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 1.5 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 5E
Chapter:
Problem:
Follow the method of Examples to write the solution set of the given homogeneous system in parametric vector form...
Step-by-Step Solution
Step 1
Given system of Equations:
\[\begin{array}{l}{x_1} + 3{x_2} + {x_3} = 0\\\\ - 4{x_1} - 9{x_2} + 2{x_3} = 0\\\\ - 3{x_2} - 6{x_3} = 0\end{array}\]We have to solve the given system of equations and write the solution in parametric vector form.
Step 2: Augmented matrix form
We simply merge the coefficient matrix A and constant vector $\bf{b}$ to find the augmented matrix.\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}A&{\bf{b}}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\{ - 4}&{ - 9}&2&0\\0&{ - 3}&{ - 6}&0\end{array}} \right]\end{array}\]
Step 3: ${R_2} \to {R_2} + 4{R_1}$
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\{ - 4}&{ - 9}&2&0\\0&{ - 3}&{ - 6}&0\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\{ - 4 + 4}&{ - 9 + 4\left( 3 \right)}&{2 + 4\left( 1 \right)}&{0 + 4\left( 0 \right)}\\0&{ - 3}&{ - 6}&0\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\0&3&6&0\\0&{ - 3}&{ - 6}&0\end{array}} \right]\end{array}\]
Step 4: ${R_3} \to {R_3} + {R_2}$
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\0&3&6&0\\0&{ - 3}&{ - 6}&0\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\0&3&6&0\\{0 + 0}&{ - 3 + 3}&{ - 6 + 6}&{0 + 0}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\0&3&6&0\\0&0&0&0\end{array}} \right]\end{array}\]
Step 5: Scale Row-2
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\0&3&6&0\\0&0&0&0\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\0&{3/3}&{6/3}&0\\0&0&0&0\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\0&1&2&0\\0&0&0&0\end{array}} \right]\end{array}\]
Step 6: ${R_1} \to {R_1} - 3{R_2}$
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\0&1&2&0\\0&0&0&0\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&{3 - 3}&{1 - 3\left( 2 \right)}&0\\0&1&2&0\\0&0&0&0\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&0&{ - 5}&0\\0&1&2&0\\0&0&0&0\end{array}} \right]\end{array}\]
Step 7: General Solution
Since the last row is free, we can take $x_3$ as a free variable and write solution in terms of $x_3$.\[\begin{array}{l}{x_1} - 5{x_3} = 0 \Rightarrow {x_1} = 5{x_3}\\\\{x_2} + 2{x_3} = 0 \Rightarrow {x_2} = - 2{x_3}\end{array}\]
Step 8: Vector form of Solution
\[\begin{array}{l}{\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}{5{x_3}}\\{ - 2{x_3}}\\{{x_3}}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\1\end{array}} \right]{x_3}\end{array}\]
Step 9: ANSWER
\[{\bf{x}} = \left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\1\end{array}} \right]{x_3}\]
Given system of Equations:
\[\begin{array}{l}{x_1} + 3{x_2} + {x_3} = 0\\\\ - 4{x_1} - 9{x_2} + 2{x_3} = 0\\\\ - 3{x_2} - 6{x_3} = 0\end{array}\]We have to solve the given system of equations and write the solution in parametric vector form.
Step 2: Augmented matrix form
We simply merge the coefficient matrix A and constant vector $\bf{b}$ to find the augmented matrix.\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}A&{\bf{b}}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\{ - 4}&{ - 9}&2&0\\0&{ - 3}&{ - 6}&0\end{array}} \right]\end{array}\]
Step 3: ${R_2} \to {R_2} + 4{R_1}$
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\{ - 4}&{ - 9}&2&0\\0&{ - 3}&{ - 6}&0\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\{ - 4 + 4}&{ - 9 + 4\left( 3 \right)}&{2 + 4\left( 1 \right)}&{0 + 4\left( 0 \right)}\\0&{ - 3}&{ - 6}&0\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\0&3&6&0\\0&{ - 3}&{ - 6}&0\end{array}} \right]\end{array}\]
Step 4: ${R_3} \to {R_3} + {R_2}$
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\0&3&6&0\\0&{ - 3}&{ - 6}&0\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\0&3&6&0\\{0 + 0}&{ - 3 + 3}&{ - 6 + 6}&{0 + 0}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\0&3&6&0\\0&0&0&0\end{array}} \right]\end{array}\]
Step 5: Scale Row-2
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\0&3&6&0\\0&0&0&0\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\0&{3/3}&{6/3}&0\\0&0&0&0\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\0&1&2&0\\0&0&0&0\end{array}} \right]\end{array}\]
Step 6: ${R_1} \to {R_1} - 3{R_2}$
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&3&1&0\\0&1&2&0\\0&0&0&0\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&{3 - 3}&{1 - 3\left( 2 \right)}&0\\0&1&2&0\\0&0&0&0\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&0&{ - 5}&0\\0&1&2&0\\0&0&0&0\end{array}} \right]\end{array}\]
Step 7: General Solution
Since the last row is free, we can take $x_3$ as a free variable and write solution in terms of $x_3$.\[\begin{array}{l}{x_1} - 5{x_3} = 0 \Rightarrow {x_1} = 5{x_3}\\\\{x_2} + 2{x_3} = 0 \Rightarrow {x_2} = - 2{x_3}\end{array}\]
Step 8: Vector form of Solution
\[\begin{array}{l}{\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}{5{x_3}}\\{ - 2{x_3}}\\{{x_3}}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\1\end{array}} \right]{x_3}\end{array}\]
Step 9: ANSWER
\[{\bf{x}} = \left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\1\end{array}} \right]{x_3}\]