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 ahmed2009 November 24th, 2015 04:45 AM

Find determinant and linearly dependent

hello all .
I have Question about the determinant and linearly dependent and it's :

Let :

http://i.imgur.com/u4ge2Cx.png

a) Find det(B) in terms of k;
b) For what value(s) of k are the column vectors of B linearly dependent;
c) For k = 0, find det(B) and .

can some one help me in this question and thank you all .

 Country Boy December 1st, 2015 10:36 AM

If, as you appear to be saying, you do not know what a "determinant" is or what "linearly dependent" means, where in the world did you get this problem? If you know how to find a determinant, it should be easy to do part (a). You will get a formula of the form "a+ bk" where a and b are specific numbers. The columns are "linearly dependent" if and only if that determinant is 0. That is, you need to solve "a+ bk= 0" for k.

For k= 0, $\displaystyle B= \begin{bmatrix}1 & 2 & 3 \\ 2 & -3 & 4 \\ 3 & 4 & -4\end{bmatrix}$. Now, do you know $\displaystyle B^t$ and $\displaystyle B^{-1}$ are?

 ahmed2009 December 3rd, 2015 03:51 AM

Quote:
 Originally Posted by Country Boy (Post 465001) If, as you appear to be saying, you do not know what a "determinant" is or what "linearly dependent" means, where in the world did you get this problem? If you know how to find a determinant, it should be easy to do part (a). You will get a formula of the form "a+ bk" where a and b are specific numbers. The columns are "linearly dependent" if and only if that determinant is 0. That is, you need to solve "a+ bk= 0" for k. For k= 0, $\displaystyle B= \begin{bmatrix}1 & 2 & 3 \\ 2 & -3 & 4 \\ 3 & 4 & -4\end{bmatrix}$. Now, do you know $\displaystyle B^t$ and $\displaystyle B^{-1}$ are?
Yes I know part 3 I just want to check , but part 2 I don't really know it , thank you so much for help .

 Country Boy December 3rd, 2015 08:38 AM

The three column vectors are $\displaystyle \begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}$, $\displaystyle \begin{bmatrix}2 \\ k- 3 \\ 4\end{bmatrix}$, and $\displaystyle \begin{bmatrix}3 \\ 4 \\ k- 4 \end{bmatrix}$ so to be linearly independent, you must be able to show that the only "a, b, and c" such that $\displaystyle a\begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}+ b\begin{bmatrix}2 \\ k- 3 \\ 4\end{bmatrix}+ c\begin{bmatrix}3 \\ 4 \\ k- 4\end{bmatrix}= \begin{bmatrix}a+ 2b+ 3c \\ 2a+ (k- 3)b+ 4c \\ 3a+ 4b+ (k- 4)c\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$

So we have the three equations a+ 2b+ 3c= 0, 2a+ (k-3)b+ 4c= 0, 3a+ 4b+ (k- 4)c= 0. Obviously, a= b= c= 0 is a solution. The vectors are linearly dependent if there exist other solutions with at least one of a, b, c non-zero. For what values of k is that true? One way to answer that is to actually try to solve the equations and see if at some point there is a reason why, for some values of k, you cannot get a unique solution- as, for example, if you would have to divide by 0.

 ahmed2009 December 11th, 2015 07:15 AM

Quote:
 Originally Posted by Country Boy (Post 467327) The three column vectors are $\displaystyle \begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}$, $\displaystyle \begin{bmatrix}2 \\ k- 3 \\ 4\end{bmatrix}$, and $\displaystyle \begin{bmatrix}3 \\ 4 \\ k- 4 \end{bmatrix}$ so to be linearly independent, you must be able to show that the only "a, b, and c" such that $\displaystyle a\begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}+ b\begin{bmatrix}2 \\ k- 3 \\ 4\end{bmatrix}+ c\begin{bmatrix}3 \\ 4 \\ k- 4\end{bmatrix}= \begin{bmatrix}a+ 2b+ 3c \\ 2a+ (k- 3)b+ 4c \\ 3a+ 4b+ (k- 4)c\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$ So we have the three equations a+ 2b+ 3c= 0, 2a+ (k-3)b+ 4c= 0, 3a+ 4b+ (k- 4)c= 0. Obviously, a= b= c= 0 is a solution. The vectors are linearly dependent if there exist other solutions with at least one of a, b, c non-zero. For what values of k is that true? One way to answer that is to actually try to solve the equations and see if at some point there is a reason why, for some values of k, you cannot get a unique solution- as, for example, if you would have to divide by 0.
Thanks for help .

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