Thread: Determine values of the constant k for which vectors are linearly independent?

1. Determine values of the constant k for which vectors are linearly independent?

Problem:

Determine all values of the constant k for which the given set of vectors is linearly independent in R4.

{(1,0,1,k),(-1,0,k,1),(2,0,1,3)}

Attempt:

So far, I have the system of equations:

x-y-2z=0
x+ky+z=0
kx+y+3z=0

Now, the determinant of the coefficient matrix yields 2(k+1)(2-k).

Here is my question: How does the determinant of the coefficient matrix allow me to deduce that "the system has only the trivial solution, and the vectors are linearly dependent if and only if k=-1 or 2"?

Thanks for the help!

2. Originally Posted by divinelogos
Problem:

Determine all values of the constant k for which the given set of vectors is linearly independent in R4.

{(1,0,1,k),(-1,0,k,1),(2,0,1,3)}

Attempt:

So far, I have the system of equations:

x-y-2z=0
x+ky+z=0
kx+y+3z=0

Now, the determinant of the coefficient matrix yields 2(k+1)(2-k).

Here is my question: How does the determinant of the coefficient matrix allow me to deduce that "the system has only the trivial solution, and the vectors are linearly dependent if and only if k=-1 or 2"?

Thanks for the help!
Take the determinant and find the values for k which make the determinant not equal to 0

3. Originally Posted by dwsmith
Take the determinant and find the values for k which make the determinant not equal to 0
That's what I did. Can you answer the question though?

4. Originally Posted by divinelogos
That's what I did. Can you answer the question though?
What does it mean if the determinant is 0?

5. Originally Posted by divinelogos
Now, the determinant of the coefficient matrix yields 2(k+1)(2-k).

Here is my question: How does the determinant of the coefficient matrix allow me to deduce that "the system has only the trivial solution, and the vectors are linearly dependent if and only if k=-1 or 2"?
I got 2k(k + 1) for the determinant.

I'll let someone else answer the other question. I keep getting definitions mixed up as I am typing. Briefly if the determinant is 0 then there are no coefficients x, y, z that solve the system. Since x, y, z do not exist then the vectors are linearly independent. If the determinant is non-zero then there is a solution for x, y, z, so coefficients exist to make the linear combination sum to 0. Thus the vectors are linearly dependent.

-Dan

6. Originally Posted by dwsmith
What does it mean if the determinant is 0?
I will answer my own question then. If the determinant is zero, then one or more of the column vectors can be written as a linear combination of one or more column vectors; therefore, the column space is lin. dep.

7. Another way of looking at it- because the right side of the system of equations is all "0"s, x= y= z= 0 is an obvious solution to the system of equations. If the determinant of the coefficients is 0, there is NOT a unique solution so there exist non-zero solutions, contradicting the definition of "indepdendent".

,

,

,

,

,

,

,

,

,

,

,

,

,

find the values of k so that s is linearly dependent

Click on a term to search for related topics.