2.7.1: Exercises 2.7
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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 \(\PageIndex{1}\) - \(\PageIndex{4}\), matrices \(A\) and \(B\) are given. Compute \((AB)^{-1}\) and \(B^{-1}A^{-1}\).
\(A=\left[\begin{array}{cc}{1}&{2}\\{1}&{1}\end{array}\right],\quad B=\left[\begin{array}{cc}{3}&{5}\\{2}&{5}\end{array}\right]\)
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\((AB)^{-1}=\left[\begin{array}{cc}{-2}&{3}\\{1}&{-1.4}\end{array}\right]\)
\(A=\left[\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}\right],\quad B=\left[\begin{array}{cc}{7}&{1}\\{2}&{1}\end{array}\right]\)
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\((AB)^{-1}=\left[\begin{array}{cc}{-7/10}&{3/10}\\{29/10}&{-11/10}\end{array}\right]\)
\(A=\left[\begin{array}{cc}{2}&{5}\\{3}&{8}\end{array}\right],\quad B=\left[\begin{array}{cc}{1}&{-1}\\{1}&{4}\end{array}\right]\)
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\((AB)^{-1}=\left[\begin{array}{cc}{29/5}&{-18/5}\\{-11/5}&{7/5}\end{array}\right]\)
\(A=\left[\begin{array}{cc}{2}&{4}\\{2}&{5}\end{array}\right],\quad B=\left[\begin{array}{cc}{2}&{2}\\{6}&{5}\end{array}\right]\)
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\((AB)^{-1}=\left[\begin{array}{cc}{-29/4}&{6}\\{17/2}&{-7}\end{array}\right]\)
In Exercises \(\PageIndex{5}\) - \(\PageIndex{8}\), a \(2\times 2\) matrix \(A\) is given. Compute \(A^{-1}\) and \((A^{-1})^{-1}\) using Theorem 2.6.3.
\(A=\left[\begin{array}{cc}{-3}&{5}\\{1}&{-2}\end{array}\right]\)
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\(A^{-1}=\left[\begin{array}{cc}{-2}&{-5}\\{-1}&{-3}\end{array}\right],\quad (A^{-1})^{-1}=\left[\begin{array}{cc}{-3}&{5}\\{1}&{-2}\end{array}\right]\)
\(A=\left[\begin{array}{cc}{3}&{5}\\{2}&{4}\end{array}\right]\)
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\(A^{-1}=\left[\begin{array}{cc}{2}&{-5/2}\\{-1}&{3/2}\end{array}\right],\quad (A^{-1})^{-1}=\left[\begin{array}{cc}{3}&{5}\\{2}&{4}\end{array}\right]\)
\(A=\left[\begin{array}{cc}{2}&{7}\\{1}&{3}\end{array}\right]\)
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\(A^{-1}=\left[\begin{array}{cc}{-3}&{7}\\{1}&{-2}\end{array}\right],\quad (A^{-1})^{-1}=\left[\begin{array}{cc}{2}&{7}\\{1}&{3}\end{array}\right]\)
\(A=\left[\begin{array}{cc}{9}&{0}\\{7}&{9}\end{array}\right]\)
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\(A^{-1}=\left[\begin{array}{cc}{1/9}&{0}\\{-7/81}&{1/9}\end{array}\right],\quad (A^{-1})^{-1}=\left[\begin{array}{cc}{9}&{0}\\{7}&{9}\end{array}\right]\)
Find \(2\times 2\) matrices \(A\) and \(B\) that are each invertible, but \(A + B\) is not.
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Solutions will vary.
Create a random \(6\times 6\) matrix \(A\), then have a calculator or computer compute \(AA^{−1}\). Was the identity matrix returned exactly? Comment on your results.
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Likely some entries that should be 0 will not be exactly 0, but rather very small values
Use a calculator or computer to compute \(AA^{-1}\), where
\[A=\left[\begin{array}{cccc}{1}&{2}&{3}&{4}\\{1}&{4}&{9}&{16}\\{1}&{8}&{27}&{64}\\{1}&{16}&{81}&{256}\end{array}\right]. \nonumber \]
Was the identity matrix returned exactly. Comment on your results.
- Answer
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Likely some entries that should be 0 will not be exactly 0, but rather very small values.