Menu Close

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-

Tinku Tara 0 Comments
Pin




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] { ((1),((−2)) ) , (((−3)),(6) )} [B] { ((1),(1) ) , ((2),(2) )} [C] { ((2),(1) ) , ((0),(1) )} [D] { ((1),(2) ) , ((4),(8) ) }
$$\mathrm{Which}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{sets}\:\mathrm{of} \\ $$$$\mathrm{vectors}\:\mathrm{is}\:\mathrm{a}\:\mathrm{basis}\:\mathrm{for}\:\mathbb{R}^{\mathrm{2}} \\ $$$$\left[\mathrm{A}\right]\:\left\{\begin{pmatrix}{\mathrm{1}}\\{−\mathrm{2}}\end{pmatrix}\:,\:\begin{pmatrix}{−\mathrm{3}}\\{\mathrm{6}}\end{pmatrix}\right\} \\ $$$$\left[\mathrm{B}\right]\:\left\{\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{pmatrix}\:,\begin{pmatrix}{\mathrm{2}}\\{\mathrm{2}}\end{pmatrix}\right\} \\ $$$$\left[\mathrm{C}\right]\:\left\{\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}\:,\begin{pmatrix}{\mathrm{0}}\\{\mathrm{1}}\end{pmatrix}\right\} \\ $$$$\left[\mathrm{D}\right]\:\left\{\begin{pmatrix}{\mathrm{1}}\\{\mathrm{2}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{4}}\\{\mathrm{8}}\end{pmatrix}\:\right\} \\ $$
Commented by Rio Michael last updated on 11/Mar/20
please explain
$$\mathrm{please}\:\mathrm{explain} \\ $$
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×a^⇀ =b^⇀ set A (−3)× ((1),((−2)) )= (((−3)),(6) ) no basis set B 2× ((1),(1) )= ((2),(2) ) no basis set C ∄r: r× ((2),(1) )= ((0),(1) ) basis set D 4× ((1),(2) )= ((4),(8) ) no basis {a^⇀ , b^⇀ } is a basis means using two real parameters p, q you can “reach” any point of R^2 with OX^(⇀) =p×a^⇀ +q×b^⇀ set C ((x),(y) )=p× ((2),(1) )+q× ((0),(1) )= (((2p)),((p+q)) ) { ((x=2p)),((y=p+q)) :} ⇔ { ((p=(x/2))),((q=y−(x/2))) :} ⇒ ⇒ for any given point ((x),(y) ) we can find a unique pair (p, q)
$$\mathrm{the}\:\mathrm{vectors}\:\mathrm{must}\:\mathrm{be}\:\mathrm{linear}\:\mathrm{independent} \\ $$$$\mathrm{this}\:\mathrm{means}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{real}\:\mathrm{number}\:{r}\:\mathrm{with} \\ $$$${r}×\overset{\rightharpoonup} {{a}}=\overset{\rightharpoonup} {{b}} \\ $$$$\mathrm{set}\:\mathrm{A}\:\left(−\mathrm{3}\right)×\begin{pmatrix}{\mathrm{1}}\\{−\mathrm{2}}\end{pmatrix}=\begin{pmatrix}{−\mathrm{3}}\\{\mathrm{6}}\end{pmatrix}\:\mathrm{no}\:\mathrm{basis} \\ $$$$\mathrm{set}\:\mathrm{B}\:\mathrm{2}×\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{2}}\\{\mathrm{2}}\end{pmatrix}\:\mathrm{no}\:\mathrm{basis} \\ $$$$\mathrm{set}\:\mathrm{C}\:\nexists{r}:\:{r}×\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{1}}\end{pmatrix}\:\mathrm{basis} \\ $$$$\mathrm{set}\:\mathrm{D}\:\mathrm{4}×\begin{pmatrix}{\mathrm{1}}\\{\mathrm{2}}\end{pmatrix}=\begin{pmatrix}{\mathrm{4}}\\{\mathrm{8}}\end{pmatrix}\:\mathrm{no}\:\mathrm{basis} \\ $$$$\left\{\overset{\rightharpoonup} {{a}},\:\overset{\rightharpoonup} {{b}}\right\}\:\mathrm{is}\:\mathrm{a}\:\mathrm{basis}\:\mathrm{means}\:\mathrm{using}\:\mathrm{two}\:\mathrm{real}\:\mathrm{parameters} \\ $$$${p},\:{q}\:\mathrm{you}\:\mathrm{can}\:“\mathrm{reach}''\:\mathrm{any}\:\mathrm{point}\:\mathrm{of}\:\mathbb{R}^{\mathrm{2}} \:\mathrm{with} \\ $$$$\overset{\rightharpoonup} {{OX}}={p}×\overset{\rightharpoonup} {{a}}+{q}×\overset{\rightharpoonup} {{b}} \\ $$$$\mathrm{set}\:\mathrm{C} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}={p}×\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}+{q}×\begin{pmatrix}{\mathrm{0}}\\{\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{2}{p}}\\{{p}+{q}}\end{pmatrix} \\ $$$$\begin{cases}{{x}=\mathrm{2}{p}}\\{{y}={p}+{q}}\end{cases}\:\Leftrightarrow\:\begin{cases}{{p}=\frac{{x}}{\mathrm{2}}}\\{{q}={y}−\frac{{x}}{\mathrm{2}}}\end{cases}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{for}\:\mathrm{any}\:\mathrm{given}\:\mathrm{point}\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:\mathrm{we}\:\mathrm{can}\:\mathrm{find}\:\mathrm{a}\:\mathrm{unique} \\ $$$$\mathrm{pair}\:\left({p},\:{q}\right) \\ $$
Commented by Rio Michael last updated on 11/Mar/20
Sir if the vectors ((a_1 ),(b_1 ),(c_1 ) ) and ((a_2 ),(b_2 ),(c_3 ) ) are linearly independent, then determinant ((a_1 ,b_1 ,c_1 ),(a_2 ,b_2 ,c_2 ))≠ 0 right?
$$\mathrm{Sir}\:\mathrm{if}\:\mathrm{the}\:\mathrm{vectors}\:\begin{pmatrix}{{a}_{\mathrm{1}} }\\{{b}_{\mathrm{1}} }\\{{c}_{\mathrm{1}} }\end{pmatrix}\:\:\:\mathrm{and}\:\begin{pmatrix}{{a}_{\mathrm{2}} }\\{{b}_{\mathrm{2}} }\\{{c}_{\mathrm{3}} }\end{pmatrix}\:\:\mathrm{are}\:\mathrm{linearly} \\ $$$$\mathrm{independent},\:\:\mathrm{then}\:\:\begin{vmatrix}{{a}_{\mathrm{1}} }&{{b}_{\mathrm{1}} }&{{c}_{\mathrm{1}} }\\{{a}_{\mathrm{2}} }&{{b}_{\mathrm{2}} }&{{c}_{\mathrm{2}} }\end{vmatrix}\neq\:\mathrm{0}\:\mathrm{right}? \\ $$
Commented by Rio Michael last updated on 11/Mar/20
thanks sir i got it
$$\mathrm{thanks}\:\mathrm{sir}\:\mathrm{i}\:\mathrm{got}\:\mathrm{it} \\ $$
Commented by MJS last updated on 11/Mar/20
yes and no vectors are ((a_1 ),(b_1 ) ) and ((a_2 ),(b_2 ) ) because we′re in R^2 and determinants only exist for n×n matrices if determinant ((a_1 ,a_2 ),(b_1 ,b_2 ))≠0 the vectors are independent
$$\mathrm{yes}\:\mathrm{and}\:\mathrm{no} \\ $$$$\mathrm{vectors}\:\mathrm{are}\:\begin{pmatrix}{{a}_{\mathrm{1}} }\\{{b}_{\mathrm{1}} }\end{pmatrix}\:\mathrm{and}\:\begin{pmatrix}{{a}_{\mathrm{2}} }\\{{b}_{\mathrm{2}} }\end{pmatrix}\:\mathrm{because}\:\mathrm{we}'\mathrm{re}\:\mathrm{in} \\ $$$$\mathbb{R}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{determinants}\:\mathrm{only}\:\mathrm{exist}\:\mathrm{for}\:{n}×{n} \\ $$$$\mathrm{matrices} \\ $$$$\mathrm{if}\:\begin{vmatrix}{{a}_{\mathrm{1}} }&{{a}_{\mathrm{2}} }\\{{b}_{\mathrm{1}} }&{{b}_{\mathrm{2}} }\end{vmatrix}\neq\mathrm{0}\:\mathrm{the}\:\mathrm{vectors}\:\mathrm{are}\:\mathrm{independent} \\ $$
Commented by Rio Michael last updated on 12/Mar/20
thanks
$${thanks} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *