Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the following system:

$$\left\{\begin{array}{cccccccc} 2x&+&3y&+&z&-&3v&=&2 \\ x&-&y&+&2z&+&v&=&0\\ 3x&+&2y&+&3z&-&2v&=&-2 \end{array}\right.$$

I have to show if the system does or doesn't have solutions using multidimensional vectors. I notice that it has more unknowns than equations so it is an undetermined system. If I form the matrix , I notice that the determinant is different from zero so this three vectors are linearly indipendent.Now what do I do to show if they have a solution or not?

Note: I have to use only determinants and linearly independent/dependent vector theory to show it.

share|improve this question
    
Since the appropriate matrix is $3\times 4$, the determinant is undefined. What do you mean by "determinant is different from zero"? –  Dennis Gulko Dec 15 '12 at 11:49
    
No, I form the matrix with the vectors ( 2 1 3) (3 -1 2) and (1 2 3) ,and here the determinant is diff from zero...I know I have to form another matrix including 2 0 -2 and find the determinant ,but I don't know how to form it.. –  Beyonce45 Dec 15 '12 at 11:55
    
@Beyonce45: So, what about the $v$'s?? –  Babak S. Dec 15 '12 at 11:56
    
@Beyonce45: I edited your question just because I think it is easier to read this way. Feel free to revert my changes if you think otherwise. –  Dennis Gulko Dec 15 '12 at 12:02

2 Answers 2

Edit:
What you need to do is to row-reduce the extended matrix: $$\left[\left.\begin{array}{cccc}2&3&1&-3\\1&-1&2&1\\3&2&3&-2\end{array}\right|\begin{array}{c}2\\0\\-2\end{array}\right]\underset{R_3-R_2}{\overset{R_2-R_1}{\longrightarrow}}\left[\left.\begin{array}{cccc}2&3&1&-3\\1&-1&2&1\\0&0&0&0\end{array}\right|\begin{array}{c}2\\0\\-4\end{array}\right]$$ And to check whether there are any $0$-rows equal to non-zero or not. If there are rows of the form $[\begin{array}{c}0&0&0&0\end{array}|\begin{array}{c}a\end{array}]$ for some $a\neq 0$, then there are no solutions. Else, since the system is undetermined, there will be infinitely many solutions.
As we can see here, the last row is $[\begin{array}{c}0&0&0&0\end{array}|\begin{array}{c}-4\end{array}]$, hence there are no solutions.

share|improve this answer
    
Exactly! Dennis. –  Babak S. Dec 15 '12 at 11:57
    
Thanks, but In my textbook the answer is : No solutions :S –  Beyonce45 Dec 15 '12 at 12:01
    
I didn't say there are. I just showed how to check whether there are any. –  Dennis Gulko Dec 15 '12 at 12:03
    
ok,thanks you again –  Beyonce45 Dec 15 '12 at 12:03
    
Can you take it from here? or do you need more details? –  Dennis Gulko Dec 15 '12 at 12:04

I agree with Dennis, but she is saying that she has to prove it using linearly dependent and independent vectors. I don't think that this is possible in this case, because the vectors in the matrix form will never be linearly independent, because the determinant is always zero. ( To find the determinant in the matrix add zeros in the fourth row to convert it to a $4\times 4$ matrix)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.