91Ó°ÊÓ

The multiplication of matrix\(A\)of the order\(m \times n\)and vector x gives a new vector defined as\(A{\bf{x}}\)or b.

This concept is defined by the rule of transformation \(T\left( {\bf{x}} \right)\). The matrix transformation is denoted as \({\bf{x}}| \to A{\bf{x}}\).

Consider the transformation\(T\left( {\bf{x}} \right) = A{\bf{x}} = 0\).

So,\(T\left( {\bf{x}} \right) = A{\bf{x}} = 0\)can be represented as shown below:

\(\left[ {\begin{array}{*{20}{c}}1&{ - 4}&7&{ - 5}\\0&1&{ - 4}&3\\2&{ - 6}&6&{ - 4}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\)

Write the augmented matrix\(\left[ {\begin{array}{*{20}{c}}A&0\end{array}} \right]\)as shown below:

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

Use the \({x_1}\) term from the first equation to eliminate the \(2{x_1}\) term from the third equation. Add \( - 2\) times row one to row three.

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

Use the \({x_2}\) term from the second equation to eliminate the \(2{x_2}\) term from the third equation. Add \( - 2\) times row two to row three.

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

Use the \({x_2}\) term from the second equation to eliminate the \( - 4{x_2}\) term from the first equation. Add 4 times row two to row one.

\(\left[ {\begin{array}{*{20}{c}}1&{ - 4}&7&{ - 5}&0\\0&1&{ - 4}&3&0\\0&0&0&0&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&{ - 9}&7&0\\0&1&{ - 4}&3&0\\0&0&0&0&0\end{array}} \right]\)

To obtain the solution, convert the augmented matrix into the system of equations.

Write the obtained matrix, \(\left[ {\begin{array}{*{20}{c}}1&0&{ - 9}&7&0\\0&1&{ - 4}&3&0\\0&0&0&0&0\end{array}} \right]\),in the equation notation.

\(\begin{array}{c}{x_1} + 0\left( {{x_2}} \right) - 9{x_3} + 7{x_4} = 0\\0\left( {{x_1}} \right) + {x_2} - 4{x_3} + 3{x_4} = 0\end{array}\)

From the above equations, \({x_1}\) and \({x_2}\) are the pivot positions. So, \({x_1}\) and \({x_2}\) are basic variables, and \({x_3}\) and \({x_4}\) are free variables.

Let, \({x_3} = s\) and \({x_4} = t\).

Substitute the values \({x_3} = s\) and \({x_4} = t\) in the equation \({x_1} - 9{x_3} + 7{x_4} = 0\) to obtain the general solution.

\(\begin{aligned}{c}{x_1} - 9\left( s \right) + 7\left( t \right) &= 0\\{x_1} &= 9s - 7t\end{aligned}\)

Substitute the value \({x_3} = s\) and \({x_4} = t\) in the equation \({x_2} - 4{x_3} + 3{x_4} = 0\) to obtain the general solution.

\(\begin{aligned}{c}{x_2} - 4\left( s \right) + 3\left( t \right) &= 0\\{x_2} &= 4s - 3t\end{aligned}\)

Obtain the vector in the parametric form by using \({x_1} = 9s - 7t\), \({x_2} = 4s - 3t\), \({x_3} = s\), and \({x_4} = t\).

\(\begin{aligned}{c}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}{9s - 7t}\\{4s - 3t}\\s\\t\end{array}} \right]\\ &= s\left[ {\begin{array}{*{20}{c}}9\\4\\1\\0\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}{ - 7}\\{ - 3}\\0\\1\end{array}} \right]\end{aligned}\)

Or it can be written as \(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = {x_3}\left[ {\begin{array}{*{20}{c}}9\\4\\1\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}{ - 7}\\{ - 3}\\0\\1\end{array}} \right]\).

So, the solution in the parametric vector form is \(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = {x_3}\left[ {\begin{array}{*{20}{c}}9\\4\\1\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}{ - 7}\\{ - 3}\\0\\1\end{array}} \right]\).