3.3.1: Exercises 3.3
- Page ID
- 70403
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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 \(\PageIndex{1}\) - \(\PageIndex{8}\), find the determinant of the \(2\times 2\) matrix.
\(\left[\begin{array}{cc}{10}&{7}\\{8}&{9}\end{array}\right]\)
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\(34\)
\(\left[\begin{array}{cc}{6}&{-1}\\{-7}&{8}\end{array}\right]\)
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\(41\)
\(\left[\begin{array}{cc}{-1}&{-7}\\{-5}&{9}\end{array}\right]\)
- Answer
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\(-44\)
\(\left[\begin{array}{cc}{-10}&{-1}\\{-4}&{7}\end{array}\right]\)
- Answer
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\(-74\)
\(\left[\begin{array}{cc}{8}&{10}\\{2}&{-3}\end{array}\right]\)
- Answer
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\(-44\)
\(\left[\begin{array}{cc}{10}&{-10}\\{-10}&{0}\end{array}\right]\)
- Answer
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\(-100\)
\(\left[\begin{array}{cc}{1}&{-3}\\{7}&{7}\end{array}\right]\)
- Answer
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\(28\)
\(\left[\begin{array}{cc}{-4}&{-5}\\{-1}&{-4}\end{array}\right]\)
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\(11\)
In Exercises \(\PageIndex{9}\) - \(\PageIndex{12}\), a matrix \(A\) is given.
- Construct the submatrices used to compute the minors \(A_{1,1}\), \(A_{1,2}\), and \(A_{1,3}\).
- Find the cofactors \(C_{1,1}\), \(C_{1,2}\), and \(C_{1,3}\).
\(\left[\begin{array}{ccc}{-7}&{-3}&{10}\\{3}&{7}&{6}\\{1}&{6}&{10}\end{array}\right]\)
- Answer
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- The submatrices are \(\left[\begin{array}{cc}{7}&{6}\\{6}&{10}\end{array}\right]\), \(\left[\begin{array}{cc}{3}&{6}\\{1}&{10}\end{array}\right]\), and \(\left[\begin{array}{cc}{3}&{7}\\{1}&{6}\end{array}\right]\), respectively.
- \(C_{1,2} = 34\), \(C_{1,2} = −24\), \(C_{1,3} = 11\)
\(\left[\begin{array}{ccc}{-2}&{-9}&{6}\\{-10}&{-6}&{8}\\{0}&{-3}&{-2}\end{array}\right]\)
- Answer
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- The submatrices are \(\left[\begin{array}{cc}{-6}&{8}\\{-3}&{-2}\end{array}\right]\), \(\left[\begin{array}{cc}{-10}&{8}\\{0}&{-2}\end{array}\right]\), and \(\left[\begin{array}{cc}{10}&{-6}\\{0}&{-3}\end{array}\right]\), respectively.
- \(C_{1,2} = 36\), \(C_{1,2} = −20\), \(C_{1,3} = -30\)
\(\left[\begin{array}{ccc}{-5}&{-3}&{3}\\{-3}&{3}&{10}\\{-9}&{3}&{9}\end{array}\right]\)
- Answer
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- The submatrices are \(\left[\begin{array}{cc}{3}&{10}\\{3}&{9}\end{array}\right]\), \(\left[\begin{array}{cc}{-3}&{10}\\{-9}&{9}\end{array}\right]\), and \(\left[\begin{array}{cc}{-3}&{3}\\{-9}&{3}\end{array}\right]\), respectively.
- \(C_{1,2} = -3\), \(C_{1,2} = −63\), \(C_{1,3} = 18\)
\(\left[\begin{array}{ccc}{-6}&{-4}&{6}\\{-8}&{0}&{0}\\{-10}&{8}&{-1}\end{array}\right]\)
- Answer
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- The submatrices are \(\left[\begin{array}{ccc}{-6}&{-4}&{6}\\{-8}&{0}&{0}\\{-10}&{8}&{-1}\end{array}\right]\), \(\left[\begin{array}{cc}{-8}&{0}\\{-10}&{-1}\end{array}\right]\), and \(\left[\begin{array}{cc}{-8}&{0}\\{-10}&{8}\end{array}\right]\), respectively.
- \(C_{1,2} = 0\), \(C_{1,2} = −8\), \(C_{1,3} = -64\)
In Exercises \(\PageIndex{13}\) – \(\PageIndex{24}\), find the determinant of the given matrix using cofactor expansion along the first row.
\(\left[\begin{array}{ccc}{3}&{2}&{3}\\{-6}&{1}&{-10}\\{-8}&{-9}&{-9}\end{array}\right]\)
- Answer
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\(-59\)
\(\left[\begin{array}{ccc}{8}&{-9}&{-2}\\{-9}&{9}&{-7}\\{5}&{-1}&{9}\end{array}\right]\)
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\(250\)
\(\left[\begin{array}{ccc}{-4}&{3}&{-4}\\{-4}&{-5}&{3}\\{3}&{-4}&{5}\end{array}\right]\)
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\(15\)
\(\left[\begin{array}{ccc}{1}&{-2}&{1}\\{5}&{5}&{4}\\{4}&{0}&{0}\end{array}\right]\)
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\(-52\)
\(\left[\begin{array}{ccc}{1}&{-4}&{1}\\{0}&{3}&{0}\\{1}&{2}&{2}\end{array}\right]\)
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\(3\)
\(\left[\begin{array}{ccc}{3}&{-1}&{0}\\{-3}&{0}&{-4}\\{0}&{-1}&{-4}\end{array}\right]\)
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\(0\)
\(\left[\begin{array}{ccc}{-5}&{0}&{-4}\\{2}&{4}&{-1}\\{-5}&{0}&{-4}\end{array}\right]\)
- Answer
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\(0\)
\(\left[\begin{array}{ccc}{1}&{0}&{0}\\{0}&{1}&{0}\\{-1}&{1}&{1}\end{array}\right]\)
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\(1\)
\(\left[\begin{array}{cccc}{0}&{0}&{-1}&{-1}\\{1}&{1}&{0}&{1}\\{1}&{1}&{-1}&{0}\\{-1}&{0}&{1}&{0}\end{array}\right]\)
- Answer
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\(0\)
\(\left[\begin{array}{cccc}{-1}&{0}&{0}&{-1}\\{-1}&{0}&{0}&{1}\\{1}&{1}&{1}&{0}\\{1}&{0}&{-1}&{-1}\end{array}\right]\)
- Answer
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\(2\)
\(\left[\begin{array}{cccc}{-5}&{1}&{0}&{0}\\{-3}&{-5}&{2}&{5}\\{-2}&{4}&{-3}&{4}\\{5}&{4}&{-3}&{3}\end{array}\right]\)
- Answer
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\(-113\)
\(\left[\begin{array}{cccc}{2}&{-1}&{4}&{4}\\{3}&{-3}&{3}&{2}\\{0}&{4}&{-5}&{1}\\{-2}&{-5}&{-2}&{-5}\end{array}\right]\)
- Answer
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\(179\)
Let \(A\) be a \(2\times 2\) matrix;
\[A=\left[\begin{array}{cc}{a}&{b}\\{c}&{d}\end{array}\right]. \nonumber \]
Show why \(\text{det}(A)=ad-bc\) by computing the cofactor expansion of \(A\) along the first row.
- Answer
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Hint: \(C_{1,1}=d\).