Show that (1,0,1),(−2,1,4),(0,3,1) forms the basis for in RR^3.

klasyvea

klasyvea

Answered question

2022-11-04

Show that (1,0,1),(−2,1,4),(0,3,1) forms the basis for R 3
I want to show that [ 1 0 1 ] , [ 2 1 4 ] , [ 0 3 1 ] forms the basis for R 3 . First I tried to show that the vectors are linearly independent
α 1 [ 1 0 1 ] + α 2 [ 2 1 4 ] + α 3 [ 0 3 1 ] = [ 0 0 0 ]
But I tried to continue from this on wards but I get that α 3 = 18 α 3 , which doesn't make sense so I'm a bit lost.
α 1 [ 1 0 1 ] + α 2 [ 2 1 4 ] + α 3 [ 0 3 1 ] = [ 0 0 0 ]

Answer & Explanation

siriceboynu1

siriceboynu1

Beginner2022-11-05Added 12 answers

α 1 [ 1 0 1 ] + α 2 [ 2 1 4 ] + α 3 [ 0 3 1 ] = [ 0 0 0 ]
You got a system of equations:
{ α 1 2 α 2 = 0 α 2 + 3 α 3 = 0 α 1 + 4 α 2 + α 3 = 0 { α 1 = 2 α 2 α 3 = 1 3 α 2 2 α 2 + 4 α 2 1 3 α 2 = 0 α 2 = α 1 = α 3 = 0
Barrett Osborn

Barrett Osborn

Beginner2022-11-06Added 3 answers

Another equivalent way to first answer is to show that the determinant of
[ 1 2 0 0 1 3 1 4 1 ]
is not zero. Doing this will give you:
det [ 1 2 0 0 1 3 1 4 1 ] = 1 ( 1 12 ) + 1 ( 6 + 0 ) = 11 6 = 17 0

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