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 \(B = \left\{ \begin{bmatrix} 1 \\ 0 \\ 3 \end{bmatrix} \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 0\\ 1 \end{bmatrix} \right\} is [\overrightarrow{v}]_B = (a, b, c) = (3, 3, 5)\)