Question

The vectors a=2i−3j+k,b=4i+3j−2k and c=2i−4j+xk are linearly
dependent. Find the value of x.

Answer 1

\[\vec{a}=2\vec{i}-3\vec{j}+\vec{k}\\
\vec{b}=4\vec{i}+3\vec{j}-2\vec{k}\\
\vec{c}=2\vec{i}-4\vec{j}+x\vec{k}\]
For these vectors to be linearly dependent, there must be some non-zero constant n_i (i = 1,2,3) such that the following is satisfied:
\[n_1\vec{a} + n_2\vec{b}+n_3\vec{c}=\vec{0}\\
\small n_1\left(2\vec{i}-3\vec{j}+\vec{k}\right) + n_2\left(4\vec{i}+3\vec{j}-2\vec{k}\right)+n_3\left(2\vec{i}-4\vec{j}+x\vec{k}\right)=0\vec{i}+0\vec{j}+0\vec{k}\\
\small \left(2n_1+4n_2+2n_3\right)\vec{i}+\left(-3n_1+3n_2-4n_3\right)\vec{j}+\left(n_1-2n_2+xn_3\right)\vec{k}=0\vec{i}+0\vec{j}+0\vec{k}\]
This gives you the system
\[\begin{cases}\;\;\;2n_1+4n_2+2n_3=0\\
-3n_1+3n_2-4n_3=0\\
\;\;\;\;\;n_1-2n_2+xn_3=0\end{cases}\]
Does that help?

Answer 2

yes thanks but i dont know hwo to solve the simultaneous equation

Answer 3

Do you know how to use matrices to describe the system? You are in linear algebra, correct?

Answer 4

no this is high school. is it determinant ? oh i do know how to setup Ax=0

Answer 5

The matrix equation is given by Ax=B, where B = 0 in this case.
\[\left[\begin{matrix}2&4&2\\-3&3&-4\\1&-2&x\end{matrix}\right]\left[\begin{matrix}n_1\\n_2\\n_3\end{matrix}\right]=\left[\begin{matrix}0\\0\\0\end{matrix}\right]\]

Answer 6

yeah need find det(a)=/=0 right

Answer 7

wait det(a)=0 i mean

Answer 8

Well, if the determinant of A is zero, then its inverse doesn't exist. To solve for each n, you need to have A's inverse.

Answer 9

Do you know how to find the inverse of a 3x3 matrix?

Answer 10

linear dependent if det=/=0 independet if =0

Answer 11

no need calculator

Answer 12

for "linear dependent if det=/=0 independet if =0" why does it work if you put the vectors as rows as well?

Answer 13

Oh, disregard that, then. I wasn't aware of the "linear independence implies det=0" thing you mentioned. The way I wrote it helped me to find the determinant on paper, but I think it works in a similar way. You're going to have to know how to find the determinant by hand, though. A calculator won't let you put in x.

Answer 14

i understand when you put as columns, if det=0, singular .: more than one solution and when det=/=0 invertible .: only trivial solution

Answer 15

Right, I'm stuck in the mindset of solving equations in that way. Also, the "column" way of writing the vectors is just the transpose of the matrix I used, so their determinants are the same. Anyway, I get the determinant of A to be 18x - 28.

Answer 16

can you explain why linear dependent if det=/=0 independet if =0 ??

Answer 17

it works for row vectors as well because determinant the same ?

Answer 18

whats transpose matrix

Answer 19

The determinant of a square matrix is the same as that of its transpose. ?

Answer 20

I just referred to this wiki page on linear independence: http://en.wikipedia.org/wiki/Linear_independence#Alternative_method_using_determinants
The transpose of a matrix is a reflection of the matrix on itself, in a way. For example, the transpose of A would be denoted A^t, and
\[A^T=\left[\begin{matrix}2&-3&1\\4&3&-2\\2&-4&x\end{matrix}\right]\]
And yes, the determinant of A is the same as the determinant of A^T.
But that's not important. What is important is that the system is linearly dependent if the determinant is zero.

Answer 21

*NOT zero, I mean. Sorry.

Answer 22

And another correction: My determinant should be 18x-26.

Answer 23

okayokay wait so. for a system to be dependent, det=0? or =/=0

Answer 24

are you sure "since the determinant is non-zero, the vectors (1, 1) and (−3, 2) are linearly independent." doesnt that say det=/=0, independent

Answer 25

Agh, this wording is tripping me up. Should be
For the system associated with matrix A to be linearly DEPENDENT, det(A) = 0. And that's final. I hope.

Answer 26

yes thatswhat i thought

Answer 27

can you explain why?

Answer 28

I'm not sure how to explain it informally. Normally, I'd try to prove it for myself using what I've learned in linear algebra, and I don't think it'd be easy to express in a single post. For now, I would just accept it as fact for this particular problem, then investigate it later.

The last thing you have to do is find the determinant, which is 18x-26. Do you know how to do that?

Answer 29

nope. only learnt det for 2x2. teacher said to just use calculator for larger matrices

Answer 30

but can you solve the 3 equations using gaussian elimination?

Answer 31

need to do it by hand like that

Answer 32

Yes, I think you could. You have to show that at least one of the n's isn't necessarily 0.

Answer 33

i mean do you know hwo to do it becausei dont