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If there are vectors \(\mathbf{v} = \begin{pmatrix} 1\\2\\1 \end{pmatrix}, \mathbf{w} = \begin{pmatrix} 1\\4 \\5 \end{pmatrix}\) and \(\mathbf{x} = \begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix}.\) Find coefficients a, b, and c, not all 0, such that \(a\begin{pmatrix} 1\\2\\1 \end{pmatrix}+b \begin{pmatrix} 1\\4 \\5 \end{pmatrix} + c\begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\)and answer with \(\dfrac{a-b}{c} \).

 Nov 3, 2019
 #1
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One solution is a = 6, b = 2, c = 4, so (a - b)/c = (6 - 2)/4 = 1.

 Nov 3, 2019
 #2
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Actually, a = 7k, b = 4k, c = k, so (a - b)/c = (7k - 4k)/k = 3.

 Nov 3, 2019
 #3
avatar+24364 
+2

If there are vectors \(\mathbf{v} = \begin{pmatrix} 1\\2\\1 \end{pmatrix}\)\(\mathbf{w} = \begin{pmatrix} 1\\4 \\5 \end{pmatrix}\),  and \( \mathbf{x} = \begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix}\).
Find coefficients a, b, and c, not all 0, such that

\(a\begin{pmatrix} 1\\2\\1 \end{pmatrix}+b \begin{pmatrix} 1\\4 \\5 \end{pmatrix} + c\begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\)
and answer with \(\dfrac{a-b}{c}\).

 

\(\begin{array}{|lrcll|} \hline &\begin{pmatrix} 1&1&-1\\2&4&6\\1&5&15 \end{pmatrix} \cdot \begin{pmatrix} a\\b\\c \end{pmatrix} &=& \begin{pmatrix} 0\\0\\0 \end{pmatrix} \\\\ (2): &2a +4b+6c &=& 0 \quad | \quad : 2 \\ &a +2b+3c &=& 0 \\\\ &\begin{pmatrix} 1&1&-1\\1&2&3\\1&5&15 \end{pmatrix} \cdot \begin{pmatrix} a\\b\\c \end{pmatrix} &=& \begin{pmatrix} 0\\0\\0 \end{pmatrix} \\\\ (3) = (3) - (1) : &\begin{pmatrix} 1&1&-1\\1&2&3\\0&4&16 \end{pmatrix} \cdot \begin{pmatrix} a\\b\\c \end{pmatrix} &=& \begin{pmatrix} 0\\0\\0 \end{pmatrix} \\\\ (3) &0\cdot a +4b+16c &=& 0 \quad | \quad : 4 \\ & b+4c &=& 0 \\\\ &\begin{pmatrix} 1&1&-1\\1&2&3\\0&1&4 \end{pmatrix} \cdot \begin{pmatrix} a\\b\\c \end{pmatrix} &=& \begin{pmatrix} 0\\0\\0 \end{pmatrix} \\\\ (2) = (2) - (1) : &\begin{pmatrix} 1&1&-1\\0&1&4\\0&1&4 \end{pmatrix} \cdot \begin{pmatrix} a\\b\\c \end{pmatrix} &=& \begin{pmatrix} 0\\0\\0 \end{pmatrix} \\\\ (3) = (3) - (2) : &\begin{pmatrix} 1&1&-1\\0&1&4\\0&0&0 \end{pmatrix} \cdot \begin{pmatrix} a\\b\\c \end{pmatrix} &=& \begin{pmatrix} 0\\0\\0 \end{pmatrix} \\ \hline \end{array}\)

 

We set \(c=k\):

\(\begin{array}{|lrcll|} \hline (2): & 0\cdot a+1\cdot b+4 c &=& 0 \quad | \quad c=k \\ & b+4k &=& 0 \\ & \mathbf{ b } &=& \mathbf{ -4k } \\\\ (1): & 1\cdot a+1\cdot b- 1 \cdot c &=& 0 \quad | \quad c=k,\ b=-4k \\ & a-4k-k &=& 0 \\ & a-5k &=& 0 \\ & \mathbf{ a } &=& \mathbf{ 5k } \\ \hline \end{array} \)

 

proof:

\(\begin{array}{|rcll|} \hline a\begin{pmatrix} 1\\2\\1 \end{pmatrix}+b \begin{pmatrix} 1\\4 \\5 \end{pmatrix} + c\begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix} &=& \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \quad | \quad a=5k,\ b=-4k,\ c=k \\\\ \hline 5k\begin{pmatrix} 1\\2\\1 \end{pmatrix}-4k \begin{pmatrix} 1\\4 \\5 \end{pmatrix} + k\begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix} &=& \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \\\\ 5k-4k-k &=& 0 \checkmark \\ 10k-16k+6k &=& 0 \checkmark \\ 5k - 20k+15k &=& 0 \checkmark \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{a+b}{c} } \\\\ &=& \dfrac{5k-(-4k)}{k} \\\\ &=& \dfrac{5k+4k}{k} \\\\ &=& \dfrac{9k}{k} \\\\ &=& \mathbf{9} \\ \hline \end{array} \)

 

laugh

 Nov 4, 2019
 #4
avatar+7763 
+1

\(a\mathbf{v} + b\mathbf{w} + c\mathbf{x} = \vec{0}\\ \begin{pmatrix} 1&1&-1\\2&4&6\\1&5&15 \end{pmatrix}\cdot\begin{pmatrix} a\\b\\c \end{pmatrix} = \begin{pmatrix} 0\\0\\0 \end{pmatrix}\\ \begin{pmatrix} 1&1&-1\\0&2&8\\0&4&16 \end{pmatrix}\cdot\begin{pmatrix} a\\b\\c \end{pmatrix} = \begin{pmatrix} 0\\0\\0 \end{pmatrix}\\ \begin{pmatrix} 1&1&-1\\0&1&4\\0&0&0 \end{pmatrix}\cdot\begin{pmatrix} a\\b\\c \end{pmatrix} = \begin{pmatrix} 0\\0\\0 \end{pmatrix}\\ \text{Let }c = k\\ b +4k = 0\\ \boxed{b = -4k}\\ a - 4k - k = 0\\ \boxed{a = 5k}\\ \dfrac{a - b}c = \dfrac{5k-(-4k)}{k} = 9\)

.
 Nov 4, 2019

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