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Write-the-vector-v-1-2-3-as-a-linear-combination-of-vectors-u-1-1-1-1-u-2-1-2-3-and-u-3-2-1-1-




Question Number 118411 by bramlexs22 last updated on 17/Oct/20
Write the vector v=(1,−2,3) as a  linear combination of vectors  u_1 =(1,1,1) ,u_2 =(1,2,3) and u_3 =(2,−1,1)
$${Write}\:{the}\:{vector}\:{v}=\left(\mathrm{1},−\mathrm{2},\mathrm{3}\right)\:{as}\:{a} \\ $$$${linear}\:{combination}\:{of}\:{vectors} \\ $$$${u}_{\mathrm{1}} =\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)\:,{u}_{\mathrm{2}} =\left(\mathrm{1},\mathrm{2},\mathrm{3}\right)\:{and}\:{u}_{\mathrm{3}} =\left(\mathrm{2},−\mathrm{1},\mathrm{1}\right) \\ $$
Answered by benjo_mathlover last updated on 17/Oct/20
linear combination of vectors  u_1 ,u_2  and u_3  , so v = pu_1 +qu_2 +ru_3    (((    1)),((−2)),((    3)) ) =  ((p),(p),(p) ) +  (((  q)),((2q)),((3q)) ) +  (((  2r)),((−r)),((   r)) )   { ((p+q+2r = 1)),((p+2q−r = −2)),((p+3q+r = 3)) :} or  { ((p+q+2r = 1)),((      q−3r = −3)),((              5r = 10)) :}  This unique solution of the triangular  system is  { ((p = −6)),((q = 3)),((r = 2)) :}. Thus v = −6u_1 +3u_2 +2u_3
$${linear}\:{combination}\:{of}\:{vectors} \\ $$$${u}_{\mathrm{1}} ,{u}_{\mathrm{2}} \:{and}\:{u}_{\mathrm{3}} \:,\:{so}\:{v}\:=\:{pu}_{\mathrm{1}} +{qu}_{\mathrm{2}} +{ru}_{\mathrm{3}} \\ $$$$\begin{pmatrix}{\:\:\:\:\mathrm{1}}\\{−\mathrm{2}}\\{\:\:\:\:\mathrm{3}}\end{pmatrix}\:=\:\begin{pmatrix}{{p}}\\{{p}}\\{{p}}\end{pmatrix}\:+\:\begin{pmatrix}{\:\:{q}}\\{\mathrm{2}{q}}\\{\mathrm{3}{q}}\end{pmatrix}\:+\:\begin{pmatrix}{\:\:\mathrm{2}{r}}\\{−{r}}\\{\:\:\:{r}}\end{pmatrix} \\ $$$$\begin{cases}{{p}+{q}+\mathrm{2}{r}\:=\:\mathrm{1}}\\{{p}+\mathrm{2}{q}−{r}\:=\:−\mathrm{2}}\\{{p}+\mathrm{3}{q}+{r}\:=\:\mathrm{3}}\end{cases}\:{or}\:\begin{cases}{{p}+{q}+\mathrm{2}{r}\:=\:\mathrm{1}}\\{\:\:\:\:\:\:{q}−\mathrm{3}{r}\:=\:−\mathrm{3}}\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}{r}\:=\:\mathrm{10}}\end{cases} \\ $$$${This}\:{unique}\:{solution}\:{of}\:{the}\:{triangular} \\ $$$${system}\:{is}\:\begin{cases}{{p}\:=\:−\mathrm{6}}\\{{q}\:=\:\mathrm{3}}\\{{r}\:=\:\mathrm{2}}\end{cases}.\:{Thus}\:{v}\:=\:−\mathrm{6}{u}_{\mathrm{1}} +\mathrm{3}{u}_{\mathrm{2}} +\mathrm{2}{u}_{\mathrm{3}} \\ $$$$ \\ $$

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