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Q1E

Expert-verifiedFound in: Page 1

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**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.**

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

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

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 operation **on 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 operation **on 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 operation **on 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 \({x_2} = 3\).

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