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 1.3 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 5E
Chapter:
Problem:
Write a system of equations that is equivalent to the given vector equation...
Step-by-Step Solution
Step 1
Given System of Equations in vector form:
\[{x_1}\left[ {\begin{array}{*{20}{c}}6\\{ - 1}\\5\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}{ - 3}\\4\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\]We have to convert the vector form form into equivalent system of equations
Step 2: Simplify
Use the definitions of scalar multiplication and vector addition to rewrite the vector equation. Multiply the scalar with vector
\[\begin{array}{l}{x_1}\left[ {\begin{array}{*{20}{c}}6\\{ - 1}\\5\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}{ - 3}\\4\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\\\\\left[ {\begin{array}{*{20}{c}}{6 \times {x_1}}\\{ - 1 \times {x_1}}\\{5 \times {x_1}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 3 \times {x_2}}\\{4 \times {x_2}}\\{0 \times {x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\\\\\left[ {\begin{array}{*{20}{c}}{6{x_1}}\\{ - {x_1}}\\{5{x_1}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 3{x_2}}\\{4{x_2}}\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\end{array}\]
Step 3: Add the two vectors
\[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{6{x_1}}\\{ - {x_1}}\\{5{x_1}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 3{x_2}}\\{4{x_2}}\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\\\\\left[ {\begin{array}{*{20}{c}}{6{x_1} + \left( { - 3{x_2}} \right)}\\{ - {x_1} + 4{x_2}}\\{5{x_1} + 0}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\\\\\left[ {\begin{array}{*{20}{c}}{6{x_1} - 3{x_2}}\\{ - {x_1} + 4{x_2}}\\{5{x_1}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\end{array}\]
Step 4: ANSWER
The pivot elements in the original matrix are marked as shown below. \[\left\{ \begin{array}{l}6{x_1} - 3{x_2} = 1\\ - {x_1} + 4{x_2} = - 7\\5{x_1} = - 5\end{array} \right.\]
Given System of Equations in vector form:
\[{x_1}\left[ {\begin{array}{*{20}{c}}6\\{ - 1}\\5\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}{ - 3}\\4\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\]We have to convert the vector form form into equivalent system of equations
Step 2: Simplify
Use the definitions of scalar multiplication and vector addition to rewrite the vector equation. Multiply the scalar with vector
\[\begin{array}{l}{x_1}\left[ {\begin{array}{*{20}{c}}6\\{ - 1}\\5\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}{ - 3}\\4\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\\\\\left[ {\begin{array}{*{20}{c}}{6 \times {x_1}}\\{ - 1 \times {x_1}}\\{5 \times {x_1}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 3 \times {x_2}}\\{4 \times {x_2}}\\{0 \times {x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\\\\\left[ {\begin{array}{*{20}{c}}{6{x_1}}\\{ - {x_1}}\\{5{x_1}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 3{x_2}}\\{4{x_2}}\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\end{array}\]
Step 3: Add the two vectors
\[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{6{x_1}}\\{ - {x_1}}\\{5{x_1}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 3{x_2}}\\{4{x_2}}\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\\\\\left[ {\begin{array}{*{20}{c}}{6{x_1} + \left( { - 3{x_2}} \right)}\\{ - {x_1} + 4{x_2}}\\{5{x_1} + 0}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\\\\\left[ {\begin{array}{*{20}{c}}{6{x_1} - 3{x_2}}\\{ - {x_1} + 4{x_2}}\\{5{x_1}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\end{array}\]
Step 4: ANSWER
The pivot elements in the original matrix are marked as shown below. \[\left\{ \begin{array}{l}6{x_1} - 3{x_2} = 1\\ - {x_1} + 4{x_2} = - 7\\5{x_1} = - 5\end{array} \right.\]