2.1.1: Ejercicios 2.1
- Page ID
- 116390
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Matrices\(A\) y\(B\) se dan a continuación. En Ejercicios\(\PageIndex{1}\) -\(\PageIndex{6}\), simplificar la expresión dada.
\[A=\left[\begin{array}{cc}{1}&{-1}\\{7}&{4}\end{array}\right]\qquad B=\left[\begin{array}{cc}{-3}&{2}\\{5}&{9}\end{array}\right]\nonumber \]
\(A+B\)
- Contestar
-
\(\left[\begin{array}{cc}{-2}&{1}\\{12}&{13}\end{array}\right]\)
\(2A-3B\)
- Contestar
-
\(\left[\begin{array}{cc}{11}&{-8}\\{-1}&{-19}\end{array}\right]\)
\(3A-A\)
- Contestar
-
\(\left[\begin{array}{cc}{2}&{-2}\\{14}&{8}\end{array}\right]\)
\(4B-2A\)
- Contestar
-
\(\left[\begin{array}{cc}{-14}&{10}\\{6}&{28}\end{array}\right]\)
\(3(A-B)+B\)
- Contestar
-
\(\left[\begin{array}{cc}{9}&{-7}\\{11}&{-6}\end{array}\right]\)
\(2(A-B)-(A-3B)\)
- Contestar
-
\(\left[\begin{array}{cc}{-2}&{1}\\{12}&{13}\end{array}\right]\)
Matrices\(A\) y\(B\) se dan a continuación. En Ejercicios\(\PageIndex{7}\) -\(\PageIndex{10}\), simplificar la expresión dada.
\[A=\left[\begin{array}{c}{3}\\{5}\end{array}\right]\qquad B=\left[\begin{array}{c}{-2}\\{4}\end{array}\right]\nonumber \]
\(4B-2A\)
- Contestar
-
\(\left[\begin{array}{c}{-14}\\{6}\end{array}\right]\)
\(-2A+3B\)
- Contestar
-
\(\left[\begin{array}{c}{-12}\\{2}\end{array}\right]\)
\(-2A-3A\)
- Contestar
-
\(\left[\begin{array}{c}{-15}\\{-25}\end{array}\right]\)
\(-B+3B-2B\)
- Contestar
-
\(\left[\begin{array}{c}{0}\\{0}\end{array}\right]\)
Matrices\(A\) y\(B\) se dan a continuación. En Ejercicios\(\PageIndex{11}\) -\(\PageIndex{14}\), encuentra\(X\) que satisface la ecuación.
\[A=\left[\begin{array}{cc}{3}&{-1}\\{2}&{5}\end{array}\right]\qquad B=\left[\begin{array}{cc}{1}&{7}\\{3}&{-4}\end{array}\right]\nonumber \]
\(2A+X=B\)
- Contestar
-
\(X=\left[\begin{array}{cc}{-5}&{9}\\{-1}&{-14}\end{array}\right]\)
\(A-X=3B\)
- Contestar
-
\(X=\left[\begin{array}{cc}{0}&{-22}\\{-7}&{17}\end{array}\right]\)
\(3A+2X=-1B\)
- Contestar
-
\(X=\left[\begin{array}{cc}{-5}&{-2}\\{-9/2}&{-11/2}\end{array}\right]\)
\(A-\frac{1}{2}X=-B\)
- Contestar
-
\(X=\left[\begin{array}{cc}{8}&{12}\\{10}&{2}\end{array}\right]\)
En Ejercicios\(\PageIndex{15}\) -\(\PageIndex{21}\), encontrar valores para los escalares\(a\) y\(b\) que satisfagan la ecuación dada.
\(a\left[\begin{array}{c}{1}\\{2}\end{array}\right]+b\left[\begin{array}{c}{-1}\\{5}\end{array}\right]=\left[\begin{array}{c}{1}\\{9}\end{array}\right]\)
- Contestar
-
\(a = 2\), \(b = 1\)
\(a\left[\begin{array}{c}{-3}\\{1}\end{array}\right]+b\left[\begin{array}{c}{8}\\{4}\end{array}\right]=\left[\begin{array}{c}{7}\\{1}\end{array}\right]\)
- Contestar
-
\(a = -1\), \(b = 1/2\)
\(a\left[\begin{array}{c}{4}\\{-2}\end{array}\right]+b\left[\begin{array}{c}{-6}\\{3}\end{array}\right]=\left[\begin{array}{c}{10}\\{-5}\end{array}\right]\)
- Contestar
-
\(a = 5/2 + 3/2b\)
\(a\left[\begin{array}{c}{1}\\{1}\end{array}\right]+b\left[\begin{array}{c}{-1}\\{3}\end{array}\right]=\left[\begin{array}{c}{5}\\{5}\end{array}\right]\)
- Contestar
-
\(a = 5\), \(b = 0\)
\(a\left[\begin{array}{c}{1}\\{3}\end{array}\right]+b\left[\begin{array}{c}{-3}\\{-9}\end{array}\right]=\left[\begin{array}{c}{4}\\{-12}\end{array}\right]\)
- Contestar
-
No hay solución.
\(a\left[\begin{array}{c}{1}\\{2}\\{3}\end{array}\right]+b\left[\begin{array}{c}{1}\\{1}\\{2}\end{array}\right]=\left[\begin{array}{c}{0}\\{-1}\\{-1}\end{array}\right]\)
- Contestar
-
\(a=-1\), \(b=1\)
\(a\left[\begin{array}{c}{1}\\{0}\\{1}\end{array}\right]+b\left[\begin{array}{c}{5}\\{1}\\{2}\end{array}\right]=\left[\begin{array}{c}{3}\\{4}\\{7}\end{array}\right]\)
- Contestar
-
No hay solución.