Which-one-of-the-following-sets-of-vectors-is-a-basis-for-R-2-A-1-2-3-6-B-1-1-2-2-C-2-1-0-1-D-1-2-4-8-

Question Number 84316 by Rio Michael last updated on 11/Mar/20

Which one of the following sets of

vectors is a basis for R^{2}

[A] {(\begin{matrix} 1 \\ - 2 \end{matrix}), (\begin{matrix} - 3 \\ 6 \end{matrix})}

[B] {(\begin{matrix} 1 \\ 1 \end{matrix}), (\begin{matrix} 2 \\ 2 \end{matrix})}

[C] {(\begin{matrix} 2 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \end{matrix})}

[D] {(\begin{matrix} 1 \\ 2 \end{matrix}), (\begin{matrix} 4 \\ 8 \end{matrix})}

Commented by Rio Michael last updated on 11/Mar/20

please explain

Answered by MJS last updated on 11/Mar/20

the vectors must be linear independent

this means {there}^{'} s no real number r with

r \times \overset{⇀}{a} = \overset{⇀}{b}

set A (- 3) \times (\begin{matrix} 1 \\ - 2 \end{matrix}) = (\begin{matrix} - 3 \\ 6 \end{matrix}) no basis

set B 2 \times (\begin{matrix} 1 \\ 1 \end{matrix}) = (\begin{matrix} 2 \\ 2 \end{matrix}) no basis

set C ∄ r : r \times (\begin{matrix} 2 \\ 1 \end{matrix}) = (\begin{matrix} 0 \\ 1 \end{matrix}) basis

set D 4 \times (\begin{matrix} 1 \\ 2 \end{matrix}) = (\begin{matrix} 4 \\ 8 \end{matrix}) no basis

{\overset{⇀}{a}, \overset{⇀}{b}} is a basis means using two real parameters

p, q you can “ {reach}^{″} any point of R^{2} with

\overset{⇀}{O X} = p \times \overset{⇀}{a} + q \times \overset{⇀}{b}

set C

(\begin{matrix} x \\ y \end{matrix}) = p \times (\begin{matrix} 2 \\ 1 \end{matrix}) + q \times (\begin{matrix} 0 \\ 1 \end{matrix}) = (\begin{matrix} 2 p \\ p + q \end{matrix})

{\begin{cases} x = 2 p \\ y = p + q \end{cases} \Leftrightarrow {\begin{cases} p = \frac{x}{2} \\ q = y - \frac{x}{2} \end{cases} \Rightarrow

\Rightarrow for any given point (\begin{matrix} x \\ y \end{matrix}) we can find a unique

pair (p, q)

Commented by Rio Michael last updated on 11/Mar/20

Sir if the vectors (\begin{matrix} a_{1} \\ b_{1} \\ c_{1} \end{matrix}) and (\begin{matrix} a_{2} \\ b_{2} \\ c_{3} \end{matrix}) are linearly

independent, then | \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix} | \neq 0 right ?

Commented by Rio Michael last updated on 11/Mar/20

thanks sir i got it

Commented by MJS last updated on 11/Mar/20

yes and no

vectors are (\begin{matrix} a_{1} \\ b_{1} \end{matrix}) and (\begin{matrix} a_{2} \\ b_{2} \end{matrix}) because {we}^{'} re in

R^{2} and determinants only exist for n \times n

matrices

if | \begin{matrix} a_{1} & a_{2} \\ b_{1} & b_{2} \end{matrix} | \neq 0 the vectors are independent

Commented by Rio Michael last updated on 12/Mar/20

t h a n k s