91Ó°ÊÓ

Anaugmented matrix for a system of equations is a matrix of numbers in which eachrowrepresents the constants from one equation, and eachcolumn represents all thecoefficients for a single variable

The augmented matrix \(\left( {\begin{array}{*{20}{c}}A&0\end{array}} \right)\) for the given system of equations \({x_1} + 3{x_2} + {x_3} = 0, - 4{x_1} - 9{x_2} + 2{x_3} = 0\) and \( - 3{x_2} - 6{x_3} = 0\) is represented as:

\(\left[ {\begin{array}{*{20}{c}}1&3&1&0\\{ - 4}&{ - 9}&2&0\\0&{ - 3}&{ - 6}&0\end{array}} \right]\)

Perform an elementary row operation to produce the first augmented matrix.

Perform the sum of \(4\) times row 1 and row 2 at row 2.

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

Perform an elementary row operation to produce the second augmented matrix.

Perform the sum of \(1\) times row 2 and row 3 at row 3.

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

Perform an elementary row operationto produce the third augmented matrix.

Perform the sum of \( - 3\) times row 2 and row 1 at row 1.

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

Perform an elementary row operation to produce the fourth augmented matrix.

Multiply row 2 by \(\frac{1}{3}\).

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

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 \(\left[ {\begin{array}{*{20}{c}}1&0&{ - 5}&0\\0&1&2&0\\0&0&0&0\end{array}} \right]\) into the equation notation.

\(\begin{array}{c}{x_1} - 5{x_3} = 0\\{x_2} + 2{x_3} = 0\end{array}\)

Thus, \({x_1} = 5{x_3},{x_2} = - 2{x_3}\), and \({x_3}\) is a free variable.

The general solution of \(Ax = 0\) in the parametric vector form can be represented as:

\(\begin{array}{c}x = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{5{x_3}}\\{ - 2{x_3}}\\{{x_3}}\end{array}} \right]\\ = {x_3}\left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\1\end{array}} \right]\end{array}\)

Thus, the general solution in the parametric vector form is \(x = {x_3}\left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\1\end{array}} \right].\)