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 October 18th, 2015, 06:17 AM #1 Senior Member     Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty Linear Dependence Given a set of vectors, I wish to determine as to whether the vectors are linear independent or linearly dependent. $\displaystyle \begin{bmatrix} 1\\-1\\1\\0 \end{bmatrix}$ , $\displaystyle \begin{bmatrix} 1\\1\\0\\1 \end{bmatrix}$ , $\displaystyle \begin{bmatrix} 1\\0\\1\\-1 \end{bmatrix}$ , $\displaystyle \begin{bmatrix} 0\\1\\-1\\1 \end{bmatrix}$ Augmented matrix: $\displaystyle \begin{bmatrix} 1&1&1&0 \\ -1&1&0&1 \\ 1&0&1&-1 \\ 0&1&-1&1 \end{bmatrix}\begin{bmatrix} 0\\0\\0\\0 \end{bmatrix}$ , visually not an augmented matrix but you get the idea. Reduced to reduced row echelon form: $\displaystyle \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix}\begin{bmatrix} 0\\0\\0\\0 \end{bmatrix}$ The system is consistent but all variables have the solution which is the trivial solution therefore this set of vectors are linearly independent. Is this correct??
 October 19th, 2015, 04:57 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 Personally, I wouldn't use matrices to do this. A set of four vectors, $\displaystyle \{v_1, v_2, v_3, v_4 \}$ is said to be "linearly dependent" if and only if there exist four numbers, a, b, c, and d, such that $\displaystyle av_1+ bv_2+ cv_3+ dv_4= 0$. So we want to see if we can find a, b, c, d such that $\displaystyle a\begin{bmatrix}1 \\ -1 \\ 1 \\ 0\end{bmatrix}+ b\begin{bmatrix} 1 \\ 1 \\ 0 \\ 1\end{bmatrix}+ c\begin{bmatrix}1 \\ 0 \\ 1 \\ -1\end{bmatrix}+ d\begin{bmatrix}0 \\ 1 \\ -1 \\ 1\end{bmatrix}$$\displaystyle = \begin{bmatrix}a+ b+ c \\ -a+ b+ d \\ a+ c- d \\ b- c+ d \end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$ So we have the four equations a+ b+ c= 0, -a+ b+ d= 0, a+ c- d= 0, and b- c+ d= 0. Adding the first two equations, 2b+ c+ d= 0. Subtracting the third equation from the first b+ 2d= 0. From that, b= -2d so the previous equation becomes -4d+ c+ d= -3d+ c= 0 or c= 3d. Replacing c by 3d, the equations are a+ b+ 3d= 0, -a+ b+ d= 0, a+ 2d= 0, and b- 2d= 0. So a= -2d and b= 2d. The first equation, a+ b+ 3d= 0 becomes -2d+ 2d+ 3d= -d= 0 so d= 0. Then a= -2d= 0 and b= 2d= 0. Finally, from a+ c- d= 0, c= 0. Since a, b, c, and d are all 0, the matrices are linearly independent. That is essentially what you have done. Thanks from hyperbola

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