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

$$\mathrm{Which}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{sets}\mathrm{of}$$$$\mathrm{vectors}\mathrm{is}\mathrm{a}\mathrm{basis}\mathrm{for}{\mathbb{R}}^2$$$$\left[\mathrm{A}\right]\{\left(\begin{array}{c}1\\ -2\end{array}\right),\left(\begin{array}{c}-3\\ 6\end{array}\right)\}$$$$\left[\mathrm{B}\right]\{\left(\begin{array}{c}1\\ 1\end{array}\right),\left(\begin{array}{c}2\\ 2\end{array}\right)\}$$$$\left[\mathrm{C}\right]\{\left(\begin{array}{c}2\\ 1\end{array}\right),\left(\begin{array}{c}0\\ 1\end{array}\right)\}$$$$\left[\mathrm{D}\right]\{\left(\begin{array}{c}1\\ 2\end{array}\right),\left(\begin{array}{c}4\\ 8\end{array}\right)\}$$

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

$$\mathrm{please}\mathrm{explain}$$

Answered by MJS last updated on 11/Mar/20

$$\mathrm{the}\mathrm{vectors}\mathrm{must}\mathrm{be}\mathrm{linear}\mathrm{independent}$$$$\mathrm{this}\mathrm{means}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{no}\mathrm{real}\mathrm{number}r\mathrm{with}$$$$r\times \stackrel{\rightharpoonup }{a}=\stackrel{\rightharpoonup }{b}$$$$\mathrm{set}\mathrm{A}(-3)\times \left(\begin{array}{c}1\\ -2\end{array}\right)=\left(\begin{array}{c}-3\\ 6\end{array}\right)\mathrm{no}\mathrm{basis}$$$$\mathrm{set}\mathrm{B}2\times \left(\begin{array}{c}1\\ 1\end{array}\right)=\left(\begin{array}{c}2\\ 2\end{array}\right)\mathrm{no}\mathrm{basis}$$$$\mathrm{set}\mathrm{C}\nexists r:r\times \left(\begin{array}{c}2\\ 1\end{array}\right)=\left(\begin{array}{c}0\\ 1\end{array}\right)\mathrm{basis}$$$$\mathrm{set}\mathrm{D}4\times \left(\begin{array}{c}1\\ 2\end{array}\right)=\left(\begin{array}{c}4\\ 8\end{array}\right)\mathrm{no}\mathrm{basis}$$$$\{\stackrel{\rightharpoonup }{a},\stackrel{\rightharpoonup }{b}\}\mathrm{is}\mathrm{a}\mathrm{basis}\mathrm{means}\mathrm{using}\mathrm{two}\mathrm{real}\mathrm{parameters}$$$$p,q\mathrm{you}\mathrm{can}‘‘{\mathrm{reach}}^{″}\mathrm{any}\mathrm{point}\mathrm{of}{\mathbb{R}}^2\mathrm{with}$$$$\stackrel{\rightharpoonup }{OX}=p\times \stackrel{\rightharpoonup }{a}+q\times \stackrel{\rightharpoonup }{b}$$$$\mathrm{set}\mathrm{C}$$$$\left(\begin{array}{c}x\\ y\end{array}\right)=p\times \left(\begin{array}{c}2\\ 1\end{array}\right)+q\times \left(\begin{array}{c}0\\ 1\end{array}\right)=\left(\begin{array}{c}2p\\ p+q\end{array}\right)$$$$\{\begin{array}{l}x=2p\\ y=p+q\end{array}\iff \{\begin{array}{l}p=\frac{x}{2}\\ q=y-\frac{x}{2}\end{array}\Rightarrow$$$$\Rightarrow \mathrm{for}\mathrm{any}\mathrm{given}\mathrm{point}\left(\begin{array}{c}x\\ y\end{array}\right)\mathrm{we}\mathrm{can}\mathrm{find}\mathrm{a}\mathrm{unique}$$$$\mathrm{pair}(p,q)$$

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

$$\mathrm{Sir}\mathrm{if}\mathrm{the}\mathrm{vectors}\left(\begin{array}{c}{a}_1\\ b_1\\ c_1\end{array}\right)\mathrm{and}\left(\begin{array}{c}{a}_2\\ b_2\\ c_3\end{array}\right)\mathrm{are}\mathrm{linearly}$$$$\mathrm{independent},\mathrm{then}\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|\ne 0\mathrm{right}?$$

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

$$\mathrm{thanks}\mathrm{sir}\mathrm{i}\mathrm{got}\mathrm{it}$$

Commented by MJS last updated on 11/Mar/20

$$\mathrm{yes}\mathrm{and}\mathrm{no}$$$$\mathrm{vectors}\mathrm{are}\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)\mathrm{and}\left(\begin{array}{c}{a}_2\\ b_2\end{array}\right)\mathrm{because}{\mathrm{we}}^{\prime }\mathrm{re}\mathrm{in}$$$${\mathbb{R}}^2\mathrm{and}\mathrm{determinants}\mathrm{only}\mathrm{exist}\mathrm{for}n\times n$$$$\mathrm{matrices}$$$$\mathrm{if}\left|\begin{array}{cc}{a}_1& a_2\\ b_1& b_2\end{array}\right|\ne 0\mathrm{the}\mathrm{vectors}\mathrm{are}\mathrm{independent}$$

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

$$thanks$$

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