Answer to Question #222295 in Linear Algebra for Daniel

Question

Determine whether u=(2,8,2) is a linear combination of u₁=(2,-2,0) u₂=(3,0,-3) and u₃=(-2,0,-1) or not

Answer 1

$solution\\ given \space \\ u=(2,8,2) \space \\ \space u_1=(2,-2,0), \space u_2=(3,0,-3) \space and \space u_3=(-2,0,-1) \space \\ now \space we \space will \space be \space check \space u \space is \space a \space linear \space combination \space of \space u_1, \space u_2 \space and \space u_3 \space or \space not,\\ first \space of \space all \space we \space write \space a \space linear \space combination \space form \space of \space u \space in \space terms \space of \space \space u_1, \space u_2 \space and \space u_3\\ such \space that\\ u=a.u_1+b. \space u_2 \space +c.u_3\\ (2,8,2) \space =a.(2,-2,0)+b. \space (3,0,-3) \space +c.(-2,0,-1)...(A) \space \\ now \space we \space solve \space this \space expression \space to \space find \space a \space ,b \space and \space c \space \\ if \space we \space will \space be \space get \space value \space of \space a \space ,b \space and \space c \space \space then \space u \space is \space a \space linear \space combination \space of \space u_1, \space u_2 \space and \space u_3 \space otherwise \space not\\ ---------------------------------------\\ again\\ (2,8,2) \space =a.(2,-2,0)+b. \space (3,0,-3) \space +c.(-2,0,-1) \space \\ (2,8,2) \space =(2a+3b-2c,-2a,-3b-c)\\ hence\\ 2a+3b-2c=2...(1)\\ -2a=8...(2)\\ -3b-c=2...(3)\\ from \space (2)\\ a=-4...(4)\\ put \space value \space of \space a \space in \space (1)\\ 3b-2c=10...(5)\\ now \space solve \space (5) \space and \space (3), \space we \space get \space \\ b=\frac{2}{3}, \space and \space c=-4...(6) ---------------------------------------\\ \space we \space get \space \\ a=-4, \space b=\frac{2}{3}, \space and \space c=-4\\ hence \space we \space say \space \\ \space u \space is \space a \space linear \space combination \space of \space u_1, \space u_2 \space and \space u_3 \space \\ put \space value \space of \space a \space ,b \space and \space c \space \space in \space eq(A)\\ we \space get\\ (2,8,2) \space =(-4).(2,-2,0)+(\frac{2}{3}). \space (3,0,-3) \space +(-4).(-2,0,-1)\\$