# Prove the determinant is non-zero (linear independence w/o row-echelon form)

#### Mampac

##### New member
Hi there,

In R100 we are given three vectors:
v1 = (1, 2, 3, 4, . . . , 100),
v2 = (2, 3, 4, 5, . . . , 100, 0),
v3 = (3, 4, 5, 6, . . . , 100, 0, 0).
Have to show these are linearly independent, without any row-echelon matrix.

I know that the rows are dependent if $$\displaystyle detA = 0$$, so I've gotta prove the opposite.

I see only 2 ways of finding the determinant -- out of the ones we've covered so far:
1) I'm hesitating to use the Triangle method, since we have unknowns. Can I form a matrix, by rows, then get a21, a31, a32 nullified and say that the product of the diagonal is nonzero? I'm hesitating since it's not a square matrix in this case. Even here I get 1 * -1 * 0 = 0.
2) As about Laplace expansion... I just can't find a valid entry to use this theorem on, even when I perform row-operations. Everything fails. If I sequester 3 columns from the ones known and create a matrix I always get a zero determinant.

By the way, when creating a matrix, should I drop the last zeros in v2 and v3, or should I append more to v1 and v2?

It seems to me there's a mistake in the subject and it should've been "show whether independent", because to me these rows are dependent as ****.

#### Jomo

##### Elite Member
What is the determinant of an upper/lower triangular matrix?

• nasi112

#### lex

##### Full Member
• Jomo

#### Jomo

##### Elite Member
(There is no square matrix here, so determinants are not of use).
Oops!
I was going to say look at aV1+bV2+V3 =0, but determinant of a upper/lower triangular square matrix would have done the trick nicely--if in fact we had a square matrix. I'll be in the corner for while thinking about how I made such a simple error.

BTW, lex, welcome to the forum!

#### lex

##### Full Member
It's all too easily done. I wouldn't give it too much thought!

• Jomo

#### Mampac

##### New member
You need to start with a definition of linear independence.
View attachment 26600
Write out the three equations for the last three entries.
Solve for x, y, z (starting with the final equation, then the second-last...)
Then look at the definition of linear independence and decide.

(There is no square matrix here, so determinants are not of use).
only a couple of hours later i realized (pointed out by a classmate) that the first vector expands to ...98, 99, 100) in the end. I thought that the dimensions of the vectors are different (such that v1 has 98 vector components, v2 has 99 vector components, and v3 has all the 100 components). that's why i was so confused and thought about appending 0s.

it was required from me to not use any row-echelon thingies (well, as a matter of fact, the subject states "Show that these vectors are linearly independent. This can be done without any row-echelon matrix", but it seems to me i'll lose some points this way).
after the aforementioned discovery i did understand that if a bring this matrix formed by vectors to row-echelon form i don't get a zero-row at the bottom thus it's linearly independent. You said it's no square matrix and I agree, but can I now separate the last three entries of each vector to form one? then it's already in a mirrored triangle form, thus the determinant is -(100 × 100 × 100) 0 thus the whole set is independent?

• lex

#### lex

##### Full Member
it's no square matrix and I agree, but can I now separate the last three entries of each vector to form one? then it's already in a mirrored triangle form, thus the determinant is -(100 × 100 × 100) 0 thus the whole set is independent?
Certainly so. If the only solution of the last three equations is x=0, y=0, z=0 then the only solution of the entire system is x=0, y=0, z=0.
Well done. Everyone is happy!

• Mampac