Answer to Question #197760 in Linear Algebra for melvin

Question

Express v=(1,-2,5) in R³ as a linear combination of the vectors U₁=(1,1,1), U₂=(1,2,3), U₃=(2,-1,1)

Answer 1

Let $v=xu_1+yu_2+zu_3.$ Then

$(1,-2,5)=x(1,1,1)+y(1,2,3)+z(2,-1,1)=(x+y+2z,x+2y-z,x+3y+z),$

and hence we get the following system

$\begin{cases} x+y+2z=1\\ x+2y-z=-2\\ x+3y+z=5 \end{cases}$

which is equivalent (after substracting from the second row the first row, from the third row the second row) to

$\begin{cases} x+y+2z=1\\ y-3z=-3\\ y+2z=7 \end{cases}$

and (after substracting from the third row the second row) to

$\begin{cases} x+y+2z=1\\ y-3z=-3\\ 5z=10 \end{cases}$

It follows that $z=2,\ y =3z-3=6-3=3,$ and $x=1-y-2z=1-3-4=-6.$

We conclude that $v=-6u_1+3u_2+2u_3.$