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 5E from Chapter 1.4 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 5E
Chapter:
Problem:
Use the definition of Ax to write the matrix equation as a vector equation, or vice versa...
Step-by-Step Solution
Step 1
Given Matrix Equations:
\[\left[ {\begin{array}{*{20}{c}}5&1&{ - 8}&4\\{ - 2}&{ - 7}&3&{ - 5}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}5\\{ - 1}\\3\\{ - 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 8}\\{16}\end{array}} \right]\]We have to convert the given matrix form into vector form. The matrix form $A{\bf{x}} = b$ can be converted to corresponding vector form by suitable multiplication of matrices in the left hand side. \[{x_1}{{\bf{v}}_{\bf{1}}} + {x_2}{{\bf{v}}_{\bf{2}}} + {x_3}{{\bf{v}}_{\bf{3}}} + ...{x_n}{{\bf{v}}_{\bf{n}}} = b\]
Step 2
Multiply the matrices\[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}5&1&{ - 8}&4\\{ - 2}&{ - 7}&3&{ - 5}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}5\\{ - 1}\\3\\{ - 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 8}\\{16}\end{array}} \right]\\\\5\left[ {\begin{array}{*{20}{c}}5\\{ - 2}\end{array}} \right] - 1\left[ {\begin{array}{*{20}{c}}1\\{ - 7}\end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}}{ - 8}\\3\end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}}4\\{ - 5}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 8}\\{16}\end{array}} \right]\end{array}\]Here, \[\begin{array}{l}{x_1} = 5,\,\,{x_2} = - 1,\,\,{x_3} = 3\,,\,\,{x_4} = - 2\\\\{{\bf{v}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}5\\{ - 2}\end{array}} \right],\,\,{{\bf{v}}_2} = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\end{array}} \right],\,\,{{\bf{v}}_3} = \left[ {\begin{array}{*{20}{c}}{ - 8}\\3\end{array}} \right],\,{{\bf{v}}_4} = \left[ {\begin{array}{*{20}{c}}4\\{ - 5}\end{array}} \right]\end{array}\]
Step 4: ANSWERS
The Vector Form\[5\left[ {\begin{array}{*{20}{c}}5\\{ - 2}\end{array}} \right] - 1\left[ {\begin{array}{*{20}{c}}1\\{ - 7}\end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}}{ - 8}\\3\end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}}4\\{ - 5}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 8}\\{16}\end{array}} \right]\]
Given Matrix Equations:
\[\left[ {\begin{array}{*{20}{c}}5&1&{ - 8}&4\\{ - 2}&{ - 7}&3&{ - 5}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}5\\{ - 1}\\3\\{ - 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 8}\\{16}\end{array}} \right]\]We have to convert the given matrix form into vector form. The matrix form $A{\bf{x}} = b$ can be converted to corresponding vector form by suitable multiplication of matrices in the left hand side. \[{x_1}{{\bf{v}}_{\bf{1}}} + {x_2}{{\bf{v}}_{\bf{2}}} + {x_3}{{\bf{v}}_{\bf{3}}} + ...{x_n}{{\bf{v}}_{\bf{n}}} = b\]
Step 2
Multiply the matrices\[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}5&1&{ - 8}&4\\{ - 2}&{ - 7}&3&{ - 5}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}5\\{ - 1}\\3\\{ - 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 8}\\{16}\end{array}} \right]\\\\5\left[ {\begin{array}{*{20}{c}}5\\{ - 2}\end{array}} \right] - 1\left[ {\begin{array}{*{20}{c}}1\\{ - 7}\end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}}{ - 8}\\3\end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}}4\\{ - 5}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 8}\\{16}\end{array}} \right]\end{array}\]Here, \[\begin{array}{l}{x_1} = 5,\,\,{x_2} = - 1,\,\,{x_3} = 3\,,\,\,{x_4} = - 2\\\\{{\bf{v}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}5\\{ - 2}\end{array}} \right],\,\,{{\bf{v}}_2} = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\end{array}} \right],\,\,{{\bf{v}}_3} = \left[ {\begin{array}{*{20}{c}}{ - 8}\\3\end{array}} \right],\,{{\bf{v}}_4} = \left[ {\begin{array}{*{20}{c}}4\\{ - 5}\end{array}} \right]\end{array}\]
Step 4: ANSWERS
The Vector Form\[5\left[ {\begin{array}{*{20}{c}}5\\{ - 2}\end{array}} \right] - 1\left[ {\begin{array}{*{20}{c}}1\\{ - 7}\end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}}{ - 8}\\3\end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}}4\\{ - 5}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 8}\\{16}\end{array}} \right]\]