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

11.\(\left( {\begin{array}{*{20}{c}}1&{ - 4}&{ - 2}&0&3&{ - 5}\\0&0&1&0&0&{ - 1}\\0&0&0&0&1&{ - 4}\\0&0&0&0&0&0\end{array}} \right)\)

The representation of a column of a matrix is \(A = \left( {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{ \cdot \cdot \cdot }&{{a_n}}\end{array}} \right)\), and the representation of a vector is \({\bf{x}} = \left( {\begin{array}{*{20}{c}}{{x_1}}\\ \vdots \\{{x_n}}\end{array}} \right)\).

According to the definition, the entries in vector x reflect the values in a linear combination of columns of matrix A.

And the product obtained using the row-vector rule is shown below.

\(\begin{array}{c}A{\bf{x}} = \left( {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{ \cdot \cdot \cdot }&{{a_n}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\ \vdots \\{{x_n}}\end{array}} \right)\\ = {x_1}{a_1} + {x_2}{a_2} + \cdots + {x_n}{a_n}\end{array}\)

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.

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:

\(\begin{array}{c}A{\bf{x}} = 0\\\left( {\begin{array}{*{20}{c}}1&{ - 4}&{ - 2}&0&3&{ - 5}\\0&0&1&0&0&{ - 1}\\0&0&0&0&1&{ - 4}\\0&0&0&0&0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\\{{x_6}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right)\end{array}\)

\(\begin{array}{c}A{\bf{x}} = 0\\\left( {\begin{array}{*{20}{c}}1&{ - 4}&{ - 2}&0&3&{ - 5}\\0&0&1&0&0&{ - 1}\\0&0&0&0&1&{ - 4}\\0&0&0&0&0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\\{{x_6}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right)\end{array}\)

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

\(\begin{array}{c}{x_1}\left( {\begin{array}{*{20}{c}}1\\0\\0\\0\end{array}} \right) + {x_2}\left( {\begin{array}{*{20}{c}}{ - 4}\\0\\0\\0\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}{ - 2}\\1\\0\\0\end{array}} \right) + {x_4}\left( {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right) + {x_5}\left( {\begin{array}{*{20}{c}}3\\0\\1\\0\end{array}} \right) + {x_6}\left( {\begin{array}{*{20}{c}}{ - 5}\\{ - 1}\\{ - 4}\\0\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{{x_1} - 4{x_2} - 2{x_3} + \left( 0 \right){x_4} + 3{x_5} - 5{x_6}}\\{\left( 0 \right){x_1} + \left( 0 \right){x_2} + {x_3} + \left( 0 \right){x_4} + \left( 0 \right){x_5} - {x_6}}\\{\left( 0 \right){x_1} + \left( 0 \right){x_2} + \left( 0 \right){x_3} + \left( 0 \right){x_4} + {x_5} - 4{x_6}}\\{\left( 0 \right){x_1} + \left( 0 \right){x_2} + \left( 0 \right){x_3} + \left( 0 \right){x_4} + \left( 0 \right){x_5} + \left( 0 \right){x_6}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right)\end{array}\)

So, the system of equations becomes:

\(\begin{array}{c}{x_1} - 4{x_2} - 2{x_3} + 3{x_5} - 5{x_6} = 0\\{x_3} - {x_6} = 0\\{x_5} - 4{x_6} = 0.\end{array}\)

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

Let \({x_2} = a\), \({x_4} = b\), \({x_6} = c\).

Substitute \({x_6} = c\) in the equation \({x_5} - 4{x_6} = 0\) to obtain the general solution.

\(\begin{array}{c}{x_5} - 4\left( c \right) = 0\\{x_5} = 4c\end{array}\)

Substitute \({x_6} = c\) in the equation \({x_3} - {x_6} = 0\) to obtain the general solution.

\(\begin{array}{c}{x_3} - \left( c \right) = 0\\{x_3} = c\end{array}\)

Substitute \({x_2} = a\) and \({x_6} = c\) in the equation \({x_1} - 4{x_2} - 2{x_3} + 3{x_5} - 5{x_6} = 0\) to obtain the general solution.

\(\begin{array}{c}{x_1} - 4\left( a \right) - 2\left( c \right) + 3\left( {4c} \right) - 5\left( c \right) = 0\\{x_1} - 4a - 2c + 12c - 5c = 0\\{x_1} = 4a - 5c\end{array}\)

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

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

Or it can be written as shown below:

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

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