Question #227526

If X=(-1,2,0), Y= (3,1,2), Z= (4,-1,0),show that linear combination at (0,1,-1).


1
Expert's answer
2021-08-19T17:23:04-0400
W=aX+bY+cZW=aX+bY+cZ

a+3b+4c=02a+bc=12b=1\begin{matrix} -a+3b+4c=0 \\ \\ 2a+b-c=1 \\ \\ 2b=-1 \\ \end{matrix}

A=(134021110201)A=\begin{pmatrix} -1 & 3 & 4 & & 0 \\ 2 & 1 & -1 & & 1 \\ 0 & 2 & 0& & -1 \\ \end{pmatrix}

R1=R1R_1=-R_1


(134021110201)\begin{pmatrix} 1 & -3 & -4 & & 0 \\ 2 & 1 & -1 & & 1 \\ 0 & 2 & 0& & -1 \\ \end{pmatrix}

R2=R22R1R_2=R_2-2R_1


(134007710201)\begin{pmatrix} 1 & -3 & -4 & & 0 \\ 0 & 7 &7 & & 1 \\ 0 & 2 & 0& & -1 \\ \end{pmatrix}

R2=R2/7R_2=R_2/7


(13400111/70201)\begin{pmatrix} 1 & -3 & -4 & & 0 \\ 0 & 1 &1 & & 1/7 \\ 0 & 2 & 0& & -1 \\ \end{pmatrix}

R1=R1+3R2R_1=R_1+3R_2


(1013/70111/70201)\begin{pmatrix} 1 & 0 & -1 & & 3/7 \\ 0 & 1 &1 & & 1/7 \\ 0 & 2 & 0& & -1 \\ \end{pmatrix}

R3=R32R2R_3=R_3-2R_2


(1013/70111/70029/7)\begin{pmatrix} 1 & 0 & -1 & & 3/7 \\ 0 & 1 &1 & & 1/7 \\ 0 & 0 & -2 & & -9/7 \\ \end{pmatrix}

R3=R3/2R_3=-R_3/2


(1013/70111/70019/14)\begin{pmatrix} 1 & 0 & -1 & & 3/7 \\ 0 & 1 & 1 & & 1/7 \\ 0 & 0 & 1 & & 9/14 \\ \end{pmatrix}

R1=R1+R3R_1=R_1+R_3


(10015/140111/70019/14)\begin{pmatrix} 1 & 0 & 0 & & 15/14 \\ 0 & 1 & 1 & & 1/7 \\ 0 & 0 & 1 & & 9/14 \\ \end{pmatrix}

R2=R2R3R_2=R_2-R_3


(10015/140101/20019/14)\begin{pmatrix} 1 & 0 & 0 & & 15/14 \\ 0 & 1 & 0 & & -1/2 \\ 0 & 0 & 1 & & 9/14 \\ \end{pmatrix}

Then a=1514,b=12,c=914.a=\dfrac{15}{14}, b=-\dfrac{1}{2}, c=\dfrac{9}{14}.



(011)=514(120)12(312)+914(410)\begin{pmatrix} 0 \\ 1 \\ -1 \\ \end{pmatrix}=\dfrac{5}{14}\begin{pmatrix} -1 \\ 2 \\ 0 \\ \end{pmatrix}-\dfrac{1}{2}\begin{pmatrix} 3 \\ 1 \\ 2\\ \end{pmatrix}+\dfrac{9}{14}\begin{pmatrix} 4 \\ -1 \\ 0 \\ \end{pmatrix}


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