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 transformation rule \(T\left( {\bf{x}} \right)\). 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:

\(\left[ {\begin{array}{*{20}{c}}1&3&9&2\\1&0&3&{ - 4}\\0&1&2&3\\{ - 2}&3&0&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right]\)

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

\(\left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\1&0&3&{ - 4}&0\\0&1&2&3&0\\{ - 2}&3&0&5&0\end{array}} \right]\)

Use the \({x_1}\) term from the first equation to eliminate the \({x_1}\) term from the second equation. Add \( - 1\) times row one to row two.

\(\left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\1&0&3&{ - 4}&0\\0&1&2&3&0\\{ - 2}&3&0&5&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\0&{ - 3}&{ - 6}&{ - 6}&0\\0&1&2&3&0\\{ - 2}&3&0&5&0\end{array}} \right]\)

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

\(\left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\0&{ - 3}&{ - 6}&{ - 6}&0\\0&1&2&3&0\\{ - 2}&3&0&5&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\0&{ - 3}&{ - 6}&{ - 6}&0\\0&1&2&3&0\\0&9&{18}&9&0\end{array}} \right]\)

Interchange rows two and three.

\(\left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\0&{ - 3}&{ - 6}&{ - 6}&0\\0&1&2&3&0\\0&9&{18}&9&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\0&1&2&3&0\\0&{ - 3}&{ - 6}&{ - 6}&0\\0&9&{18}&9&0\end{array}} \right]\)

Use the \({x_2}\) term from the second equation to eliminate the \( - 3{x_2}\) term from the third equation. Add 3 times row two to row three.

\(\left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\0&1&2&3&0\\0&{ - 3}&{ - 6}&{ - 6}&0\\0&9&{18}&9&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\0&1&2&3&0\\0&0&0&3&0\\0&9&{18}&9&0\end{array}} \right]\)

Use the \({x_2}\) term from the second equation to eliminate the \(9{x_2}\) term from the fourth equation. Add \( - 9\) times row two to row four.

\(\left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\0&1&2&3&0\\0&0&0&3&0\\0&9&{18}&9&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\0&1&2&3&0\\0&0&0&3&0\\0&0&0&{ - 18}&0\end{array}} \right]\)

Use the \(2{x_4}\) term from the first equation to eliminate the \( - 18{x_4}\) term from the fourth equation. Add 9 times row two to row four.

\(\left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\0&1&2&3&0\\0&0&0&3&0\\0&0&0&{ - 18}&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&3&0&0\\0&1&2&0&0\\0&0&0&1&0\\0&0&0&0&0\end{array}} \right]\)

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

Write the obtained matrix, \(\left[ {\begin{array}{*{20}{c}}1&0&3&0&0\\0&1&2&0&0\\0&0&0&1&0\\0&0&0&0&0\end{array}} \right]\), in the equation notation.

\(\begin{aligned}{c}{x_1} + 0\left( {{x_2}} \right) + 3{x_3} + 0\left( {{x_4}} \right) &= 0\\0\left( {{x_1}} \right) + {x_2} + 2{x_3} + 0\left( {{x_4}} \right) &= 0\\{x_4} &= 0\end{aligned}\)

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

Let, \({x_3} = s\).

Substitute the value \({x_3} = s\) in the equation \({x_1} + 3{x_3} = 0\) to obtain the general solution.

\(\begin{aligned}{c}{x_1} + 3\left( s \right) &= 0\\{x_1} &= - 3s\end{aligned}\)

Substitute the value \({x_3} = s\) in the equation \({x_2} + 2{x_3} = 0\) to obtain the general solution.

\(\begin{aligned}{c}{x_2} + 2\left( s \right) &= 0\\{x_2} &= - 2s\end{aligned}\)

Obtain the vector in the parametric form by using \({x_1} = - 3s\), \({x_2} = - 2s\), \({x_3} = s\), and \({x_4} = 0\).

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

Or it can be written as \(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = {x_3}\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 2}\\1\\0\end{array}} \right]\).

So, the solution in the parametric vector form is \({\bf{x}} = {x_3}\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 2}\\1\\0\end{array}} \right]\).