3.2.1: Exercises 3.2
- Page ID
- 69562
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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 \(\PageIndex{1}\) - \(\PageIndex{15}\), find the trace of the given matrix.
\(\left[\begin{array}{cc}{1}&{-5}\\{9}&{5}\end{array}\right]\)
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\(6\)
\(\left[\begin{array}{cc}{-3}&{-10}\\{-6}&{4}\end{array}\right]\)
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\(1\)
\(\left[\begin{array}{cc}{7}&{5}\\{-5}&{-4}\end{array}\right]\)
- Answer
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\(3\)
\(\left[\begin{array}{cc}{-6}&{0}\\{-10}&{9}\end{array}\right]\)
- Answer
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\(3\)
\(\left[\begin{array}{ccc}{-4}&{1}&{1}\\{-2}&{0}&{0}\\{-1}&{-2}&{-5}\end{array}\right]\)
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\(-9\)
\(\left[\begin{array}{ccc}{0}&{-3}&{1}\\{5}&{-5}&{5}\\{-4}&{1}&{0}\end{array}\right]\)
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\(-5\)
\(\left[\begin{array}{ccc}{-2}&{-3}&{5}\\{5}&{2}&{0}\\{-1}&{-3}&{1}\end{array}\right]\)
- Answer
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\(1\)
\(\left[\begin{array}{ccc}{4}&{2}&{-1}\\{-4}&{1}&{4}\\{0}&{-5}&{5}\end{array}\right]\)
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\(10\)
\(\left[\begin{array}{ccc}{2}&{6}&{4}\\{-1}&{8}&{-10}\end{array}\right]\)
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Not defined; the matrix must be square.
\(\left[\begin{array}{cc}{6}&{5}\\{2}&{10}\\{3}&{3}\end{array}\right]\)
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Not defined; the matrix must be square.
\(\left[\begin{array}{cccc}{-10}&{6}&{-7}&{-9}\\{-2}&{1}&{6}&{-9}\\{0}&{4}&{-4}&{0}\\{-3}&{-9}&{3}&{-10}\end{array}\right]\)
- Answer
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\(-23\)
\(\left[\begin{array}{cccc}{5}&{2}&{2}&{2}\\{-7}&{4}&{-7}&{-3}\\{9}&{-9}&{-7}&{2}\\{-4}&{8}&{-8}&{-2}\end{array}\right]\)
- Answer
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\(0\)
\(I_{4}\)
- Answer
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\(4\)
\(I_{n}\)
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\(n\)
A matrix \(A\) that is skew symmetric.
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\(0\)
In Exercises \(\PageIndex{16}\) - \(\PageIndex{19}\), verify Theorem 3.2.1 by:
- Showing that \(\text{tr}(A)+\text{tr}(B)=\text{tr}(A+B)\) and
- Showing that \(\text{tr}(AB)=\text{tr}(BA)\).
\(A=\left[\begin{array}{cc}{1}&{-1}\\{9}&{-6}\end{array}\right],\quad B=\left[\begin{array}{cc}{-1}&{0}\\{-6}&{3}\end{array}\right]\)
- Answer
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- \(\text{tr}(A)=-5;\:\text{tr}(B)=-4;\:\text{tr}(A+B)=-9\)
- \(\text{tr}(AB)=23=\text{tr}(BA)\)
\(A=\left[\begin{array}{cc}{0}&{-8}\\{1}&{8}\end{array}\right],\quad B=\left[\begin{array}{cc}{-4}&{5}\\{-4}&{2}\end{array}\right]\)
- Answer
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- \(\text{tr}(A)=8;\:\text{tr}(B)=-2;\:\text{tr}(A+B)=6\)
- \(\text{tr}(AB)=53=\text{tr}(BA)\)
\(A=\left[\begin{array}{ccc}{-8}&{-10}&{10}\\{10}&{5}&{-6}\\{-10}&{1}&{3}\end{array}\right],\quad B=\left[\begin{array}{ccc}{-10}&{-4}&{-3}\\{-4}&{-5}&{4}\\{3}&{7}&{3}\end{array}\right]\)
- Answer
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- \(\text{tr}(A)=0;\:\text{tr}(B)=-12;\:\text{tr}(A+B)=-12\)
- \(\text{tr}(AB)=86=\text{tr}(BA)\)
\(A=\left[\begin{array}{ccc}{-10}&{7}&{5}\\{7}&{7}&{-5}\\{8}&{-9}&{2}\end{array}\right],\quad B=\left[\begin{array}{ccc}{-3}&{-4}&{9}\\{4}&{-1}&{-9}\\{-7}&{-8}&{10}\end{array}\right]\)
- Answer
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- \(\text{tr}(A)=-1;\:\text{tr}(B)=6;\:\text{tr}(A+B)=5\)
- \(\text{tr}(AB)=201=\text{tr}(BA)\)