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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\(T\left( {\bf{x}} \right) = A{\bf{x}} = b\).

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

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

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

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

Use the \({x_1}\) term from the first equation to eliminate the \(3{x_1}\) term from the third equation. Add \( - 3\) times row one to row three. Then, use the \({x_1}\) term from the first equation to eliminate the \( - 2{x_1}\) term from the second equation. Add 2 times row one to row two.

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

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. Then, multiply row three by \(\frac{1}{5}\).

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

Use the \({x_3}\) term from the third equation to eliminate the \(2{x_3}\) term from the second equation. Add \( - 2\) times row 3 to row 2.

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

Use the \({x_3}\) term from the third equation to eliminate the \( - 2{x_3}\) term from the first equation. Add 2 times row three to row one.

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

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

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

\(\begin{array}{c}{x_1} + 0\left( {{x_2}} \right) + 0\left( {{x_3}} \right) = 3\\0\left( {{x_1}} \right) + {x_2} + 0\left( {{x_3}} \right) = 1\\0\left( {{x_1}} \right) + 0\left( {{x_2}} \right) + {x_3} = 2\end{array}\)

According to the pivot positions in the obtained matrix, \({x_1}\), \({x_2}\), and \({x_3}\) are the basic variables, and there are no free variables.

Thus, the general solution is \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}3\\1\\2\end{array}} \right]\).

This solution is unique.