91Ó°ÊÓ

To express a system in the augmented matrix form, extract the coefficients of the variables and the constants and place these entries in the column of the matrix.

Thus, the augmented matrix for the given system of equations \({x_1} + 5{x_2} = 7\) and \( - 2{x_1} - 7{x_2} = - 5\) is represented as follows:

\(\left( {\begin{aligned}{*{20}{c}}1&5&7\\{ - 2}&{ - 7}&{ - 5}\end{aligned}} \right)\)

A basic principle states that row operations do not affect the solution set of a linear system.

To obtain the solution of the system of equations, you must eliminate one of the variables, \({x_1}\) or\({x_2}\).

Use the \({x_1}\) term in the first equation to eliminate the \( - 2{x_1}\) term from the second equation. Perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&5&7\\{ - 2}&{ - 7}&{ - 5}\end{aligned}} \right)\) as shown below.

Add 2 times the first row to the second row; i.e., \({R_2} \to {R_2} + 2{R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}1&5&7\\{ - 2 + \left( {2 \times 1} \right)}&{ - 7 + \left( {2 \times 5} \right)}&{ - 5 + \left( {2 \times 7} \right)}\end{aligned}} \right)\)

After performing the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}1&5&7\\0&3&9\end{aligned}} \right)\)

To obtain 1 as the coefficient of\({x_2}\), perform an elementary row operationon the matrix\(\left( {\begin{aligned}{*{20}{c}}1&5&7\\0&3&9\end{aligned}} \right)\) as shown below.

Multiply the second row by\(\frac{1}{3};\) i.e., \({R_2} \to \frac{1}{3}{R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}1&5&7\\{\frac{0}{3}}&{\frac{3}{3}}&{\frac{9}{3}}\end{aligned}} \right)\)

After performing the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}1&5&7\\0&1&3\end{aligned}} \right)\)

Use the \({x_2}\) term in the second equation to eliminate the \(5{x_2}\) term from the first equation. Perform an elementary row operationon the matrix\(\left( {\begin{aligned}{*{20}{c}}1&5&7\\0&1&3\end{aligned}} \right)\) as shown below.

Add \( - 5\) times the second row to the first row; i.e., \({R_1} \to {R_1} - 5{R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}{1 - \left( {5 \times 0} \right)}&{5 - \left( {5 \times 1} \right)}&{7 - \left( {5 \times 3} \right)}\\0&1&3\end{aligned}} \right)\)

After performing the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}1&0&{ - 8}\\0&1&3\end{aligned}} \right)\)

Again, you have to convert the augmented matrix into the system of equations to obtain the solution of the system of equations.

Write the obtained matrix \(\left( {\begin{aligned}{*{20}{c}}1&0&{ - 8}\\0&1&3\end{aligned}} \right)\)into the equation notation:

\(\begin{aligned}{c}{x_1} + 0\left( {{x_2}} \right) = - 8\\0\left( {{x_1}} \right) + {x_2} = 3\end{aligned}\)

Now, obtain the solution of the system of equations by equating \({x_1}\) to \( - 8\) and \({x_2}\) to \(3\):

\(\begin{aligned}{c}{x_1} = - 8\\{x_2} = 3\end{aligned}\)

Thus, the required values for the given system of equations are \({x_1} = - 8\) and \({x_2} = 3\).