Answer to Question #227526 in Linear Algebra for umme habiba sumi

Question

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

Answer 1

$$W=aX+bY+cZ$$

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

$$A=\begin{pmatrix} -1 & 3 & 4 & & 0 \\ 2 & 1 & -1 & & 1 \\ 0 & 2 & 0& & -1 \\ \end{pmatrix}$$

$R_1=-R_1$

$$\begin{pmatrix} 1 & -3 & -4 & & 0 \\ 2 & 1 & -1 & & 1 \\ 0 & 2 & 0& & -1 \\ \end{pmatrix}$$

$R_2=R_2-2R_1$

$$\begin{pmatrix} 1 & -3 & -4 & & 0 \\ 0 & 7 &7 & & 1 \\ 0 & 2 & 0& & -1 \\ \end{pmatrix}$$

$R_2=R_2/7$

$$\begin{pmatrix} 1 & -3 & -4 & & 0 \\ 0 & 1 &1 & & 1/7 \\ 0 & 2 & 0& & -1 \\ \end{pmatrix}$$

$R_1=R_1+3R_2$

$$\begin{pmatrix} 1 & 0 & -1 & & 3/7 \\ 0 & 1 &1 & & 1/7 \\ 0 & 2 & 0& & -1 \\ \end{pmatrix}$$

$R_3=R_3-2R_2$

$$\begin{pmatrix} 1 & 0 & -1 & & 3/7 \\ 0 & 1 &1 & & 1/7 \\ 0 & 0 & -2 & & -9/7 \\ \end{pmatrix}$$

$R_3=-R_3/2$

$$\begin{pmatrix} 1 & 0 & -1 & & 3/7 \\ 0 & 1 & 1 & & 1/7 \\ 0 & 0 & 1 & & 9/14 \\ \end{pmatrix}$$

$R_1=R_1+R_3$

$$\begin{pmatrix} 1 & 0 & 0 & & 15/14 \\ 0 & 1 & 1 & & 1/7 \\ 0 & 0 & 1 & & 9/14 \\ \end{pmatrix}$$

$R_2=R_2-R_3$

$$\begin{pmatrix} 1 & 0 & 0 & & 15/14 \\ 0 & 1 & 0 & & -1/2 \\ 0 & 0 & 1 & & 9/14 \\ \end{pmatrix}$$

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

$$\begin{pmatrix} 0 \\ 1 \\ -1 \\ \end{pmatrix}=\dfrac{5}{14}\begin{pmatrix} -1 \\ 2 \\ 0 \\ \end{pmatrix}-\dfrac{1}{2}\begin{pmatrix} 3 \\ 1 \\ 2\\ \end{pmatrix}+\dfrac{9}{14}\begin{pmatrix} 4 \\ -1 \\ 0 \\ \end{pmatrix}$$