91Ó°ÊÓ

An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant are on the other side of the equal sign), and each column represents all the coefficients for a single variable.

The system of equations in the augmented matrix form \(\left( {\begin{array}{*{20}{c}}A&{\bf{b}}\end{array}} \right)\) is shown below:

\(\left( {\begin{array}{*{20}{c}}A&{\bf{b}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&3&{ - 5}&4\\1&4&{ - 8}&7\\{ - 3}&{ - 7}&9&{ - 6}\end{array}} \right)\)

Use \({x_1}\) in the first equation to eliminate the \({x_1}\) term from the second equation. Add \( - 1\) time row one to row two.

\(\left( {\begin{array}{*{20}{c}}A&{\bf{b}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&3&{ - 5}&4\\0&{ - 1}&3&{ - 3}\\{ - 3}&{ - 7}&9&{ - 6}\end{array}} \right)\)

Use \({x_1}\) in the first equation to eliminate the \( - 3{x_1}\) term from the third equation. Add 3 times row one to row three.

\(\left( {\begin{array}{*{20}{c}}A&{\bf{b}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&3&{ - 5}&4\\0&{ - 1}&3&{ - 3}\\0&2&{ - 6}&6\end{array}} \right)\)

To obtain 1 as the coefficient of \(2{x_2}\) in the third equation, multiply it by \(\frac{1}{2}\).

\(\left( {\begin{array}{*{20}{c}}A&{\bf{b}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&3&{ - 5}&4\\0&{ - 1}&3&{ - 3}\\0&1&{ - 3}&3\end{array}} \right)\)

Use \( - {x_2}\) in the second equation to eliminate the \({x_2}\) term from the third equation. Add row two to row one.

\(\left( {\begin{array}{*{20}{c}}A&{\bf{b}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0&4&{ - 5}\\0&1&{ - 3}&3\\0&0&0&0\end{array}} \right)\)

There are three columns in matrix A, which means there should be three entries in vector x.

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

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

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

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

The system of equations is shown below:

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

From the above equations, \({x_1}\), and \({x_2}\) are the pivot positions. So, \({x_1}\) and \({x_2}\) are basic variables, and \({x_3}\) is a free variable.

Let, \({x_1} = r\), \({x_2} = s\), \({x_3} = t\).

Substitute \({x_3} = t\) in the equation \({x_1} + 4{x_3} = - 5\) to obtain the general solution.

Substitute \({x_3} = t\) in the equation \({x_2} - 3{x_3} = 3\) to obtain the general solution.

\(\begin{array}{c}{x_2} - 3\left( t \right) = 3\\{x_2} = 3 + 3t\end{array}\)

Obtain the vector in the parametric form by using \({x_1} = - 5 - 4t\), \({x_2} = 3 + 3t\), and \({x_3} = t\).

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

Or it can be written as shown below:

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

Thus, the solution in the parametric vector form is \(\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 5}\\3\\0\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}{ - 4}\\3\\1\end{array}} \right)\).

Geometrically, the line passes through the vector \(\left( {\begin{array}{*{20}{c}}{ - 5}\\3\\0\end{array}} \right)\) and is parallel to the system \(\left( {\begin{array}{*{20}{c}}{ - 4}\\3\\1\end{array}} \right)\).

Ifa line passes through vector\({\bf{a}}\)and is parallel to vector b,then the parametric equation of the line for the homogeneous equation is represented as\({\bf{x}} = t{\bf{b}}\), where\(t\)is a parameter.

From Exercise 6, the parametric equation is\(\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = {x_3}\left( {\begin{array}{*{20}{c}}{ - 4}\\3\\1\end{array}} \right)\).

The comparison of the obtained equation with the equation in Exercise 6 shows that the line passes through the vector \(\left( {\begin{array}{*{20}{c}}{ - 5}\\3\\0\end{array}} \right)\) and is parallel to the system.

Thus, the line passes through the vector \(\left( {\begin{array}{*{20}{c}}{ - 5}\\3\\0\end{array}} \right)\) and is parallel to the system in Exercise 5.