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: 1

Answer

a) $Rank A=2$ b) $dim.Nul.A=2$ c) $Basis\,for\, Col.A\,=\begin{bmatrix}1\\-1\\5\end{bmatrix},\begin{bmatrix}-4\\2\\-6\end{bmatrix}$ d) $Basis\,for\,Row.A=\begin{pmatrix}[1&0& -1&5] , [0&-2&5&-6]\end{pmatrix}$ e) $Basis\,for\,Nul.A=\begin{bmatrix}1\\\frac{5}{2}\\1\\0\end{bmatrix},\begin{bmatrix}-5\\-3\\0\\1\end{bmatrix}$

Work Step by Step

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