# Show Linearly Independent

• Oct 18th 2009, 02:10 PM
VkL
Show Linearly Independent
Consider the set of vectors in the real vector space R3

1 0 1
3 1 p
3 -p -1

For what value(s) p is the set of vectors linearly independent?

Ok, so. I think I'm on the right path.

For A set of vectors to be Linearly independent, the determinent cannot = 0

So i found the Det of that matrix to be p^2 - 3p - 4 and got that p cannot equal 4, or -1. Because that would make the det = zero.

So is he answer all real numbers besides 4 and - 1?
• Oct 18th 2009, 02:28 PM
HallsofIvy
Yes, that is completely correct.

Note that a more fundamental definition of "independent vectors" is that a linear combination of them is 0 only if all coefficients are 0.

Here such a linear combination would be a(1 0 1)+ b(3 1 p)+ c(3 - p -1)= (0 0 0) (or a(1 3 3)+ b(0 1 -p)+ c(1 p -1)= (0 0 0)- it was not clear whether your vectors were the rows or columns) which gives the equation a+ 3b+ 3c= 0, b- pc= 0, a+ pb- c= 0. Find the values of p for which those equations have only a= b= c= 0 as solution. Of course, that occurs when the determinant of the coefficient matrix is non-zero which leads right back to your solution.
• Oct 18th 2009, 03:19 PM
VkL
Yes, how you explained it is the general def. But thanks a lot for confirming.