Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 4 - Vector Spaces - 4.6 Exercises - Page 238: 4

Answer

a) $Rank A=3, $ b)$dim.Nul A=3$ c)$Basis\,for\, Col.A\,=\begin{bmatrix}1\\1\\1\\1\\1 \end{bmatrix},\begin{bmatrix}1\\2\\{ - 1}\\{ - 3}\\{ - 2} \end{bmatrix},\begin{bmatrix}7\\{10}\\1\\{ - 5}\\0 \end{bmatrix}$ d) $Basis\,for\,Row A=\begin{pmatrix}[1&1&-3&7&9&-9] , [0&1&-1&3&4&-3],[0&0&0&1&-1&-2]\end{pmatrix}$ e) $Basis\, for\, Nul A=\begin{bmatrix}2 \\1 \\1 \\0 \\0 \\0\end{bmatrix},\begin{bmatrix}-9 \\-7 \\0 \\1 \\1 \\0\end{bmatrix},\begin{bmatrix}-2 \\-3 \\0 \\2 \\0 \\1\end{bmatrix}$

Work Step by Step

Taking the matrix A as row equivalent to matrix B; $A=\begin{bmatrix}1&1&{ - 3}&7&9&{ - 9}\\1&2&{ - 4}&{10}&{13}&{ - 12}\\1&{ - 1}&{ - 1}&1&1&{ - 3}\\1&{ - 3}&1&{ - 5}&{ - 7}&3\\1&{ - 2}&0&0&{ - 5}&{ - 4} \end{bmatrix}$ $B =\begin{bmatrix}1&1&{ - 3}&7&9&{ - 9}\\0&1&{ - 1}&3&4&{ - 3}\\0&0&0&1&{ - 1}&{ - 2}\\0&0&0&0&0&0\\0&0&0&0&0&0 \end{bmatrix}$ a) $Rank A=3, $It is he number of pivot columns in B and A b)$.dim.Nul A=3$ $dim. Nul A=n-Rank A$ but , $n=6$ $dim.Nul A=6-3=3$ c)$Basis\,for\, Col.A\,=\begin{bmatrix}1\\1\\1\\1\\1 \end{bmatrix},\begin{bmatrix}1\\2\\{ - 1}\\{ - 3}\\{ - 2} \end{bmatrix},\begin{bmatrix}7\\{10}\\1\\{ - 5}\\0 \end{bmatrix}$ Basis for Col.A=the pivot columns of matrix A as seen in Matrix B. the pivotal columns in B are column 1 , Column 2, and Column 4. d) $Basis\,for\,Row A=\begin{pmatrix}[1&1&-3&7&9&-9] , [0&1&-1&3&4&-3],[0&0&0&1&-1&-2]\end{pmatrix}$ The basis for Row A=The non-zero rows as given by Matrix B e)writing matrix B in equation form as $\begin{pmatrix}x_{1}-2 x_{3}+9 x_{5}+2 x_{6}=0 \\x_{2}-x_{3}+7 x_{5}+3 x_{6}=0 \\x_{4}-x_{5}-2 x_{6}=0\end{pmatrix}$ with respect to the free variables $x_{3},x_{5},x_{6}$: $\begin{pmatrix}{{x_1} = 2{x_3} - 9{x_5} - 2{x_6}}\\{{x_2} = {x_3} - 7{x_5} - 3{x_6}}\\{{x_3} = {x_3}}\\{{x_4} = {x_5} + 2{x_6}}\\{{x_5} = {x_5}}\\{{x_6} = {x_6}}\end{pmatrix}$ we can write the solution in parametric form; $x_{3}\begin{bmatrix}2 \\1 \\1 \\0 \\0 \\0 \end{bmatrix}+x_{5}\begin{bmatrix}-9 \\-7 \\0 \\1 \\1 \\0\end{bmatrix}+x_{6}\begin{bmatrix}-2 \\-3 \\0 \\2 \\0 \\1\end{bmatrix}$ $Basis\, for\, Nul A=\begin{bmatrix}2 \\1 \\1 \\0 \\0 \\0\end{bmatrix},\begin{bmatrix}-9 \\-7 \\0 \\1 \\1 \\0\end{bmatrix},\begin{bmatrix}-2 \\-3 \\0 \\2 \\0 \\1\end{bmatrix}$
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