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Q10E

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Linear Algebra and its Applications
Found in: Page 1
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Find the general solutions of the systems whose augmented matrices are given in Exercises 10.

10. \(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 1}&3\\3&{ - 6}&{ - 2}&2\end{array}} \right]\)

The general solutions are \(\left\{ \begin{array}{l}{x_1} = - 4 + 2{x_2}\\{x_2}{\rm{ is free}}\\{x_3} = - 7\end{array} \right.\).

See the step by step solution

Step by Step Solution

Step 1: Identify pivot position

To identify the pivot and the pivot position, observe the leftmost column of the matrix (nonzero column), that is, the pivot column. At the top of this column, 1 is the pivot.

Step 2: Apply row operation

To obtain the pivot position, make the second term of the pivot column 0.

Use \({x_1}\) and \( - 2{x_2}\) terms in the first equation to eliminate \(3{x_1}\) and \( - 6{x_2}\) terms from the second equation. Add \( - 3\) times row 1 to row 2.

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 1}&3\\0&0&1&{ - 7}\end{array}} \right]\)

Step 3: Apply row operation

To eliminate the \( - {x_3}\) term from the first equation, add rows 1 and 2.

\(\left( {\begin{array}{*{20}{c}}1&{ - 2}&0&{ - 4}\\0&0&1&{ - 7}\end{array}} \right)\)

Step 4: Mark the pivot positions in the matrix

Mark the nonzero leading entries in columns 1 and 3.

Step 5: Convert the matrix into the equation

To obtain the general solution of the system of equations, you must convert the augmented matrix into the system of equations.

Write the obtained matrix into the equation notation.

Step 6: Obtain the general solutions of the system of equations

According to the pivot positions in the obtained matrix, \({x_1}\) and \({x_3}\) are the basic variables. Also, \({x_2}\) is the free variable. By using the free variable, the basic variables can be solved.

Thus, the general solutions are shown below:

\(\left\{ \begin{array}{l}{x_1} = - 4 + 2{x_2}\\{x_2}{\rm{ is free}}\\{x_3} = - 7\end{array} \right.\)

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