Give a correct answer for given question (A) Argue why { (1,0,3), (2,3,1), (0,0,1) } is a coordinate system ( bases ) for R^3 ?(B) Find the coordinates of (7, 6, 16) relative to the set in part (A)

Give a correct answer for given question

(A) Argue why $${ (1,0,3), (2,3,1), (0,0,1) }$$ is a coordinate system ( bases ) for $$R^3$$ ?

(B) Find the coordinates of $$(7, 6, 16)$$ relative to the set in part (A)

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Given $$B = \left\{ \begin{bmatrix} 1 \\ 0 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 \\ 3\\ 1 \end{bmatrix}, \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \right\}$$

By definition, we say that tha vector $$\left\{ v_1, v_2, ..., v_n \right\}$$ are linearly dependent if there exists scalars $$\displaystyle\alpha_{{1}},\alpha_{{2}},\ldots,\alpha_{{n}}$$ not all of them zero such that $$\displaystyle\alpha_{{1}}{v}_{{1}}+\alpha_{{2}}{v}_{{2}}+\ldots+\alpha_{{n}}{v}_{{n}}=\overline{{0}}.$$

We say that the vectors $$\left\{ v_1, v_2, ..., v_n \right\}$$ are linearly independent if $$\displaystyle\alpha_{{1}}{v}_{{1}}+\alpha_{{2}}{v}_{{2}}+\ldots+\alpha_{{n}}{v}_{{n}}=\overline{{0}}$$ then

The set of vectors $$\left\{ v_1, v_2, ..., v_n \right\}$$ is said to be a basis for vector space V if i) set of vectors $$\left\{ v_1, v_2, ..., v_n \right\}$$ is linearly independent ii) span $$\left\{ v_1, v_2, ..., v_n \right\}=V$$ If B = $$\left\{ v_1, v_2, ..., v_n \right\}$$ is a basis in a vector space V than every vector $$\overrightarrow{v}\in{V}$$ can be uniquely expressed as a linear combination of basis vectors $$\displaystyle{b}_{{1}},{b}_{{2}},\ldots,{b}_{{n}}.$$ i.e there exists unique scalsrs $$\displaystyle\alpha_{{1}},\alpha_{{2}},\ldots,\alpha_{{n}}$$ such that, $$\overrightarrow{v}=\alpha_{{1}}{b}_{{1}}+,\alpha_{{2}}{b}_{{2}}+\ldots+\alpha_{{n}}{b}_{{n}}.$$ The coordinates of the vector $$\overrightarrow{v}$$ relative to the basis $$\displaystyle{B}$$ is the sequence of co-ordinates, i.e. $$[v]_B = (\alpha_1, \alpha_2, .., \alpha_n)$$

Consider $$A = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 3 & 1 & 1 \end{bmatrix}$$ Applying $$\displaystyle{R}_{{3}}\rightarrow{R}_{{3}}-{3}{R}_{{1}},$$ we get
$$R_3 \rightarrow R_3 - 3 \ R_1,\ we \ get\ A \sim \begin{bmatrix} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 0 & -5 & 1 \end{bmatrix}$$

Applying $$\displaystyle{R}_{{3}}\rightarrow{5}{R}_{{2}}+{3}{R}_{{3}},$$ we get $$\sim \begin{bmatrix} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}$$ (because in eacelon form, first, second and third columns have pivot elements)

$$\displaystyle\therefore$$ The set of vectors $$\left\{ \begin{bmatrix} 1 \\ 0 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 \\ 3\\ 1 \end{bmatrix}, \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \right\}$$ is linearly independent dim $$(R^3)=3 \ dim\ (\left\{ \begin{bmatrix} 1 \\ 0 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 \\ 3\\ 1 \end{bmatrix}, \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \right\}) = 3 \ \Rightarrow \left\{ \begin{bmatrix} 1 \\ 0 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 \\ 3\\ 1 \end{bmatrix}, \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \right\} = R^3 \$$

$$\therefore$$Basis for $$R^3\ is \left\{ \begin{bmatrix} 1 \\ 0 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 \\ 3\\ 1 \end{bmatrix}, \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \right\}$$ b. Let $$\begin{bmatrix}7 \\ 6 \\ 16 \end{bmatrix} = (a)\begin{bmatrix}1 \\ 0 \\ 3 \end{bmatrix} + (b) \begin{bmatrix}2 \\ 3 \\ 1 \end{bmatrix} + (c) \begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix} \ \Rightarrow a + 2b = 7 \rightarrow (i) \ 3b = 6 \ \Rightarrow b = 2 \rightarrow (ii) \ 3a + b + c = 16 \rightarrow (iii)$$

From (i), $$\displaystyle{a}={7}-{2}{b}={7}-{4}={3}.$$

From (ii), $$3a + b + c = 16 \ \Rightarrow c = 16 - 3a - b = 16 - 9 - 2 = 5 \ \therefore \overrightarrow{v} = \begin{bmatrix}7 \\ 6 \\ 16 \end{bmatrix} = (3)\begin{bmatrix}1 \\ 0 \\ 3 \end{bmatrix} + (2)\begin{bmatrix}2 \\ 3 \\ 1 \end{bmatrix} + (5)\begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix} \$$

$$\therefore$$The coordinates of the vector $$\overrightarrow{v} = \begin{bmatrix}7 \\ 6 \\ 16 \end{bmatrix}$$ raletive to the basis $$B = \left\{ \begin{bmatrix} 1 \\ 0 \\ 3 \end{bmatrix} \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 0\\ 1 \end{bmatrix} \right\} is [\overrightarrow{v}]_B = (a, b, c) = (3, 3, 5)$$