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 November 24th, 2015, 04:45 AM #1 Member   Joined: Nov 2015 From: saudi arabia Posts: 30 Thanks: 5 Find determinant and linearly dependent hello all . I have Question about the determinant and linearly dependent and it's : Let : 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 . December 1st, 2015, 10:36 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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? Thanks from ahmed2009 December 3rd, 2015, 03:51 AM   #3
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 Originally Posted by Country Boy 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 . December 3rd, 2015, 08:38 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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 from ahmed2009 December 11th, 2015, 07:15 AM   #5
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 Originally Posted by Country Boy 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 . Tags dependent, determinant, find, linearly Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Ganesh Ujwal Linear Algebra 0 January 2nd, 2015 09:53 PM Cyberdollar Algebra 1 June 15th, 2014 07:28 AM Helenf12 Linear Algebra 6 March 4th, 2013 05:25 AM tmlfan_179027 Linear Algebra 2 February 19th, 2010 02:18 PM tinynerdi Linear Algebra 1 January 21st, 2010 03:11 PM

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