Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 4E from Chapter 4.6 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 4E
Chapter:
Problem:
0
Step-by-Step Solution
Step 1
Given Matrices:\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{r}}1&1&{ - 3}&7&9&{ - 9}\\1&2&{ - 4}&{10}&{13}&{ - 12}\\1&{ - 1}&{ - 1}&1&1&{ - 3}\\1&{ - 3}&1&{ - 5}&{ - 7}&3\\1&{ - 2}&0&0&{ - 5}&{ - 4}\end{array}} \right],\\\\B = \left[ {\begin{array}{*{20}{r}}1&1&{ - 3}&7&9&{ - 9}\\0&1&{ - 1}&3&4&{ - 3}\\0&0&0&1&{ - 1}&{ - 2}\\0&0&0&0&0&0\\0&0&0&0&0&0\end{array}} \right]\end{array}\]We are given that the matrix A is row equivalent to B. We have to find the following entities:\[\begin{array}{l}\left( a \right)\,\,{\rm{rank}}\,\,A\\\left( b \right)\dim \,{\rm{NulA}}\\\left( c \right){\rm{Basis}}\,({\rm{Col}}\,A)\\\left( d \right){\rm{Basis}}\,({\rm{Row}}\,A)\\\left( e \right){\rm{Basis}}\,({\rm{NulA}})\end{array}\]
Step 2: $\left( a \right)\dim \,{\rm{rank A}}$
The dimension of rank(A) will be the number of pivot columns in A. The number of pivot columns in B are 3 so:\[{\rm{rank}}\,\,A\,\, = 3\]
Step 3: $\left( b \right)\dim \,{\rm{NulA}}$
From the Rank-Nullity Theorem the relation between rank of matrix, dimension of Null space and number of columns is given by:\[{\rm{rank}}\,\,A + \dim \,{\rm{NulA}} = n\]Using this we have:\[\begin{array}{l}\dim \,{\rm{NulA}} = n - {\rm{rank}}\,\,A\\ = 6 - 3\\ = 3\end{array}\]So, dimension of Null space of A is 3.
Step 4: $ \rm{(c)} {\rm{Basis}}\,({\rm{Col}}\,A)$
The basis of column space of A will be the pivot columns of matrix A. The matrix B is:\[B = \left[ {\begin{array}{*{20}{r}}1&1&{ - 3}&7&9&{ - 9}\\0&1&{ - 1}&3&4&{ - 3}\\0&0&0&1&{ - 1}&{ - 2}\\0&0&0&0&0&0\\0&0&0&0&0&0\end{array}} \right]\]As we can see that pivot columns of B are first two columns, thus basis of column space of A will be given by corresponding columns of A.\[{\rm{ Col }}A{\rm{ = }}\left\{ {\left[ {\begin{array}{*{20}{l}}1\\1\\1\\1\\1\end{array}} \right],\left[ {\begin{array}{*{20}{c}}1\\2\\{ - 1}\\{ - 3}\\{ - 2}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}7\\{10}\\1\\{ - 5}\\0\end{array}} \right]} \right\}\]
Step 5: $ \rm{(d)} \ {\rm{Basis}}\,({\rm{Row}}\,A)$
The basis of row space of A will be the pivot rows of matrix B. The matrix B is:\[B = \left[ {\begin{array}{*{20}{r}}1&1&{ - 3}&7&9&{ - 9}\\0&1&{ - 1}&3&4&{ - 3}\\0&0&0&1&{ - 1}&{ - 2}\\0&0&0&0&0&0\\0&0&0&0&0&0\end{array}} \right]\]As we can see that pivot rows of B are first three rows, thus basis of row space of A will be given by the rows of B.\[{\rm{Basis}}{\mkern 1mu} ({\rm{Row}}{\mkern 1mu} A) = \left\{ \begin{array}{l}(1,1, - 3,7,9, - 9),\\(0,1, - 1,3,4, - 3),\\(0,0,0,1, - 1, - 2)\end{array} \right\}\]
Step 6: $ \rm{(e)}{\rm{Basis}}\,({\rm{Nul}}\,A)$
To find the basis for Null space of A, we will solve the system of equations given by $A{\bf{x}} = 0$ (or $B{\bf{x}} = 0$). The solution will be written in parametric form. The basis for Null space will be given by the coefficient vectors of free variables. \[\begin{array}{*{20}{l}}{B{\bf{x}} = 0}\\{\left[ {\begin{array}{*{20}{r}}1&1&{ - 3}&7&9&{ - 9}\\0&1&{ - 1}&3&4&{ - 3}\\0&0&0&1&{ - 1}&{ - 2}\\0&0&0&0&0&0\\0&0&0&0&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\\{{x_6}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\\0\end{array}} \right]}\end{array}\]The Augmented matrix:\[M = \left[ {\begin{array}{*{20}{r}}1&1&{ - 3}&7&9&{ - 9}\\0&1&{ - 1}&3&4&{ - 3}\\0&0&0&1&{ - 1}&{ - 2}\\0&0&0&0&0&0\\0&0&0&0&0&0\end{array}\begin{array}{*{20}{c}}{}\\{}\\{}\\{}\\{}\end{array}\begin{array}{*{20}{c}}0\\0\\0\\0\\0\end{array}} \right]\]
Step 7: System of Equations
The system of equations are
\[\begin{array}{r}x_{1}-2 x_{3}+9 x_{5}+2 x_{6}=0 \\x_{2}-x_{3}+7 x_{5}+3 x_{6}=0 \\x_{4}-x_{5}-2 x_{6}=0\end{array}\]Using $x_3$, $x_5$ and $x_6$ as free variables
\[\begin{array}{*{20}{l}}{{x_1} = 2{x_3} - 9{x_5} - 2{x_6}}\\{{x_2} = {x_3} - 7{x_5} - 3{x_6}}\\{{x_3} = {x_3}}\\{{x_4} = {x_5} + 2{x_6}}\\{{x_5} = {x_5}}\\{{x_6} = {x_6}}\end{array}\]
Step 8: System of Equations
We can write the solution in parametric form:\[\left[\begin{array}{l}x_{1} \\x_{2} \\x_{3} \\x_{4} \\x_{5} \\x_{6}\end{array}\right]=\left[\begin{array}{l}2 \\1 \\1 \\0 \\0 \\0\end{array}\right] x_{3}+\left[\begin{array}{c}-9 \\-7 \\0 \\1 \\1 \\0\end{array}\right] x_{5}+\left[\begin{array}{c}-2 \\-3 \\0 \\2 \\0 \\1\end{array}\right] x_{6}\]
Null Space
So, the Null space is
\[{\bf{NulA}} = \left\{ {\left[ {\begin{array}{*{20}{l}}2\\1\\1\\0\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 9}\\{ - 7}\\0\\1\\1\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 3}\\0\\2\\0\\1\end{array}} \right]} \right\}\]
Given Matrices:\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{r}}1&1&{ - 3}&7&9&{ - 9}\\1&2&{ - 4}&{10}&{13}&{ - 12}\\1&{ - 1}&{ - 1}&1&1&{ - 3}\\1&{ - 3}&1&{ - 5}&{ - 7}&3\\1&{ - 2}&0&0&{ - 5}&{ - 4}\end{array}} \right],\\\\B = \left[ {\begin{array}{*{20}{r}}1&1&{ - 3}&7&9&{ - 9}\\0&1&{ - 1}&3&4&{ - 3}\\0&0&0&1&{ - 1}&{ - 2}\\0&0&0&0&0&0\\0&0&0&0&0&0\end{array}} \right]\end{array}\]We are given that the matrix A is row equivalent to B. We have to find the following entities:\[\begin{array}{l}\left( a \right)\,\,{\rm{rank}}\,\,A\\\left( b \right)\dim \,{\rm{NulA}}\\\left( c \right){\rm{Basis}}\,({\rm{Col}}\,A)\\\left( d \right){\rm{Basis}}\,({\rm{Row}}\,A)\\\left( e \right){\rm{Basis}}\,({\rm{NulA}})\end{array}\]
Step 2: $\left( a \right)\dim \,{\rm{rank A}}$
The dimension of rank(A) will be the number of pivot columns in A. The number of pivot columns in B are 3 so:\[{\rm{rank}}\,\,A\,\, = 3\]
Step 3: $\left( b \right)\dim \,{\rm{NulA}}$
From the Rank-Nullity Theorem the relation between rank of matrix, dimension of Null space and number of columns is given by:\[{\rm{rank}}\,\,A + \dim \,{\rm{NulA}} = n\]Using this we have:\[\begin{array}{l}\dim \,{\rm{NulA}} = n - {\rm{rank}}\,\,A\\ = 6 - 3\\ = 3\end{array}\]So, dimension of Null space of A is 3.
Step 4: $ \rm{(c)} {\rm{Basis}}\,({\rm{Col}}\,A)$
The basis of column space of A will be the pivot columns of matrix A. The matrix B is:\[B = \left[ {\begin{array}{*{20}{r}}1&1&{ - 3}&7&9&{ - 9}\\0&1&{ - 1}&3&4&{ - 3}\\0&0&0&1&{ - 1}&{ - 2}\\0&0&0&0&0&0\\0&0&0&0&0&0\end{array}} \right]\]As we can see that pivot columns of B are first two columns, thus basis of column space of A will be given by corresponding columns of A.\[{\rm{ Col }}A{\rm{ = }}\left\{ {\left[ {\begin{array}{*{20}{l}}1\\1\\1\\1\\1\end{array}} \right],\left[ {\begin{array}{*{20}{c}}1\\2\\{ - 1}\\{ - 3}\\{ - 2}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}7\\{10}\\1\\{ - 5}\\0\end{array}} \right]} \right\}\]
Step 5: $ \rm{(d)} \ {\rm{Basis}}\,({\rm{Row}}\,A)$
The basis of row space of A will be the pivot rows of matrix B. The matrix B is:\[B = \left[ {\begin{array}{*{20}{r}}1&1&{ - 3}&7&9&{ - 9}\\0&1&{ - 1}&3&4&{ - 3}\\0&0&0&1&{ - 1}&{ - 2}\\0&0&0&0&0&0\\0&0&0&0&0&0\end{array}} \right]\]As we can see that pivot rows of B are first three rows, thus basis of row space of A will be given by the rows of B.\[{\rm{Basis}}{\mkern 1mu} ({\rm{Row}}{\mkern 1mu} A) = \left\{ \begin{array}{l}(1,1, - 3,7,9, - 9),\\(0,1, - 1,3,4, - 3),\\(0,0,0,1, - 1, - 2)\end{array} \right\}\]
Step 6: $ \rm{(e)}{\rm{Basis}}\,({\rm{Nul}}\,A)$
To find the basis for Null space of A, we will solve the system of equations given by $A{\bf{x}} = 0$ (or $B{\bf{x}} = 0$). The solution will be written in parametric form. The basis for Null space will be given by the coefficient vectors of free variables. \[\begin{array}{*{20}{l}}{B{\bf{x}} = 0}\\{\left[ {\begin{array}{*{20}{r}}1&1&{ - 3}&7&9&{ - 9}\\0&1&{ - 1}&3&4&{ - 3}\\0&0&0&1&{ - 1}&{ - 2}\\0&0&0&0&0&0\\0&0&0&0&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\\{{x_6}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\\0\end{array}} \right]}\end{array}\]The Augmented matrix:\[M = \left[ {\begin{array}{*{20}{r}}1&1&{ - 3}&7&9&{ - 9}\\0&1&{ - 1}&3&4&{ - 3}\\0&0&0&1&{ - 1}&{ - 2}\\0&0&0&0&0&0\\0&0&0&0&0&0\end{array}\begin{array}{*{20}{c}}{}\\{}\\{}\\{}\\{}\end{array}\begin{array}{*{20}{c}}0\\0\\0\\0\\0\end{array}} \right]\]
Step 7: System of Equations
The system of equations are
\[\begin{array}{r}x_{1}-2 x_{3}+9 x_{5}+2 x_{6}=0 \\x_{2}-x_{3}+7 x_{5}+3 x_{6}=0 \\x_{4}-x_{5}-2 x_{6}=0\end{array}\]Using $x_3$, $x_5$ and $x_6$ as free variables
\[\begin{array}{*{20}{l}}{{x_1} = 2{x_3} - 9{x_5} - 2{x_6}}\\{{x_2} = {x_3} - 7{x_5} - 3{x_6}}\\{{x_3} = {x_3}}\\{{x_4} = {x_5} + 2{x_6}}\\{{x_5} = {x_5}}\\{{x_6} = {x_6}}\end{array}\]
Step 8: System of Equations
We can write the solution in parametric form:\[\left[\begin{array}{l}x_{1} \\x_{2} \\x_{3} \\x_{4} \\x_{5} \\x_{6}\end{array}\right]=\left[\begin{array}{l}2 \\1 \\1 \\0 \\0 \\0\end{array}\right] x_{3}+\left[\begin{array}{c}-9 \\-7 \\0 \\1 \\1 \\0\end{array}\right] x_{5}+\left[\begin{array}{c}-2 \\-3 \\0 \\2 \\0 \\1\end{array}\right] x_{6}\]
Null Space
So, the Null space is
\[{\bf{NulA}} = \left\{ {\left[ {\begin{array}{*{20}{l}}2\\1\\1\\0\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 9}\\{ - 7}\\0\\1\\1\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 3}\\0\\2\\0\\1\end{array}} \right]} \right\}\]