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.

4. Find the point of intersection of the lines \({x_1} - 5{x_2} = 1\) and on the line \(3{x_1} - 7{x_2} = 5\).

Short answer

The point of intersection is \(\left( {{x_1},{x_2}} \right) \equiv \left( {\frac{9}{4},\frac{1}{4}} \right)\).

Step-by-step solution

Convert the given system of equations into an augmented matrix

To express a system in theaugmented 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 \({x_1} - 5{x_2} = 1\) and \(3{x_1} - 7{x_2} = 5\) is represented as

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

Apply row operation

To get the point of intersection of the system of equations, you are required to eliminate one of the variables, \({x_1}\) or \({x_2}\).

Use the \({x_1}\) term in the first equation to eliminate the \(3{x_1}\) term from the second equation.

A basic principle states that row operations do not affect the solution set of a linear system. Perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&{ - 5}&1\\3&{ - 7}&5\end{aligned}} \right)\) as shown below.

Add \( - 3\) times row 1 to row 2; i.e., \({R_2} \to {R_2} - 3{R_1}\).

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

After the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 5}&1\\0&8&2\end{aligned}} \right)\)

Apply row operation

To obtain 1 as the coefficient of \({x_2}\), perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&{ - 5}&1\\0&8&2\end{aligned}} \right)\) as shown below.

Multiply row 2 by \(\frac{1}{8};\) i.e., \({R_2} \to \frac{1}{8}{R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 5}&1\\{\frac{0}{8}}&{\frac{8}{8}}&{\frac{2}{8}}\end{aligned}} \right)\)

After the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 5}&1\\0&1&{\frac{1}{4}}\end{aligned}} \right)\)

Apply row operation

Use the \({x_1}\) term in the second equation to eliminate the \( - 5{x_2}\) term from the first equation. Perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&{ - 5}&1\\0&1&{\frac{1}{4}}\end{aligned}} \right)\) as shown below.

Add 5 times row 2 to row 1; i.e., \({R_1} \to {R_1} + 5{R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}{1 + 5\left( 0 \right)}&{ - 5 + 5\left( 1 \right)}&{1 + 5\left( {\frac{1}{4}} \right)}\\0&1&{\frac{1}{4}}\end{aligned}} \right)\)

After the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}1&0&{\frac{9}{4}}\\0&1&{\frac{1}{4}}\end{aligned}} \right)\)

Convert the matrix into the equation

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

Write the obtained matrix \(\left( {\begin{aligned}{*{20}{c}}1&0&{\frac{9}{4}}\\0&1&{\frac{1}{4}}\end{aligned}} \right)\)into the equation notation:

\(\begin{aligned}{c}{x_1} + 0\left( {{x_2}} \right) = \frac{9}{4}\\0\left( {{x_1}} \right) + {x_2} = \frac{1}{4}\end{aligned}\)

Obtain the solution of the system of equations

Now, obtain the point of intersection of the system of equations by equating \({x_1}\) to \(\frac{9}{4}\) and \({x_2}\) to \(\frac{1}{4}\):

\(\begin{aligned}{c}{x_1} = \frac{9}{4}\\{x_2} = \frac{1}{4}\end{aligned}\)

Thus, the point of intersection is \(\left( {{x_1},{x_2}} \right) \equiv \left( {\frac{9}{4},\frac{1}{4}} \right).\)