Question

Solve each system in Exercises 1–4 by using elementary row operations on the equations or on the augmented matrix. Follow the systematic elimination procedure.

1. \(\begin{aligned}{c}{x_1} + 5{x_2} = 7\\ - 2{x_1} - 7{x_2} = - 5\end{aligned}\)

Short answer

The values are \({x_1} = - 8\) and\({x_2} = 3.\)

Step-by-step solution

Convert the given system into an augmented matrix

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:

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

Apply an elementary row operation

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

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

Apply an elementary row operation

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

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

Apply an elementary row operation

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

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

Convert the matrix into the equation

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:

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

Obtain the solution of the system of equations

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 \({x_2} = 3\).