9 , − 6 ] \vec{u_{2}}=[5, 9, -6] u 2 = [ 5 , 9 , − 6 ]
u 3 ⃗ = [ 5 , − 7 , h ] \vec{u_{3}}=[5, -7, h] u 3 = [ 5 , − 7 , h ]
Now the three vectors will be linearly independent when for three scalars, a , b a, b a , b and c c c , we can write,
a u 1 ⃗ + b u 2 ⃗ + c u 3 ⃗ = 0 a\vec{u_{1}}+b\vec{u_{2}}+c\vec{u_{3}}=0 a u 1 + b u 2 + c u 3 = 0 if and only iff when a = b = c = 0 a=b=c=0 a = b = c = 0
Therefore,
a u 1 ⃗ + b u 2 ⃗ + c u 3 ⃗ = a [ 1 , − 3 , − 2 ] + b [ 5 , 9 , − 6 ] + c [ 5 , − 7 , h ] = 0 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 u 1 + b u 2 + c 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 [ 1 , − 3 , − 2 ] + b [ 5 , 9 , − 6 ] + c [ 5 , − 7 , h ] = [ 0 , 0 , 0 ]
[ a , − 3 a , − 2 a ] + [ 5 b , 9 b , − 6 b ] + [ 5 c , − 7 c , h c ] = [ 0 , 0 , 0 ] [a, -3a, -2a]+[5b, 9b, -6b]+[5c, -7c, hc]=[0, 0, 0] [ a , − 3 a , − 2 a ] + [ 5 b , 9 b , − 6 b ] + [ 5 c , − 7 c , h c ] = [ 0 , 0 , 0 ]
a + 5 b + 5 c = 0 − 3 a + 9 b − 7 c = 0 − 2 a − 6 b + h c = 0 a+5b+5c=0\\
-3a+9b-7c=0\\
-2a-6b+hc=0\\ a + 5 b + 5 c = 0 − 3 a + 9 b − 7 c = 0 − 2 a − 6 b + h c = 0
Solving these equations,
since for the linearly independent, a = b = c = 0 a=b=c=0 a = b = c = 0 , which requires,
h + 26 3 ≠ 0 h+\frac{26}{3}\neq0 h + 3 26 = 0
h ≠ 26 3 h\neq\frac{26}{3} h = 3 26
Hence other than this value, the three vectors will be linearly independent.
Comments