+76 Consider the vectors \(\mathbf{v} = \begin{pmatrix} 1\\2\\1 \end{pmatrix}, \mathbf{w} = \begin{pmatrix} 1\\4 \\5 \end{pmatrix}\), and \(\mathbf{x} = \begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix}.\)
If the vectors \(\mathbf{v}, \mathbf{w}\), and \(\mathbf{x}\) are linearly independent, answer with ? (a question mark.) If they aren't, 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 \(\dfrac{a-b}{c}\).
I have tried this multiple times and I got a=0, b=0, and c=0 but it was incorrect. I tried distributing the vectors and solving for a, b, and c with the 3 equations. Can someone please tell me if I am doing anything wrong? Thanks!
+26387 Consider the vectors \(\mathbf{v} = \begin{pmatrix} 1\\2\\1 \end{pmatrix}\), \(\mathbf{w} = \begin{pmatrix} 1\\4 \\5 \end{pmatrix}\), and \(\mathbf{x} = \begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix}\).
If the vectors \(\mathbf{v}\), \(\mathbf{w}\) and \(\mathbf{x}\) are linearly independent, answer with ? (a question mark.)
If they aren't, 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 \(\dfrac{a-b}{c}\).
\(\begin{array}{|rcll|} \hline \begin{array}{|rcll|} \hline a\begin{pmatrix} 1\\2\\1 \end{pmatrix}+b \begin{pmatrix} 1\\4 \\5 \end{pmatrix} + c\begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix} &=& \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \\ \hline \end{array}\\\\ \left(\begin{array}{@{}ccc|c@{}} 1 & 1 & -1& 0 \\ 2 & 4 & 6 & 0\\ 1 & 5 & 15 & 0 \end{array}\right)_{\frac{R_2}{2}\rightarrow R_2} \Rightarrow \left(\begin{array}{@{}ccc|c@{}} 1 & 1 & -1& 0 \\ 1 & 2 & 3 & 0\\ 1 & 5 & 15 & 0 \end{array}\right)_{R_3-R_1 \rightarrow R_3} \Rightarrow \\ \Rightarrow \left(\begin{array}{@{}ccc|c@{}} 1 & 1 & -1& 0 \\ 1 & 2 & 3 & 0\\ 0 & 4 & 16 & 0 \end{array}\right)_{\frac{R_3}{4}\rightarrow R_3} \Rightarrow \left(\begin{array}{@{}ccc|c@{}} 1 & 1 & -1& 0 \\ 1 & 2 & 3 & 0\\ 0 & 1 & 4 & 0 \end{array}\right)_{R_2-R_1 \rightarrow R_2} \Rightarrow \\ \Rightarrow \left(\begin{array}{@{}ccc|c@{}} 1 & 1 & -1& 0 \\ 0 & 1 & 4 & 0\\ 0 & 1 & 4 & 0 \end{array}\right)_{R_3-R_2\rightarrow R_3} \Rightarrow \left(\begin{array}{@{}ccc|c@{}} 1 & 1 & -1& 0 \\ 0 & 1 & 4 & 0\\ 0 & 0 & 0 & 0 \end{array}\right)_{R_1-R_2 \rightarrow R_1} \Rightarrow \\ \Rightarrow \left(\begin{array}{@{}ccc|c@{}} 1 & 0 & -5& 0 \\ 0 & 1 & 4 & 0\\ 0 & 0 & 0 & 0 \end{array}\right) \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \left(\begin{array}{@{}ccc|c@{}} 1 & 0 & -5& 0 \\ 0 & 1 & 4 & 0\\ 0 & 0 & 0 & 0 \end{array}\right) \\ & \begin{array}{rcll} a -5c &=& 0 \\ b+4c &=& 0 \\ c &=& k \\ \end{array}\quad \Rightarrow \quad \begin{array}{rcll} a -5k &=& 0 \\ b+4k &=& 0 \\ c &=& k \\ \end{array}\quad \Rightarrow \quad \begin{array}{rcll} \mathbf{a} &=& \mathbf{5k} \\ \mathbf{b} &=& \mathbf{-4k} \\ \mathbf{c} &=& \mathbf{k} \\ \end{array} \\ \hline \end{array} \)
\((a,\ b,\ c) = (5k,\ -4k,\ k) \Rightarrow \text{Nontrivial}\Rightarrow \mathbf{\text{Linearly dependent}}\)
\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{a-b}{c}} &=& \dfrac{5k-(-4k)}{k} \\\\ \dfrac{a-b}{c} &=& \dfrac{9k}{k} \\\\ \mathbf{\dfrac{a-b}{c}} &=& \mathbf{9} \\ \hline \end{array}\)
+26387 Consider the vectors \(\mathbf{v} = \begin{pmatrix} 1\\2\\1 \end{pmatrix}\), \(\mathbf{w} = \begin{pmatrix} 1\\4 \\5 \end{pmatrix}\), and \(\mathbf{x} = \begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix}\).
If the vectors \(\mathbf{v}\), \(\mathbf{w}\) and \(\mathbf{x}\) are linearly independent, answer with ? (a question mark.)
If they aren't, 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 \(\dfrac{a-b}{c}\).
\(\begin{array}{|rcll|} \hline \begin{array}{|rcll|} \hline a\begin{pmatrix} 1\\2\\1 \end{pmatrix}+b \begin{pmatrix} 1\\4 \\5 \end{pmatrix} + c\begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix} &=& \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \\ \hline \end{array}\\\\ \left(\begin{array}{@{}ccc|c@{}} 1 & 1 & -1& 0 \\ 2 & 4 & 6 & 0\\ 1 & 5 & 15 & 0 \end{array}\right)_{\frac{R_2}{2}\rightarrow R_2} \Rightarrow \left(\begin{array}{@{}ccc|c@{}} 1 & 1 & -1& 0 \\ 1 & 2 & 3 & 0\\ 1 & 5 & 15 & 0 \end{array}\right)_{R_3-R_1 \rightarrow R_3} \Rightarrow \\ \Rightarrow \left(\begin{array}{@{}ccc|c@{}} 1 & 1 & -1& 0 \\ 1 & 2 & 3 & 0\\ 0 & 4 & 16 & 0 \end{array}\right)_{\frac{R_3}{4}\rightarrow R_3} \Rightarrow \left(\begin{array}{@{}ccc|c@{}} 1 & 1 & -1& 0 \\ 1 & 2 & 3 & 0\\ 0 & 1 & 4 & 0 \end{array}\right)_{R_2-R_1 \rightarrow R_2} \Rightarrow \\ \Rightarrow \left(\begin{array}{@{}ccc|c@{}} 1 & 1 & -1& 0 \\ 0 & 1 & 4 & 0\\ 0 & 1 & 4 & 0 \end{array}\right)_{R_3-R_2\rightarrow R_3} \Rightarrow \left(\begin{array}{@{}ccc|c@{}} 1 & 1 & -1& 0 \\ 0 & 1 & 4 & 0\\ 0 & 0 & 0 & 0 \end{array}\right)_{R_1-R_2 \rightarrow R_1} \Rightarrow \\ \Rightarrow \left(\begin{array}{@{}ccc|c@{}} 1 & 0 & -5& 0 \\ 0 & 1 & 4 & 0\\ 0 & 0 & 0 & 0 \end{array}\right) \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \left(\begin{array}{@{}ccc|c@{}} 1 & 0 & -5& 0 \\ 0 & 1 & 4 & 0\\ 0 & 0 & 0 & 0 \end{array}\right) \\ & \begin{array}{rcll} a -5c &=& 0 \\ b+4c &=& 0 \\ c &=& k \\ \end{array}\quad \Rightarrow \quad \begin{array}{rcll} a -5k &=& 0 \\ b+4k &=& 0 \\ c &=& k \\ \end{array}\quad \Rightarrow \quad \begin{array}{rcll} \mathbf{a} &=& \mathbf{5k} \\ \mathbf{b} &=& \mathbf{-4k} \\ \mathbf{c} &=& \mathbf{k} \\ \end{array} \\ \hline \end{array} \)
\((a,\ b,\ c) = (5k,\ -4k,\ k) \Rightarrow \text{Nontrivial}\Rightarrow \mathbf{\text{Linearly dependent}}\)
\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{a-b}{c}} &=& \dfrac{5k-(-4k)}{k} \\\\ \dfrac{a-b}{c} &=& \dfrac{9k}{k} \\\\ \mathbf{\dfrac{a-b}{c}} &=& \mathbf{9} \\ \hline \end{array}\)
