Question

Consider each matrix in Exercises 5 and 6 as the augmented matrix of a linear system. State in words the next two elementary row operations that should be performed in the process of solving the system.

5. \(\left( {\begin{aligned}{*{20}{c}}1&{ - 4}&5&0&7\\0&1&{ - 3}&0&6\\0&0&1&0&2\\0&0&0&1&{ - 5}\end{aligned}} \right)\)

Short answer

The first row operation should replace the second row with its sum with 3 times the third row; i.e., \({R_2} \to {R_2} + 3{R_3}\), and the second row operation should replace the first row with its sum with –5 times the third row; i.e., \({R_1} \to {R_1} - 5{R_3}\).

Step-by-step solution

Description of the given augmented matrix

The given augmented matrix of a linear system is provided below:

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 4}&5&0&7\\0&1&{ - 3}&0&6\\0&0&1&0&2\\0&0&0&1&{ - 5}\end{aligned}} \right)\)

Since the above matrix has five columns, you can conclude that the given augmented matrix of a linear system consists of four unknown variables, say\({x_1},\,\,{x_2},\,\,{x_3}\,,\)and \({x_4}\). To obtain the solution of the system, you need to convert the given augmented matrix into the upper or lowertriangular matrix.

The system is already in a “triangular” form.

Elementary row operation 1

From the matrix, the fourth equation does not contain \({x_1}\), \({x_2}\), and \({x_3}\) variables. Moreover, the third equation does not contain \({x_1}\), \({x_2}\), and \({x_4}\) variables. So, from the third equation, the value of \({x_3}\) can be obtained, and from the fourth equation, the value of \({x_4}\) can be obtained.

Thus, the third equation can be written as \({x_3} = 2\). The fourth equation can be written as \({x_4} = - 5\).

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

Use the \({x_3}\) term in the third equation to eliminate the variable \( - 3{x_3}\) in the second equation. Perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&{ - 4}&5&0&7\\0&1&{ - 3}&0&6\\0&0&1&0&2\\0&0&0&1&{ - 5}\end{aligned}} \right)\) as shown below.

The first row operation should replace the second row with its sum with 3 times the third row; i.e., \({R_2} \to {R_2} + 3{R_3}\).

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 4}&5&0&7\\{0 + 3\left( 0 \right)}&{1 + 3\left( 0 \right)}&{ - 3 + 3\left( 1 \right)}&{0 + 3\left( 0 \right)}&{6 + 3\left( 2 \right)}\\0&0&1&0&2\\0&0&0&1&{ - 5}\end{aligned}} \right)\)

After performing the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 4}&5&0&7\\0&1&0&0&{12}\\0&0&1&0&2\\0&0&0&1&{ - 5}\end{aligned}} \right)\)

Elementary row operation 2

Use the \({x_3}\) term in the third equation to eliminate the variable \(5{x_3}\) in the first equation. Perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&{ - 4}&5&0&7\\0&1&0&0&{12}\\0&0&1&0&2\\0&0&0&1&{ - 5}\end{aligned}} \right)\) as shown below.

Therefore, the second row operation should replace the first row with its sum with –5 times the third row; i.e., \({R_1} \to {R_1} - 5{R_3}\).

\(\left( {\begin{aligned}{*{20}{c}}{1 - 5\left( 0 \right)}&{ - 4 - 5\left( 0 \right)}&{5 - 5\left( 1 \right)}&{0 - 5\left( 0 \right)}&{7 - 5\left( 2 \right)}\\0&1&0&0&{12}\\0&0&1&0&2\\0&0&0&1&{ - 5}\end{aligned}} \right)\)

After performing the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 4}&0&0&{ - 3}\\0&1&0&0&{12}\\0&0&1&0&2\\0&0&0&1&{ - 5}\end{aligned}} \right)\)

Conclusion

After these two operations, the augmented matrix becomes a triangular matrix, and the required solution can be computed by converting the matrix into equations.

Hence, the two required elementary row operations that should be performed in the process of solving the system are

(i) the first row operation replacing the second row with its sum with 3 times the third row; i.e., \({R_2} \to {R_2} + 3{R_3}\) and

(ii) the second row operation replacing the first row with its sum with –5 times the third row; i.e., \({R_1} \to {R_1} - 5{R_3}\).