Answer to Question #143191 in Linear Algebra for Ojugbele Daniel

Question #143191
Write the vector v=( 1, -2, 5) as a linear combination of the vector e1=( 1, 1, 1), e2= ( 1, 2, 3) and e3=( 2, -1, 1)
1
Expert's answer
2020-11-11T19:09:23-0500

linear combimation has the form:

v=α(e1)+β(e2)+γ(e3)v ⃗=α*(e1) ⃗+β*(e2) ⃗+γ*(e3) ⃗


(1;2;5)=α(1;1;1)+β(1;2;3)+γ(2;1;1)(1; -2; 5)=α*(1;1;1)+β*(1;2;3)+γ*(2; -1;1)

(1;2;5)=(α;α;α;)+(β;2β;3β)+(2γ;γ;γ)(1; -2; 5)=(α;α;α;)+(β;2β;3β)+(2γ;-γ;γ)

find α,β,γ:\alpha, \beta, \gamma:

system of equations:


α+β+2γ=1,(1)α+β+2γ=1, (1)α+2βγ=2,(2)α+2β-γ=-2, (2)α+3β+γ=5;(3)α+3β+γ=5; (3)


solution system:

equation (1) minus equation (2):

β+3γ=3;-\beta+3\gamma=3;

β=3γ3;\beta=3\gamma-3; (4)

equation (2) plus equation (3):

2α+5β=3;2\alpha+5\beta=3;

α=(35β)/2;α=(3-5β)/2;

α=(35(3γ3))/2;\alpha=(3-5(3\gamma-3))/2; (5)

equation (1):

(35(3γ3))/2+3γ3+2γ=1;(3−5(3γ−3))/2+3\gamma-3+2\gamma=1;

315γ+15+10γ6=2;3-15\gamma+15+10\gamma-6=2;

5γ=10;-5\gamma=-10;

γ=2;\gamma=2;

equation (4):

β=323;\beta=3*2-3;

β=3;\beta=3;

equation (1):

α=12γβ;\alpha=1-2\gamma-\beta;

α=1223;\alpha=1-2*2-3;

α=6;\alpha=-6;


v=6(e1)+3(e2)+2(e3);v ⃗=-6*(e1) ⃗+3*(e2) ⃗+2*(e3);

(1;2;5)=6(1;1;1)+3(1;2;3)+2(2;1;1);(1; -2; 5)=-6*(1;1;1)+3*(1;2;3)+2*(2; -1;1);


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