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Q10E
Expert-verifiedFind 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 \(\left\{ \begin{array}{l}{x_1} = - 4 + 2{x_2}\\{x_2}{\rm{ is free}}\\{x_3} = - 7\end{array} \right.\).
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.
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.
\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 1}&3\\0&0&1&{ - 7}\end{array}} \right]\)
To eliminate the \( - {x_3}\) term from the first equation, add rows 1 and 2.
\(\left( {\begin{array}{*{20}{c}}1&{ - 2}&0&{ - 4}\\0&0&1&{ - 7}\end{array}} \right)\)
Mark the nonzero leading entries in columns 1 and 3.
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.
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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