Answer to Question #144428 in Linear Algebra for Sourav Mondal

Question

Let B = f(a1,a2, a3) be an ordered basis of
R³ with a1 = (1, 0, -1), a2 = (1, 1, 1),
a3 = (1, 0, 0). Write the vector v = (a, b, c) as
a linear combination of the basis vectors
from B.

Answer 1

Given the ordered basis $B = \{ a_1,a_2,a_3\} = \text{\textbraceleft} \begin{pmatrix}1\\0\\-1 \end{pmatrix}, \begin{pmatrix}1\\1\\1 \end{pmatrix}, \begin{pmatrix}1\\0\\0 \end{pmatrix}\text{\textbraceright}$

then the expression of $v = \begin{pmatrix}a\\b\\c \end{pmatrix}$ as a linear combination will be of the form

$\begin{pmatrix}a\\b\\c \end{pmatrix} = x\begin{pmatrix}1\\0\\-1 \end{pmatrix} + y\begin{pmatrix}1\\1\\1 \end{pmatrix} + z\begin{pmatrix}1\\0\\0 \end{pmatrix} = \begin{pmatrix}x+y+z\\y\\-x+y \end{pmatrix} \text{ where } x,y,z \in \mathbb{R}$

$\begin{pmatrix}a\\b\\c \end{pmatrix}= \begin{pmatrix}x+y+z\\y\\-x+y \end{pmatrix} \text{then we have the system of equations below}$

$\begin{alignedat}{3} x+y+z = a \dots(1)\\ y=b \dots(2)\\-x+y=c \dots(3)\end{alignedat}$

$\text{From (2) we have y=b; inserting this into (3) we have: }\\ -x+ b=c \text{ then we have } -x=c-b\\\text{then } x=b-c \\ \text{Inserting the values of x and y into (1) we have }\\ (b-c)+b+z=a \implies z+2b-c=a\\ \implies z = a-2b+c$

$\therefore x =b-c;y=b \text{ and } z=a-2b+c$

$\text{Therefore v as the linear combination of the ordered basis is given by } \\ \boxed{v=\begin{pmatrix}a\\b\\c \end{pmatrix} = (b-c)\begin{pmatrix}1\\0\\-1 \end{pmatrix} + b\begin{pmatrix}1\\1\\1 \end{pmatrix} + (a-2b+c)\begin{pmatrix}1\\0\\0 \end{pmatrix}}$