Question

In Exercises 7-12, describe all solutions of \(Ax = 0\) in parametric vector form, where \(A\) is row equivalent to the given matrix.

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

Short answer

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

Step-by-step solution

Write the matrix in the augmented form

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.

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

Write the augmented matrix in the system of equations

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

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

Separate the variables into free and basic types

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

Obtain the value of basic variables in parametric forms

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.

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

Write the solution in the parametric form

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