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Examine the following vectors for linear dependence/ independence : $X_1$=(a,b,c) , $X_2$=(b,c,a) ; $X_3$=(c,a,b) where a+b+c $\ne$ 0
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Consider the matrix equation
$ k_1X_1+k_2X_2+k_3X_3 \\ \; \\ \; \\ \therefore k(a,b,c)+k_2(b,c,a)+k_3(c,a,b) \;=\; 0 \\ \; \\ \; \\ \therefore ak_1 + bk_2 + ck_3 \;=\; 0 \\ bk_1 + ck_2 + ak_3 \;=\; 0 \\ ck_1 + ak_2 + bk_3 \;=\; 0 $

Which can be written in matrix form as follows -

$ \left[ \begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array}\right] \left[ \begin{array}{ccc} k_1 \\ k_2 \\ k_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ \; \\ R_3 \rightarrow R_1+R_2+R_3 \; , \; R_2 \rightarrow R_1+R_2+R_3 \\ \; \\ \; \\ \left[ \begin{array}{ccc} a & b & c \\ a+b+c & a+b+c & a+b+c \\ a+b+c & a+b+c & a+b+c \end{array}\right] \left[ \begin{array}{ccc} k_1 \\ k_2 \\ k_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ \; \\ \; \\ R_3 \rightarrow R_3-R_2 \\ \; \\ \; \\ \left[ \begin{array}{ccc} a & b & c \\ a+b+c & a+b+c & a+b+c \\ 0 & 0 & 0 \end{array}\right] \left[ \begin{array}{ccc} k_1 \\ k_2 \\ k_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ \; \\ R_2 \rightarrow [email protected] \times \dfrac{a}{a+b+c} \\ \; \\ \; \\ \left[ \begin{array}{ccc} a & b & c \\ a & a & a \\ 0 & 0 & 0 \end{array}\right] \left[ \begin{array}{ccc} k_1 \\ k_2 \\ k_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ \; \\ R_2 \rightarrow R_2 - R_1 \\ \; \\ \; \\ \left[ \begin{array}{ccc} a & b & c \\ 0 & a-b & a-c \\ 0 & 0 & 0 \end{array}\right] \left[ \begin{array}{ccc} k_1 \\ k_2 \\ k_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ \therefore (a-b)k_2 + (a-c)k_3 = 0 \\ \; \\ (a-b)k_2 = (c-a)k_3 \\ \; \\ \therefore k_3 \;=\; \Bigg( \dfrac{a-b}{c-a} \Bigg) k_2 \\ \; \\ \therefore ak_1 + bk_2 + \Bigg( \dfrac{a-b}{c-a} \Bigg) k_2 \;=\; 0 \\ \; \\ \therefore \therefore ak_1 + \Bigg( b+ \dfrac{a-b}{c-a} \Bigg) k_2 \;=\; 0 \\ \; \\ \; \\ \therefore k_1 \;=\; \Bigg( \dfrac{bc-ab+a-b}{c-a} \Bigg) \Bigg( \dfrac{-1}{a} \Bigg) k_2 \\ \; \\ \; \\ k_1X_1+k_2X_2+k_3X_3 \;=\; 0 \\ \; \\ \therefore \Bigg[ \dfrac{bc-ab+a-b}{c-a} \Bigg( \dfrac{-1}{a} \Bigg) X_1 + X_2 + \Bigg( \dfrac{a-b}{c-a} \Bigg) X_3 \Bigg] k_2 \;=\; 0 \\ \; \\ \; \\ \therefore \dfrac{bc-ab+a-b}{c-a} \Bigg( \dfrac{-1}{a} \Bigg) X_1 + X_2 + \Bigg( \dfrac{a-b}{c-a} \Bigg) X_3 \;=\; 0 \\ \; \\ \; \\ \; \\ \therefore X_2 \;=\; \dfrac{bc-ab+a-b}{a(c-a)} X_1 + \Bigg( \dfrac{a-b}{c-a} \Bigg) X_3 $

Hence given vectors are linearly dependent.

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