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The vectors \(\dbinom{3}{2}\) and \(\dbinom{-4}{1}\) can be written as linear combinations of \(\mathbf{u}\) and \(\mathbf{w}\)\(\begin{align*} \dbinom{3}{2} &= 5\mathbf{u}+8\mathbf{w} \\ \dbinom{-4}{1} &= -3\mathbf{u}+\mathbf{w} . \end{align*}\)
The vector \(\dbinom{5}{-2}\) can be written as the linear combination \(a\mathbf{u}+b\mathbf{w}\). Find the ordered pair \((a,b)\).

 Feb 7, 2020
 #1
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(a,b) = (17,-4).

 Feb 7, 2020
 #2
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The vectors \(\dbinom{3}{2}\) and \(\dbinom{-4}{1}\) can be written as linear combinations of  \(u\) and \(w\):


\(\begin{align*} \dbinom{3}{2} = 5\mathbf{u}+8\mathbf{w} ,\\\\ \dbinom{-4}{1} = -3\mathbf{u}+\mathbf{w} . \end{align*}\)
 

The vector \(\dbinom{5}{-2}\) can be written as the linear combination \(a\mathbf{u}+b\mathbf{w}\).
Find the ordered pair \((a,b)\).

 

 

\(\text{Let $\dbinom{3}{2} = v_1$ } \\ \text{Let $\dbinom{-4}{1} = v_2$ }\)

 

\(\begin{array}{|lrcll|} \hline (1): & \mathbf{5\mathbf{u}+8\mathbf{w}} &=& \mathbf{v_1} \\\\ & -3\mathbf{u}+\mathbf{w} &=& v_2 \quad | \quad \times 8 \\ (2): & -24 u + 8w &=& 8v_2 \\ \hline (1)-(2): & 29u &=& v_1-8v_2 \quad | \quad v_1=\dbinom{3}{2},\ v_2 = \dbinom{-4}{1} \\ & 29u &=& \dbinom{3}{2}-8\dbinom{-4}{1} \\ & 29u &=& \dbinom{3}{2}+\dbinom{32}{-8} \\ & 29u &=& \dbinom{35}{-6} \\ & \mathbf{u} &=& \mathbf{\dfrac{1}{29} \dbinom{35}{-6}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline -3\mathbf{u}+\mathbf{w} &=& v_2 \\ w &=& v_2+3u \quad | \quad v_2 = \dbinom{-4}{1},\ \mathbf{u=\dfrac{1}{29} \dbinom{35}{-6}} \\ w &=& \dbinom{-4}{1}+\dfrac{3}{29} \dbinom{35}{-6} \\ w &=& \dfrac{29}{29}\dbinom{-4}{1}+\dfrac{3}{29} \dbinom{35}{-6} \\ w &=& \dfrac{1}{29} \left( 29\dbinom{-4}{1}+3\dbinom{35}{-6} \right) \\ w &=& \dfrac{1}{29} \dbinom{-4*29+3*35}{29-18} \\ \mathbf{w} &=& \mathbf{\dfrac{1}{29} \dbinom{-11}{11}} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline a\mathbf{u}+b\mathbf{w} &=& \dbinom{5}{-2} \quad | \quad \mathbf{u=\dfrac{1}{29} \dbinom{35}{-6}},\ \mathbf{w=\dfrac{1}{29} \dbinom{-11}{11}} \\\\ \dfrac{a}{29} \dbinom{35}{-6} + \dfrac{b}{29} \dbinom{-11}{11} &=& \dbinom{5}{-2} \quad | \quad \times 29 \\\\ a\dbinom{35}{-6} +b\dbinom{-11}{11} &=& \dbinom{5*29}{-2*29} \\\\ a\dbinom{35}{-6} +b\dbinom{-11}{11} &=& \dbinom{145}{-58} \\ \hline \end{array} \)

 

\(\begin{array}{|lrcll|} \hline (1): & 35a-11b &=& 145 \\ (2): & -6a+11b &=& -58 \\ \hline (1)+(2): & 35a-6a &=& 145 -58 \\ & 29a &=& 87 \quad | \quad :29 \\ & \mathbf{a} &=& \mathbf{3} \\ \hline & -6a+11b &=& -58 \\ & 11b &=& 6a - 58 \\ & 11b &=& 6*3 - 58 \\ & 11b &=& -40 \quad | \quad : 11 \\ & \mathbf{b} &=& \mathbf{-\dfrac{40}{11}} \\ \hline \end{array}\)

 

\((a,b) = \mathbf{\left(3,\ -\dfrac{40}{11}\right)} \)

 

laugh

 Feb 7, 2020

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