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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 
and answer with a/b. 

 Feb 8, 2019
 #1
avatar+4467 
0

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

 

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

 

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

 

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

 

\(a = \dfrac 4 7,~b = \dfrac 1 7\\ \dfrac a b = 4\)

.
 Feb 8, 2019
edited by Rom  Feb 8, 2019
 #2
avatar+21860 
+8

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 no, answer with 0.
If yes, find coefficients a and b such \(a \dbinom{1}{3} + b \dbinom{3}{2} = \dbinom{1}{2}\)
and answer with \(\dfrac{a}{b}\).

 

\(\mathbf{\vec{v}_{\perp} = \ ?} \)

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

 

\(\mathbf{\vec{w}_{\perp} = \ ?} \)

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

 

\(\mathbf{a,b = \ ?}\)

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

 

 

laugh

 Feb 8, 2019

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