Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
We have solutions for your book!
See our solution for Question 37E from Chapter 1.4 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 37E
Chapter:
Problem:
[M] In Exercises 37–40, determine if the columns of the matrix span ℝ4.
Step-by-Step Solution
Given Information
We are given with a matrix \[ \left[ \begin{array} { r r r r } { 7 } & { 2 } & { - 5 } & { 8 } \\ { - 5 } & { - 3 } & { 4 } & { - 9 } \\ { 6 } & { 10 } & { - 2 } & { 7 } \\ { - 7 } & { 9 } & { 2 } & { 15 } \end{array} \right] \] We have to find out if the columns of the matrix span $R^3$.
Step-1: Row-Reduced echelon form of the matrix
\[\begin{array}{l} A = \left[ {\begin{array}{*{20}{c}} 7&2&{ - 5}&8\\ { - 5}&{ - 3}&4&{ - 9}\\ 6&{10}&{ - 2}&7\\ { - 7}&9&2&{15} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&{\dfrac{2}{7}}&{\dfrac{{ - 5}}{7}}&{\dfrac{8}{7}}\\ { - 5}&{ - 3}&4&{ - 9}\\ 6&{10}&{ - 2}&7\\ { - 7}&9&2&{15} \end{array}} \right]::\,\,\left\{ {{R_1} = \dfrac{{{R_1}}}{7}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{\dfrac{2}{7}}&{\dfrac{{ - 5}}{7}}&{\dfrac{8}{7}}\\ 0&{ - \dfrac{{11}}{7}}&{\dfrac{3}{7}}&{ - \dfrac{{23}}{7}}\\ 0&{\dfrac{{58}}{7}}&{\dfrac{{16}}{7}}&{\dfrac{1}{7}}\\ 0&{11}&{ - 3}&{23} \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_2} = {R_2} + 5{R_1}\\ {R_3} = {R_3} - 6{R_1}\\ {R_4} = {R_4} + 7{R_1} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{\dfrac{2}{7}}&{\dfrac{{ - 5}}{7}}&{\dfrac{8}{7}}\\ 0&1&{ - \dfrac{3}{{11}}}&{\dfrac{{23}}{{11}}}\\ 0&{\dfrac{{58}}{7}}&{\dfrac{{16}}{7}}&{\dfrac{1}{7}}\\ 0&{11}&{ - 3}&{23} \end{array}} \right]::\,\,\,\left\{ {{R_2} = - \dfrac{7}{{11}}{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{r}} 1&0&{ - \dfrac{7}{{11}}}&{\dfrac{6}{{11}}}\\ 0&1&{ - \dfrac{3}{{11}}}&{\dfrac{{23}}{{11}}}\\ 0&{\dfrac{{58}}{7}}&{\dfrac{{16}}{7}}&{\dfrac{1}{7}}\\ 0&{11}&{ - 3}&{23} \end{array}} \right]::\,\,\left\{ {{R_1} = {R_1} - \dfrac{2}{7}{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - \dfrac{7}{{11}}}&{\dfrac{6}{{11}}}\\ 0&1&{ - \dfrac{3}{{11}}}&{\dfrac{{23}}{{11}}}\\ 0&0&{\dfrac{{50}}{{11}}}&{ - \dfrac{{189}}{{11}}}\\ 0&{11}&{ - 3}&{23} \end{array}} \right]::\,\,\left\{ {{R_3} = {R_3} - \dfrac{{58}}{7}{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - \dfrac{7}{{11}}}&{\dfrac{6}{{11}}}\\ 0&1&{ - \dfrac{3}{{11}}}&{\dfrac{{23}}{{11}}}\\ 0&0&{\dfrac{{50}}{{11}}}&{ - \dfrac{{189}}{{11}}}\\ 0&0&0&0 \end{array}} \right]::\,\,\left\{ {{R_4} = {R_4} - 11{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - \dfrac{7}{{11}}}&{\dfrac{6}{{11}}}\\ 0&1&{ - \dfrac{3}{{11}}}&{\dfrac{{23}}{{11}}}\\ 0&0&1&{ - \dfrac{{189}}{{50}}}\\ 0&0&0&0 \end{array}} \right]::\,\left\{ {{R_3} = \dfrac{{11}}{{50}}{R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&0&{ - \dfrac{{93}}{{50}}}\\ 0&1&{ - \dfrac{3}{{11}}}&{\dfrac{{23}}{{11}}}\\ 0&0&1&{ - \dfrac{{189}}{{50}}}\\ 0&0&0&0 \end{array}} \right]::\,\left\{ {{R_1} = {R_1} + \dfrac{7}{{50}}{R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&0&{ - \dfrac{{93}}{{50}}}\\ 0&1&0&{\dfrac{{53}}{{50}}}\\ 0&0&1&{ - \dfrac{{189}}{{50}}}\\ 0&0&0&0 \end{array}} \right]::\,\left\{ {{R_2} = {R_2} + \dfrac{3}{{11}}{R_3}} \right\} \end{array}\] We can see that the last row has no pivot element, Therefore,
The columns o A span fot not span $R^4$
We are given with a matrix \[ \left[ \begin{array} { r r r r } { 7 } & { 2 } & { - 5 } & { 8 } \\ { - 5 } & { - 3 } & { 4 } & { - 9 } \\ { 6 } & { 10 } & { - 2 } & { 7 } \\ { - 7 } & { 9 } & { 2 } & { 15 } \end{array} \right] \] We have to find out if the columns of the matrix span $R^3$.
Step-1: Row-Reduced echelon form of the matrix
\[\begin{array}{l} A = \left[ {\begin{array}{*{20}{c}} 7&2&{ - 5}&8\\ { - 5}&{ - 3}&4&{ - 9}\\ 6&{10}&{ - 2}&7\\ { - 7}&9&2&{15} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&{\dfrac{2}{7}}&{\dfrac{{ - 5}}{7}}&{\dfrac{8}{7}}\\ { - 5}&{ - 3}&4&{ - 9}\\ 6&{10}&{ - 2}&7\\ { - 7}&9&2&{15} \end{array}} \right]::\,\,\left\{ {{R_1} = \dfrac{{{R_1}}}{7}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{\dfrac{2}{7}}&{\dfrac{{ - 5}}{7}}&{\dfrac{8}{7}}\\ 0&{ - \dfrac{{11}}{7}}&{\dfrac{3}{7}}&{ - \dfrac{{23}}{7}}\\ 0&{\dfrac{{58}}{7}}&{\dfrac{{16}}{7}}&{\dfrac{1}{7}}\\ 0&{11}&{ - 3}&{23} \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_2} = {R_2} + 5{R_1}\\ {R_3} = {R_3} - 6{R_1}\\ {R_4} = {R_4} + 7{R_1} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{\dfrac{2}{7}}&{\dfrac{{ - 5}}{7}}&{\dfrac{8}{7}}\\ 0&1&{ - \dfrac{3}{{11}}}&{\dfrac{{23}}{{11}}}\\ 0&{\dfrac{{58}}{7}}&{\dfrac{{16}}{7}}&{\dfrac{1}{7}}\\ 0&{11}&{ - 3}&{23} \end{array}} \right]::\,\,\,\left\{ {{R_2} = - \dfrac{7}{{11}}{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{r}} 1&0&{ - \dfrac{7}{{11}}}&{\dfrac{6}{{11}}}\\ 0&1&{ - \dfrac{3}{{11}}}&{\dfrac{{23}}{{11}}}\\ 0&{\dfrac{{58}}{7}}&{\dfrac{{16}}{7}}&{\dfrac{1}{7}}\\ 0&{11}&{ - 3}&{23} \end{array}} \right]::\,\,\left\{ {{R_1} = {R_1} - \dfrac{2}{7}{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - \dfrac{7}{{11}}}&{\dfrac{6}{{11}}}\\ 0&1&{ - \dfrac{3}{{11}}}&{\dfrac{{23}}{{11}}}\\ 0&0&{\dfrac{{50}}{{11}}}&{ - \dfrac{{189}}{{11}}}\\ 0&{11}&{ - 3}&{23} \end{array}} \right]::\,\,\left\{ {{R_3} = {R_3} - \dfrac{{58}}{7}{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - \dfrac{7}{{11}}}&{\dfrac{6}{{11}}}\\ 0&1&{ - \dfrac{3}{{11}}}&{\dfrac{{23}}{{11}}}\\ 0&0&{\dfrac{{50}}{{11}}}&{ - \dfrac{{189}}{{11}}}\\ 0&0&0&0 \end{array}} \right]::\,\,\left\{ {{R_4} = {R_4} - 11{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&{ - \dfrac{7}{{11}}}&{\dfrac{6}{{11}}}\\ 0&1&{ - \dfrac{3}{{11}}}&{\dfrac{{23}}{{11}}}\\ 0&0&1&{ - \dfrac{{189}}{{50}}}\\ 0&0&0&0 \end{array}} \right]::\,\left\{ {{R_3} = \dfrac{{11}}{{50}}{R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&0&{ - \dfrac{{93}}{{50}}}\\ 0&1&{ - \dfrac{3}{{11}}}&{\dfrac{{23}}{{11}}}\\ 0&0&1&{ - \dfrac{{189}}{{50}}}\\ 0&0&0&0 \end{array}} \right]::\,\left\{ {{R_1} = {R_1} + \dfrac{7}{{50}}{R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&0&{ - \dfrac{{93}}{{50}}}\\ 0&1&0&{\dfrac{{53}}{{50}}}\\ 0&0&1&{ - \dfrac{{189}}{{50}}}\\ 0&0&0&0 \end{array}} \right]::\,\left\{ {{R_2} = {R_2} + \dfrac{3}{{11}}{R_3}} \right\} \end{array}\] We can see that the last row has no pivot element, Therefore,
The columns o A span fot not span $R^4$