Opening the book…
Write the solution set of the following system as a linear combination of vectors \[\left[\begin{array}{rrr}1&-1&2\\1&-2&1\\3&-4&5\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}0\\0\\0\end{array}\right]\nonumber\]
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Solution is: \(\left[\begin{array}{r}-3t \\ -t\\ t\end{array}\right],\:t\in\mathbb{R}\). A basis for the solution space is \(\left[\begin{array}{r}-3\\-1\\1\end{array}\right]\nonumber\)
Using Exercise \(\PageIndex{1}\) find the general solution to the following linear system. \[\left[\begin{array}{rrr}1&-1&2\\1&-2&1\\3&-4&5\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}1\\2\\4\end{array}\right]\nonumber\]
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Note that this has the same matrix as the above problem. Solution is: \(\left[\begin{array}{r}-3t \\ -t \\ t\end{array}\right]+\left[\begin{array}{r}0\\-1\\0\end{array}\right],\:t\in\mathbb{R}\)
Write the solution set of the following system as a linear combination of vectors \[\left[\begin{array}{rrr}0&-1&2\\1&-2&1\\1&-4&5\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}0\\0\\0\end{array}\right]\nonumber\]
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Solution is: \(\left[\begin{array}{c}3t \\ 2t \\ t\end{array}\right]\), A basis is \(\left[\begin{array}{c}3\\2\\1\end{array}\right]\)
Using Exercise \(\PageIndex{3}\) find the general solution to the following linear system. \[\left[\begin{array}{rrr}0&-1&2\\1&-2&1\\1&-4&5\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}1\\-1\\1\end{array}\right]\nonumber\]
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Solution is: \(\left[\begin{array}{c}3t \\ 2t \\ t\end{array}\right] +\left[\begin{array}{c}-3\\-1\\0\end{array}\right],\:\hat{t}\in\mathbb{R}\)
Write the solution set of the following system as a linear combination of vectors. \[\left[\begin{array}{ccc}1&-1&2\\1&-2&0\\3&-4&4\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}0\\0\\0\end{array}\right]\nonumber\]
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Solution is: \(\left[\begin{array}{r}-4t\\ -2t\\ t\end{array}\right]\). A basis is \(\left[\begin{array}{r}-4\\-2\\1\end{array}\right]\)
Using Exercise \(\PageIndex{5}\) find the general solution to the following linear system. \[\left[\begin{array}{ccc}1&-1&2\\1&-2&0\\3&-4&4\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}1\\2\\4\end{array}\right]\nonumber\]
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Solution is: \(\left[\begin{array}{r}-4t \\ -2t \\ t\end{array}\right]+\left[\begin{array}{r}0\\-1\\0\end{array}\right],\:t \in\mathbb{R}\)
Write the solution set of the following system as a linear combination of vectors \[\left[\begin{array}{ccc}0&-1&2\\1&0&1\\1&-2&5\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}0\\0\\0\end{array}\right]\nonumber\]
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Solution is: \(\left[\begin{array}{r}-t\\ 2t \\ t\end{array}\right],\:t \in\mathbb{R}\)
Using Exercise \(\PageIndex{7}\) find the general solution to the following linear system. \[\left[\begin{array}{ccc}0&-1&2\\1&0&1\\1&-2&5\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}1\\-1\\1\end{array}\right]\nonumber\]
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Solution is: \(\left[\begin{array}{r}-t \\ 2t \\ t\end{array}\right]+\left[\begin{array}{r}-1\\-1\\0\end{array}\right]\)
Write the solution set of the following system as a linear combination of vectors \[\left[\begin{array}{cccc}1&0&1&1\\1&-1&1&0\\3&-1&3&2\\3&3&0&3\end{array}\right]\left[\begin{array}{c}x\\y\\z\\w\end{array}\right]=\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]\nonumber\]
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Solution is: \(\left[\begin{array}{r}0\\ -t \\ -t \\ t\end{array}\right],\:t\in\mathbb{R}\)
Using Exercise \(\PageIndex{9}\) find the general solution to the following linear system. \[\left[\begin{array}{cccc}1&0&1&1\\1&-1&1&0\\3&-1&3&2\\3&3&0&3\end{array}\right]\left[\begin{array}{c}x\\y\\z\\w\end{array}\right]=\left[\begin{array}{c}1\\2\\4\\3\end{array}\right]\nonumber\]
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Solution is: \(\left[\begin{array}{r}0\\ -t \\ -t \\ t\end{array}\right]+\left[\begin{array}{r}2\\-1\\-1\\0\end{array}\right]\)
Write the solution set of the following system as a linear combination of vectors \[\left[\begin{array}{cccc}1&1&0&1\\2&1&1&2\\1&0&1&1\\0&0&0&0\end{array}\right]\left[\begin{array}{c}x\\y\\z\\w\end{array}\right]=\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]\nonumber\]
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Solution is: \(\left[\begin{array}{c}-s-t \\ s\\s\\t\end{array}\right],\:s,t\in\mathbb{R}\). A basis is \[\left\{\left[\begin{array}{r}-1\\1\\1\\0\end{array}\right],\:\left[\begin{array}{r}-1\\0\\0\\1\end{array}\right]\right\}\nonumber\]
Using Exercise \(\PageIndex{11}\) find the general solution to the following linear system. \[\left[\begin{array}{rrrr}1&1&0&1\\2&1&1&2\\1&0&1&1\\0&-1&1&1\end{array}\right]\left[\begin{array}{c}x\\y\\z\\w\end{array}\right]=\left[\begin{array}{r}2\\-1\\-3\\0\end{array}\right]\nonumber\]
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Solution is: \[\left[\begin{array}{r}-t\\ t\\ t\\0\end{array}\right]+\left[\begin{array}{c}-8\\5\\0\\5\end{array}\right]\nonumber\]
Write the solution set of the following system as a linear combination of vectors \[\left[\begin{array}{rrrr}1&1&0&1\\1&-1&1&0\\3&1&1&2\\3&3&0&3\end{array}\right]\left[\begin{array}{c}x\\y\\z\\w\end{array}\right]=\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]\nonumber\]
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Solution is: \[\left[\begin{array}{c}-\frac{1}{2}s-\frac{1}{2}t \\ \frac{1}{2}s-\frac{1}{2}t \\ s\\t\end{array}\right]\nonumber\] for \(s,t\in\mathbb{R}\). A basis is \[\left\{\left[\begin{array}{r}-1\\1\\2\\0\end{array}\right],\:\left[\begin{array}{r}-1\\1\\0\\1\end{array}\right]\right\}\nonumber\]
Using Exercise \(\PageIndex{13}\) find the general solution to the following linear system. \[\left[\begin{array}{rrrr}1&1&0&1\\1&-1&1&0\\3&1&1&2\\3&3&0&3\end{array}\right]\left[\begin{array}{c}x\\y\\z\\w\end{array}\right]=\left[\begin{array}{c}1\\2\\4\\3\end{array}\right]\nonumber\]
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Solution is: \[\left[\begin{array}{r}\frac{3}{2} \\ -\frac{1}{2}\\0\\0\end{array}\right]+\left[\begin{array}{c}-\frac{1}{2}s-\frac{1}{2}t \\ \frac{1}{2}s-\frac{1}{2}t \\ s\\t\end{array}\right]\nonumber\]
Write the solution set of the following system as a linear combination of vectors \[\left[\begin{array}{rrrr}1&1&0&1\\2&1&1&2\\1&0&1&1\\0&-1&1&1\end{array}\right]\left[\begin{array}{c}x\\y\\z\\w\end{array}\right]=\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]\nonumber\]
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Solution is: \(\left[\begin{array}{r}-t \\ t \\ t\\0\end{array}\right]\), a basis is \(\left[\begin{array}{c}1\\1\\1\\0\end{array}\right]\).
Using Exercise \(\PageIndex{15}\) find the general solution to the following linear system. \[\left[\begin{array}{rrrr}1&1&0&1\\2&1&1&2\\1&0&1&1\\0&-1&1&1\end{array}\right]\left[\begin{array}{c}x\\y\\z\\w\end{array}\right]=\left[\begin{array}{r}2\\-1\\-3\\1\end{array}\right]\nonumber\]
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Solution is: \(\left[\begin{array}{r}-t \\ t \\ t\\0\end{array}\right]+\left[\begin{array}{r}-9\\5\\0\\6\end{array}\right],t\in\mathbb{R}\).
Suppose \(A\vec{x}=\vec{b}\) has a solution. Explain why the solution is unique precisely when \(A\vec{x}=\vec{0}\) has only the trivial solution.
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If not, then there would be a infintely many solutions to \(A\vec{x}=\vec{0}\) and each of these added to a solution to \(A\vec{x}=\vec{b}\) would be a solution to \(A\vec{x}=\vec{b}\).
This page titled 5.9E: Exercises for Section 5.9 was last modified on Thu, 10 Apr 2025 05:36:23 GMT and is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Doli Bambhania, Fatemeh Yarahmadi, and Bill Wilson.
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