91影视

The number of columns in matrix \(A\) should be equal to the number of entries in vector x so that \(A{\bf{x}}\) can be defined.

Consider matrix \(A = \left( {\begin{array}{*{20}{c}}1&5&2&{ - 6}&9&0\\0&0&1&{ - 7}&4&{ - 8}\\0&0&0&0&1&0\\0&0&0&0&0&0\end{array}} \right)\) in the augmented form, \(\left( {\begin{array}{*{20}{c}}A&0\end{array}} \right)\), as shown below:

\(\left( {\begin{array}{*{20}{c}}A&0\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&5&2&{ - 6}&9&0&0\\0&0&1&{ - 7}&4&{ - 8}&0\\0&0&0&0&0&1&0\\0&0&0&0&0&0&0\end{array}} \right)\)

Use \({x_6}\) in the third equation to eliminate \( - 8{x_6}\) from the second equation. Add 8 times row three to row two.

\(\left( {\begin{array}{*{20}{c}}A&0\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&5&2&{ - 6}&9&0&0\\0&0&1&{ - 7}&4&0&0\\0&0&0&0&0&1&0\\0&0&0&0&0&0&0\end{array}} \right)\)

Use \({x_3}\) in the second equation to eliminate \(2{x_3}\) from the first equation. Add \( - 2\) times row two to row one.

\(\left( {\begin{array}{*{20}{c}}A&0\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&5&0&8&1&0&0\\0&0&1&{ - 7}&4&0&0\\0&0&0&0&0&1&0\\0&0&0&0&0&0&0\end{array}} \right)\)

It is given that there are six columns in the given matrix, which means there should be six entries in vector x.

Thus, the equation \(A{\bf{x}} = 0\) can be written as shown below:

\(\begin{array}{c}A{\bf{x}} = 0\\\left( {\begin{array}{*{20}{c}}1&5&0&8&1&0&0\\0&0&1&{ - 7}&4&0&0\\0&0&0&0&0&1&0\\0&0&0&0&0&0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\\{{x_6}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right)\end{array}\)

Write the above matrix equation in the system of equations as shown below:

\(\begin{array}{c}{x_1}\left( {\begin{array}{*{20}{c}}1\\0\\0\\0\end{array}} \right) + {x_2}\left( {\begin{array}{*{20}{c}}5\\0\\0\\0\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}0\\1\\0\\0\end{array}} \right) + {x_4}\left( {\begin{array}{*{20}{c}}8\\{ - 7}\\0\\0\end{array}} \right) + {x_5}\left( {\begin{array}{*{20}{c}}1\\4\\0\\0\end{array}} \right) + {x_6}\left( {\begin{array}{*{20}{c}}0\\0\\1\\0\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{{x_1} + 5{x_2} + \left( 0 \right){x_3} + 8{x_4} + {x_5} + \left( 0 \right){x_6}}\\{\left( 0 \right){x_1} + \left( 0 \right){x_2} + {x_3} - 7{x_4} + 4{x_5} + \left( 0 \right){x_6}}\\{\left( 0 \right){x_1} + \left( 0 \right){x_2} + \left( 0 \right){x_3} + \left( 0 \right){x_4} + \left( 0 \right){x_5} + {x_6}}\\{\left( 0 \right){x_1} + \left( 0 \right){x_2} + \left( 0 \right){x_3} + \left( 0 \right){x_4} + \left( 0 \right){x_5} + \left( 0 \right){x_6}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right)\end{array}\)

So, the system of equations becomes

\(\begin{array}{c}{x_1} + 5{x_2} + 8{x_4} + {x_5} = 0\\{x_3} - 7{x_4} + 4{x_5} = 0\\{x_6} = 0.\end{array}\)

From the above equations, \({x_1}\), \({x_3}\), and \({x_6}\) are the pivot positions. So, \({x_1}\), \({x_3}\), \({x_6}\) are basic variables, and \({x_2}\), \({x_4}\), \({x_5}\) are the free variables.

So, let \({x_2} = a\), \({x_4} = b\), \({x_5} = c\).

Substitute \({x_4} = b\), and \({x_5} = c\) in the equation \({x_3} - 7{x_4} + 4{x_5} = 0\) to obtain the general solution.

\(\begin{array}{c}{x_3} - 7\left( b \right) + 4\left( c \right) = 0\\{x_3} = 7b - 4c\end{array}\)

Substitute \({x_2} = a\), \({x_4} = b\), and \({x_5} = c\) in the equation \({x_1} + 5{x_2} + 8{x_4} + {x_5} = 0\) to obtain the general solution.

\(\begin{array}{c}{x_1} + 5\left( a \right) + 8\left( b \right) + \left( c \right) = 0\\{x_1} = - 5a - 8b - c\end{array}\)

Obtain the vector in the parametric form by using \({x_1} = - 5a - 8b - c\), \({x_2} = a\), \({x_3} = 7b - 4c\), \({x_4} = b\), \({x_5} = c\), and \({x_6} = 0\).

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\\{{x_6}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 5a - 8b - c}\\a\\{7b - 4c}\\b\\c\\0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 5a}\\a\\0\\0\\0\\0\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - 8b}\\0\\{7b}\\b\\0\\0\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - c}\\0\\{ - 4c}\\0\\c\\0\end{array}} \right)\\ = a\left( {\begin{array}{*{20}{c}}{ - 5}\\1\\0\\0\\0\\0\end{array}} \right) + b\left( {\begin{array}{*{20}{c}}{ - 8}\\0\\7\\1\\0\\0\end{array}} \right) + c\left( {\begin{array}{*{20}{c}}{ - 1}\\0\\{ - 4}\\0\\1\\0\end{array}} \right)\end{array}\)

Or it can be written as shown below:

\(\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\\{{x_6}}\end{array}} \right) = {x_2}\left( {\begin{array}{*{20}{c}}{ - 5}\\1\\0\\0\\0\\0\end{array}} \right) + {x_4}\left( {\begin{array}{*{20}{c}}{ - 8}\\0\\7\\1\\0\\0\end{array}} \right) + {x_5}\left( {\begin{array}{*{20}{c}}{ - 1}\\0\\{ - 4}\\0\\1\\0\end{array}} \right)\)

Thus, the solution in the parametric vector form is:\(\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\\{{x_6}}\end{array}} \right) = {x_2}\left( {\begin{array}{*{20}{c}}{ - 5}\\1\\0\\0\\0\\0\end{array}} \right) + {x_4}\left( {\begin{array}{*{20}{c}}{ - 8}\\0\\7\\1\\0\\0\end{array}} \right) + {x_5}\left( {\begin{array}{*{20}{c}}{ - 1}\\0\\{ - 4}\\0\\1\\0\end{array}} \right)\).