Question

SEEMINGLY SIMPLE LINEAR ALGEBRA QUESTION:

Which values of c give dependent columns (combinations equal zero)?

The matrix is at this link:
http://www.wolframalpha.com/input/?i=%7B%7B1%2C3%2C5%7D%2C%7B1%2C2%2C4%7D%2C%7B1%2C1%2Cc%7D%7D

Any help in solving this would be greatly appreciated!

Answer 1

Just use the (very useful) fact that the columns would be (linearly) dependent IF AND ONLY IF the matrix is singular, and a matrix is singular IF AND ONLY IF its determinant is 0.
You already have the determinant
3 - c
For what values of c is this going to be 0? :)

Answer 2

What do you mean "we already have the determinant"? I had to compute it.

Answer 3

Oh... you did link me to wolfram, and it has the determinant right there...

Answer 4

(If that sounded rude, I didn't mean it that way.)

Answer 5

c = 3

Answer 6

Oh, I just was checking if there was some trick not involving computers.

Answer 7

singular = not invertible right?

Answer 8

There you go :)
In fact
\[\left[\begin{matrix}5 \\ 4 \\ 3\end{matrix}\right]=\left[\begin{matrix}3 \\ 2 \\ 1\end{matrix}\right]+2\left[\begin{matrix}1 \\ 1 \\ 1\end{matrix}\right]\]
And since one column is a linear combination of the others, the columns are linearly dependent.
and yes, singular = not invertible

Answer 9

what's the vector equation you just showed me in my problem? or is it just a random example?

Answer 10

No, I worked it out. Actually, I worked this out before I remembered that all that was needed for the matrix to be singular was for the determinant to be zero :)
Don't worry about it, with this type of problem, getting determinant is probably easier...

Answer 11

Alight. Basically, I get it now thanks to you. :)

Thanks!

Answer 12

No problem :)
Terence out :D

Answer 13

Lol. :D