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# Consider the vectors $\mathbf{v} = \begin{pmatrix} 1\\3 \end{pmatrix}$ and $\mathbf{w} = \begin{pmatrix} 3\\2 \end{pmatrix}$. Can you write

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Consider the vectors $$f(v) = \begin{pmatrix} 1\\3 \end{pmatrix}$$ and $$f{w} = \begin{pmatrix} 3\\2 \end{pmatrix}$$. Can you write$$f{u} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$$
as a linear combination of  v and w? If no, answer with 0. If yes, find coefficients a and v such $$a \begin{pmatrix} 1 \\ 3 \end{pmatrix} + b \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix}1 \\ 2 \end{pmatrix}$$ that

Feb 8, 2019

#1
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$$\text{This post is a mess. I'm assuming you mean}\\ v= \begin{pmatrix}1\\3\end{pmatrix}, ~u = \begin{pmatrix}3\\2\end{pmatrix}\\ \text{write }u = \begin{pmatrix}1\\2\end{pmatrix} = a v + b u \text{ if possible}$$

$$\text{well we have to establish that }u \text{ and }v \text{ are linearly inedpendent}\\ \text{we put them into a matrix and check that the determinant is non-zero}\\ \left|\begin{pmatrix}1&3\\3&2\end{pmatrix}\right| = 2-9=-7\neq 0\\ \text{so the two vectors are linearly independent}$$

$$\text{we have the matrix equation }\\ \begin{pmatrix}1 &3 \\ 3 & 2\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}1\\2\end{pmatrix}$$

$$\text{using Gauss elimination}\\ \begin{pmatrix}1&3&|&1\\3&2&|&2\end{pmatrix}\\ \begin{pmatrix}1&3&|&1\\0&-7&|&-1\end{pmatrix}\\ \begin{pmatrix}1&3&|&1\\0&1&|&\dfrac 1 7\end{pmatrix}\\ \begin{pmatrix}1&0&|&\dfrac 4 7\\0&1&|&\dfrac 1 7\end{pmatrix}\\$$

$$a = \dfrac 4 7,~b = \dfrac 1 7\\ \dfrac a b = 4$$

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Feb 8, 2019
edited by Rom  Feb 8, 2019
#2
+25506
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Consider the vectors

$$\mathbf{\vec{v}} = \dbinom{1}{3}$$ and  $$\mathbf{\vec{w}} = \dbinom{3}{2}$$
Can you write $$\mathbf{\vec{u}} = \dbinom{1}{2}$$
as a linear combination of  $$\vec{v}$$ and $$\vec{w}$$ ?
If yes, find coefficients a and b such $$a \dbinom{1}{3} + b \dbinom{3}{2} = \dbinom{1}{2}$$
and answer with $$\dfrac{a}{b}$$.

$$\mathbf{\vec{v}_{\perp} = \ ?}$$

$$\begin{array}{|rcll|} \hline \vec{v} \cdot \vec{v}_{\perp} &=& 0 \\ \dbinom{1}{3} \cdot \underbrace{\dbinom{-3}{1}}_{=\vec{v}_{\perp}} &=& -3+3 = 0 \\ \hline \end{array}$$

$$\mathbf{\vec{w}_{\perp} = \ ?}$$

$$\begin{array}{|rcll|} \hline \vec{w} \cdot \vec{w}_{\perp} &=& 0 \\ \dbinom{3}{2} \cdot \underbrace{\dbinom{-2}{3}}_{=\vec{w}_{\perp}} &=& -6+6 = 0 \\ \hline \end{array}$$

$$\mathbf{a,b = \ ?}$$

$$\begin{array}{|lrclccccc|} \hline & & & & & I. & & II. \\ & a\vec{v} + b\vec{w} &=& \vec{u} & | & \cdot \vec{v}_{\perp}& | & \cdot \vec{w}_{\perp}& | \\ \hline I. & \underbrace{a\vec{v}\vec{v}_{\perp}}_{=0} + b\vec{w}\vec{v}_{\perp} &=& \vec{u}\vec{v}_{\perp} \\ & b\vec{w}\vec{v}_{\perp} &=& \vec{u}\vec{v}_{\perp} \\ & \mathbf{b} &\mathbf{=}& \mathbf{\dfrac{ \vec{u}\vec{v}_{\perp} } { \vec{w}\vec{v}_{\perp} } } \\\\ & b &=& \dfrac{ \dbinom{1}{2}\dbinom{-3}{1} } { \dbinom{3}{2}\dbinom{-3}{1} } \\\\ & b &=& \dfrac{ -3+2 } { -9+2 } \\\\ & b &=& \dfrac{ -1 } { -7} \\\\ & \mathbf{b} &\mathbf{=}& \mathbf{\dfrac{ 1 } { 7}} \\ \hline II. & a\vec{v}\vec{w}_{\perp} + \underbrace{b\vec{w}\vec{w}_{\perp}}_{=0} &=& \vec{u}\vec{w}_{\perp} \\ & a\vec{v}\vec{w}_{\perp} &=& \vec{u}\vec{w}_{\perp} \\ & \mathbf{a} &\mathbf{=}& \mathbf{\dfrac{ \vec{u}\vec{w}_{\perp} } { \vec{v}\vec{w}_{\perp} } } \\\\ & a &=& \dfrac{ \dbinom{1}{2}\dbinom{-2}{3} } { \dbinom{1}{3}\dbinom{-2}{3} } \\\\ & a &=& \dfrac{ -2+6 } { -2+9 } \\\\ & \mathbf{a} &\mathbf{=}& \mathbf{\dfrac{ 4 } { 7 }} \\ \\ \hline & \dfrac{a}{b} &=& \dfrac{\dfrac{ 4 } { 7 }}{\dfrac{ 1 } { 7}} \\\\ & &=& \dfrac{ 4 } { 7 } \cdot \dfrac{ 7 } { 1} \\\\ & \mathbf{\dfrac{a}{b}} &\mathbf{=}& \mathbf{4} \\ \hline \end{array}$$

Feb 8, 2019