Question #197760

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)


1
Expert's answer
2022-01-10T15:10:41-0500

Let v=xu1+yu2+zu3.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+2yz,x+3y+z),(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


{x+y+2z=1x+2yz=2x+3y+z=5\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


{x+y+2z=1y3z=3y+2z=7\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


{x+y+2z=1y3z=35z=10\begin{cases} x+y+2z=1\\ y-3z=-3\\ 5z=10 \end{cases}


It follows that z=2, y=3z3=63=3,z=2,\ y =3z-3=6-3=3, and x=1y2z=134=6.x=1-y-2z=1-3-4=-6.


We conclude that v=6u1+3u2+2u3.v=-6u_1+3u_2+2u_3.


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