Question

Find the general solutions of the systems whose augmented matrices are given as

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

Short answer

The general solution of the system is

\({x_1} = 5 + 7{x_2} - 6{x_4}\)

\({x_2}\)is free.

\({x_3} = - 3 + 2{x_4}\)

\({x_4}\) is free.

Step-by-step solution

Apply row operation

A basic principle states that row operations do not affect the solution set of a linear system. Perform an elementary row operation to produce the first augmented matrix.

Replace row \(3\) by adding row 1 to row 3.

\(\left[ {\begin{array}{*{20}{c}}1&{ - 7}&0&6&5\\0&0&1&{ - 2}&{ - 3}\\0&0&{ - 4}&8&{12}\end{array}} \right]\)

Apply row operation

Perform an elementary row operationto produce a second augmented matrix.

Replace row 3 by adding\(4\)times row 2 to row 3.

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

Convert the matrix into the equation

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

Write the obtained matrix into the equation notation.

\(\begin{array}{l}{x_1} - 7{x_2} + 6{x_4} = 5\\{x_3} - 2{x_4} = - 3\end{array}\)

Thus, the general solution of the system is

\({x_1} = 5 + 7{x_2} - 6{x_4}\)

\({x_2}\)is free.

\({x_3} = - 3 + 2{x_4}\)

\({x_4}\) is free.