2.4.1: Exercises 2.4
- Page ID
- 67842
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In Exercises \(\PageIndex{1}\) – \(\PageIndex{6}\), a matrix \(A\) and vectors \(\vec{b}\), \(\vec{u}\) and \(\vec{v}\) are given. Verify that \(\vec{u}\) and \(\vec{v}\) are both solutions to the equation \(A\vec{x}=\vec{b}\); that is, show that \(A\vec{u}=A\vec{v}=\vec{b}\).
\(A=\left[\begin{array}{cc}{1}&{-2}\\{-3}&{6}\end{array}\right],\)
\(\vec{b}=\left[\begin{array}{c}{0}\\{0}\end{array}\right],\:\vec{u}=\left[\begin{array}{c}{2}\\{1}\end{array}\right],\:\vec{v}=\left[\begin{array}{c}{-10}\\{-5}\end{array}\right]\)
- Answer
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Multiply \(A\vec{u}\) and \(A\vec{v}\) to verify.
\(A=\left[\begin{array}{cc}{1}&{-2}\\{-3}&{6}\end{array}\right],\)
\(\vec{b}=\left[\begin{array}{c}{2}\\{-6}\end{array}\right],\:\vec{u}=\left[\begin{array}{c}{0}\\{-1}\end{array}\right],\:\vec{v}=\left[\begin{array}{c}{2}\\{0}\end{array}\right]\)
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Multiply \(A\vec{u}\) and \(A\vec{v}\) to verify.
\(A=\left[\begin{array}{cc}{1}&{0}\\{2}&{0}\end{array}\right],\)
\(\vec{b}=\left[\begin{array}{c}{0}\\{0}\end{array}\right],\:\vec{u}=\left[\begin{array}{c}{0}\\{-1}\end{array}\right],\:\vec{v}=\left[\begin{array}{c}{0}\\{59}\end{array}\right]\)
- Answer
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Multiply \(A\vec{u}\) and \(A\vec{v}\) to verify.
\(A=\left[\begin{array}{cc}{1}&{0}\\{2}&{0}\end{array}\right],\)
\(\vec{b}=\left[\begin{array}{c}{-3}\\{-6}\end{array}\right],\:\vec{u}=\left[\begin{array}{c}{-3}\\{-1}\end{array}\right],\:\vec{v}=\left[\begin{array}{c}{-3}\\{59}\end{array}\right]\)
- Answer
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Multiply \(A\vec{u}\) and \(A\vec{v}\) to verify.
\(A=\left[\begin{array}{cccc}{0}&{-3}&{-1}&{-3}\\{-4}&{2}&{-3}&{5}\end{array}\right],\)
\(\vec{b}=\left[\begin{array}{c}{0}\\{0}\end{array}\right],\:\vec{u}=\left[\begin{array}{c}{11}\\{4}\\{-12}\\{0}\end{array}\right],\:\vec{v}=\left[\begin{array}{c}{9}\\{-12}\\{0}\\{12}\end{array}\right]\)
- Answer
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Multiply \(A\vec{u}\) and \(A\vec{v}\) to verify.
\(A=\left[\begin{array}{cccc}{0}&{-3}&{-1}&{-3}\\{-4}&{2}&{-3}&{5}\end{array}\right],\)
\(\vec{b}=\left[\begin{array}{c}{48}\\{36}\end{array}\right],\:\vec{u}=\left[\begin{array}{c}{-17}\\{-16}\\{0}\\{0}\end{array}\right],\:\vec{v}=\left[\begin{array}{c}{-8}\\{-28}\\{0}\\{12}\end{array}\right]\)
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Multiply \(A\vec{u}\) and \(A\vec{v}\) to verify.
In Exercises \(\PageIndex{7}\) - \(\PageIndex{9}\), a matrix \(A\) and vectors \(\vec{b}\), \(\vec{u}\) and \(\vec{v}\) are given. Verify that \(A\vec{u}=\vec{0}\), \(A\vec{v}=\vec{b}\) and \(A(\vec{u}+\vec{v})=\vec{b}\).
\(A=\left[\begin{array}{ccc}{2}&{-2}&{-1}\\{-1}&{1}&{-1}\\{-2}&{2}&{-1}\end{array}\right],\)
\(\vec{b}=\left[\begin{array}{c}{1}\\{1}\\{1}\end{array}\right],\:\vec{u}=\left[\begin{array}{c}{1}\\{1}\\{0}\end{array}\right],\:\vec{v}=\left[\begin{array}{c}{1}\\{1}\\{-1}\end{array}\right]\)
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Multiply \(A\vec{u}\), \(A\vec{v}\) and \(A(\vec{u}+\vec{v})\) to verify.
\(A=\left[\begin{array}{ccc}{1}&{-1}&{3}\\{3}&{-3}&{-3}\\{-1}&{1}&{1}\end{array}\right],\)
\(\vec{b}=\left[\begin{array}{c}{-1}\\{-3}\\{1}\end{array}\right],\:\vec{u}=\left[\begin{array}{c}{2}\\{2}\\{0}\end{array}\right],\:\vec{v}=\left[\begin{array}{c}{2}\\{3}\\{0}\end{array}\right]\)
- Answer
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Multiply \(A\vec{u}\), \(A\vec{v}\) and \(A(\vec{u}+\vec{v})\) to verify.
\(A=\left[\begin{array}{ccc}{2}&{0}&{0}\\{0}&{1}&{-3}\\{3}&{1}&{-3}\end{array}\right],\)
\(\vec{b}=\left[\begin{array}{c}{2}\\{-4}\\{-1}\end{array}\right],\:\vec{u}=\left[\begin{array}{c}{0}\\{6}\\{2}\end{array}\right],\:\vec{v}=\left[\begin{array}{c}{1}\\{-1}\\{1}\end{array}\right]\)
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Multiply \(A\vec{u}\), \(A\vec{v}\) and \(A(\vec{u}+\vec{v})\) to verify.
In Exercises \(\PageIndex{10}\) - \(\PageIndex{24}\), a matrix \(A\) and vector \(\vec{b}\) are given.
- Solve the equation \(A\vec{x}=\vec{0}\).
- Solve the equation \(A\vec{x}=\vec{b}\).
In each of the above, be sure to write your answer in vector format. Also, when possible, give 2 particular solutions to each equation.
\(A=\left[\begin{array}{cc}{0}&{2}\\{-1}&{3}\end{array}\right],\:\vec{b}=\left[\begin{array}{c}{-2}\\{-1}\end{array}\right]\)
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- \(\vec{x}=\left[\begin{array}{c}{0}\\{0}\end{array}\right]\)
- \(\vec{x}=\left[\begin{array}{c}{-2}\\{-1}\end{array}\right]\)
\(A=\left[\begin{array}{cc}{-4}&{-1}\\{-3}&{-2}\end{array}\right],\:\vec{b}=\left[\begin{array}{c}{1}\\{4}\end{array}\right]\)
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- \(\vec{x}=\left[\begin{array}{c}{0}\\{0}\end{array}\right]\)
- \(\vec{x}=\left[\begin{array}{c}{2/5}\\{-13/5}\end{array}\right]\)
\(A=\left[\begin{array}{cc}{1}&{-2}\\{0}&{1}\end{array}\right],\:\vec{b}=\left[\begin{array}{c}{0}\\{-5}\end{array}\right]\)
- Answer
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- \(\vec{x}=\left[\begin{array}{c}{0}\\{0}\end{array}\right]\)
- \(\vec{x}=\left[\begin{array}{c}{-10}\\{-5}\end{array}\right]\)
\(A=\left[\begin{array}{cc}{1}&{0}\\{5}&{-4}\end{array}\right],\:\vec{b}=\left[\begin{array}{c}{-2}\\{-1}\end{array}\right]\)
- Answer
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- \(\vec{x}=\left[\begin{array}{c}{0}\\{0}\end{array}\right]\)
- \(\vec{x}=\left[\begin{array}{c}{-2}\\{-9/4}\end{array}\right]\)
\(A=\left[\begin{array}{cc}{2}&{-3}\\{-4}&{6}\end{array}\right],\:\vec{b}=\left[\begin{array}{c}{1}\\{-1}\end{array}\right]\)
- Answer
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- \(\vec{x}=x_{2}\left[\begin{array}{c}{3/2}\\{1}\end{array}\right]\)
- No solution.
\(A=\left[\begin{array}{ccc}{-4}&{3}&{2}\\{-4}&{5}&{0}\end{array}\right],\:\vec{b}=\left[\begin{array}{c}{-4}\\{-4}\end{array}\right]\)
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- \(\vec{x}=x_{3}\left[\begin{array}{c}{5/4}\\{1}\\{1}\end{array}\right]\)
- \(\vec{x}=\left[\begin{array}{c}{1}\\{0}\\{0}\end{array}\right]+x_{3}\left[\begin{array}{c}{5/4}\\{1}\\{1}\end{array}\right]\)
\(A=\left[\begin{array}{ccc}{1}&{5}&{-2}\\{1}&{4}&{5}\end{array}\right],\:\vec{b}=\left[\begin{array}{c}{0}\\{1}\end{array}\right]\)
- Answer
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- \(\vec{x}=x_{3}\left[\begin{array}{c}{-33}\\{7}\\{1}\end{array}\right]\)
- \(\vec{x}=\left[\begin{array}{c}{5}\\{-1}\\{0}\end{array}\right]+x_{3}\left[\begin{array}{c}{-33}\\{7}\\{1}\end{array}\right]\)
\(A=\left[\begin{array}{ccc}{-1}&{-2}&{-2}\\{3}&{4}&{-2}\end{array}\right],\:\vec{b}=\left[\begin{array}{c}{-4}\\{-4}\end{array}\right]\)
- Answer
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- \(\vec{x}=x_{3}\left[\begin{array}{c}{14}\\{-10}\\{0}\end{array}\right]\)
- \(\vec{x}=\left[\begin{array}{c}{-4}\\{2}\end{array}\right]+x_{3}\left[\begin{array}{c}{14}\\{-10}\\{0}\end{array}\right]\)
\(A=\left[\begin{array}{ccc}{2}&{2}&{2}\\{5}&{5}&{-3}\end{array}\right],\:\vec{b}=\left[\begin{array}{c}{3}\\{-3}\end{array}\right]\)
- Answer
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- \(\vec{x}=x_{2}\left[\begin{array}{c}{-1}\\{1}\\{0}\end{array}\right]\)
- \(\vec{x}=\left[\begin{array}{c}{3/16}\\{0}\\{21/16}\end{array}\right]+x_{2}\left[\begin{array}{c}{-1}\\{1}\\{0}\end{array}\right]\)
\(A=\left[\begin{array}{cccc}{1}&{5}&{-4}&{-1}\\{1}&{0}&{-2}&{1}\end{array}\right],\:\vec{b}=\left[\begin{array}{c}{0}\\{-2}\end{array}\right]\)
- Answer
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- \(\vec{x}=x_{3}\left[\begin{array}{c}{2}\\{2/5}\\{1}\\{0}\end{array}\right]+x_{4}\left[\begin{array}{c}{-1}\\{2/5}\\{0}\\{1}\end{array}\right]\)
- \(\vec{x}=\left[\begin{array}{c}{-2}\\{2/5}\\{0}\\{0}\end{array}\right]+x_{3}\left[\begin{array}{c}{2}\\{2/5}\\{1}\\{0}\end{array}\right]+\left[\begin{array}{c}{-1}\\{2/5}\\{0}\\{1}\end{array}\right]\)
\(A=\left[\begin{array}{cccc}{-4}&{2}&{-5}&{4}\\{0}&{1}&{-1}&{5}\end{array}\right],\:\vec{b}=\left[\begin{array}{c}{-3}\\{-2}\end{array}\right]\)
- Answer
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- \(\vec{x}=x_{3}\left[\begin{array}{c}{-3/4}\\{1}\\{1}\\{0}\end{array}\right]+x_{4}\left[\begin{array}{c}{-3/2}\\{-5}\\{0}\\{1}\end{array}\right]\)
- \(\vec{x}=\left[\begin{array}{c}{-1/4}\\{-2}\\{0}\\{0}\end{array}\right]+x_{3}\left[\begin{array}{c}{-3/4}\\{1}\\{1}\\{0}\end{array}\right]+x_{4}\left[\begin{array}{c}{-3/2}\\{-5}\\{0}\\{1}\end{array}\right]\)
\(A=\left[\begin{array}{ccccc}{0}&{0}&{2}&{1}&{4}\\{-2}&{-1}&{-4}&{-1}&{5}\end{array}\right],\:\vec{b}=\left[\begin{array}{c}{3}\\{4}\end{array}\right]\)
- Answer
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- \(\vec{x}=x_{2}\left[\begin{array}{c}{-1/2}\\{1}\\{0}\\{0}\\{0}\end{array}\right]+x_{4}\left[\begin{array}{c}{1/2}\\{0}\\{-1/2}\\{1}\\{0}\end{array}\right]+x_{5}\left[\begin{array}{c}{13/2}\\{0}\\{-2}\\{0}\\{1}\end{array}\right]\)
- \(\vec{x}=\left[\begin{array}{c}{-5}\\{0}\\{3/2}\\{0}\\{0}\end{array}\right]+x_{2}\left[\begin{array}{c}{-1/2}\\{1}\\{0}\\{0}\\{0}\end{array}\right]+x_{4}\left[\begin{array}{c}{1/2}\\{0}\\{-1/2}\\{1}\\{0}\end{array}\right]+x_{5}\left[\begin{array}{c}{13/2}\\{0}\\{-2}\\{0}\\{1}\end{array}\right]\)
\(A=\left[\begin{array}{ccccc}{3}&{0}&{-2}&{-4}&{5}\\{2}&{3}&{2}&{0}&{2}\\{-5}&{0}&{4}&{0}&{5}\end{array}\right],\:\vec{b}=\left[\begin{array}{c}{-1}\\{-5}\\{4}\end{array}\right]\)
- Answer
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- \(\vec{x}=x_{4}\left[\begin{array}{c}{8}\\{-12}\\{10}\\{1}\\{0}\end{array}\right]+x_{5}\left[\begin{array}{c}{-15}\\{68/3}\\{-20}\\{0}\\{1}\end{array}\right]\)
- \(\vec{x}=\left[\begin{array}{c}{2}\\{-16/3}\\{7/2}\\{0}\\{0}\end{array}\right]+x_{4}\left[\begin{array}{c}{8}\\{-12}\\{10}\\{1}\\{0}\end{array}\right]+x_{5}\left[\begin{array}{c}{-15}\\{68/3}\\{-20}\\{0}\\{1}\end{array}\right]\)
\(A=\left[\begin{array}{ccccc}{-1}&{3}&{1}&{-3}&{4}\\{3}&{-3}&{-1}&{1}&{-4}\\{-2}&{3}&{-2}&{-3}&{1}\end{array}\right],\:\vec{b}=\left[\begin{array}{c}{1}\\{1}\\{-5}\end{array}\right]\)
- Answer
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- \(\vec{x}=x_{4}\left[\begin{array}{c}{1}\\{13/9}\\{-1/3}\\{1}\\{0}\end{array}\right]+x_{5}\left[\begin{array}{c}{0}\\{-1}\\{-1}\\{0}\\{1}\end{array}\right]\)
- \(\vec{x}=\left[\begin{array}{c}{1}\\{1/9}\\{5/3}\\{0}\\{0}\end{array}\right]+x_{4}\left[\begin{array}{c}{1}\\{13/9}\\{-1/3}\\{1}\\{0}\end{array}\right]+x_{5}\left[\begin{array}{c}{0}\\{-1}\\{-1}\\{0}\\{1}\end{array}\right]\)
\(A=\left[\begin{array}{ccccc}{-4}&{-2}&{-1}&{4}&{0}\\{5}&{-4}&{3}&{-1}&{1}\\{4}&{-5}&{3}&{1}&{-4}\end{array}\right],\:\vec{b}=\left[\begin{array}{c}{3}\\{2}\\{1}\end{array}\right]\)
- Answer
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- \(\vec{x}=x_{4}\left[\begin{array}{c}{3}\\{-1}\\{-6}\\{1}\\{0}\end{array}\right]+x_{5}\left[\begin{array}{c}{-17}\\{12}\\{44}\\{0}\\{1}\end{array}\right]\)
- \(\vec{x}=\left[\begin{array}{c}{7}\\{-6}\\{-19}\\{0}\\{0}\end{array}\right]+x_{4}\left[\begin{array}{c}{3}\\{-1}\\{-6}\\{1}\\{0}\end{array}\right]+x_{5}\left[\begin{array}{c}{-17}\\{12}\\{44}\\{0}\\{1}\end{array}\right]\)
In Exercises \(\PageIndex{25}\) – \(\PageIndex{28}\), a matrix \(A\) and vector \(\vec{b}\) are given. Solve the equation \(A\vec{x}=\vec{b}\), write the solution in vector format, and sketch the solution as the appropriate line on the Cartesian plane.
\(A=\left[\begin{array}{cc}{2}&{4}\\{-1}&{-2}\end{array}\right],\:\vec{b}=\left[\begin{array}{c}{0}\\{0}\end{array}\right]\)
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\(\vec{x}=x_{2}\left[\begin{array}{c}{-2}\\{1}\end{array}\right]=x_{2}\vec{v}\)
Figure \(\PageIndex{1}\)
\(A=\left[\begin{array}{cc}{2}&{4}\\{-1}&{-2}\end{array}\right],\:\vec{b}=\left[\begin{array}{c}{-6}\\{3}\end{array}\right]\)
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\(\vec{x}=\left[\begin{array}{c}{-3}\\{0}\end{array}\right]+x_{2}\left[\begin{array}{c}{-2}\\{1}\end{array}\right]=\vec{x_{p}}+x_{2}\vec{v}\)
Figure \(\PageIndex{2}\)
\(A=\left[\begin{array}{cc}{2}&{-5}\\{-4}&{-10}\end{array}\right],\:\vec{b}=\left[\begin{array}{c}{1}\\{2}\end{array}\right]\)
- Answer
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\(\vec{x}=\left[\begin{array}{c}{0.5}\\{0}\end{array}\right]+x_{2}\left[\begin{array}{c}{2.5}\\{1}\end{array}\right]=\vec{x_{p}}+x_{2}\vec{v}\)
Figure \(\PageIndex{3}\)
\(A=\left[\begin{array}{cc}{2}&{-5}\\{-4}&{-10}\end{array}\right],\:\vec{b}=\left[\begin{array}{c}{0}\\{0}\end{array}\right]\)
- Answer
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\(\vec{x}=x_{2}\left[\begin{array}{c}{2.5}\\{1}\end{array}\right]=x_{2}\vec{v}\)
Figure \(\PageIndex{4}\)