Solve System w/ Gaussian Elimination & Partial Pivoting (Sufyan)

[Prove It]

We write the system as an augmented matrix

$\displaystyle \begin{align*} \left[ \begin{matrix} \phantom{-}1 & 4 & \phantom{-}1 & 0 & \phantom{-}5 \\ \phantom{-}1 & 6 & -1 & 4 & \phantom{-}7 \\ -1 & 2 & -9 & 2 & -9 \\ \phantom{-}0 & 1 & \phantom{-}2 & 0 & \phantom{-}4 \end{matrix} \right] \end{align*}$

In order to solve the system we must upper triangularise the matrix. As we do, the right hand column becomes matrix $\displaystyle \begin{align*} \mathbf{g} \end{align*}$.

As the elements in column 1 have a maximum magnitude of 1, there is no need to interchange any rows. So we apply R2 - R1 to R2 and R3 + R1 to R3, giving

$\displaystyle \begin{align*} \left[ \begin{matrix} \phantom{-}1 & 4 & \phantom{-}1 & 0 & \phantom{-}5 \\ \phantom{-}0 & 2 & -2 & 4 & \phantom{-}2 \\ \phantom{-}0 & 6 & -8 & 2 & -4 \\ \phantom{-}0 & 1 & \phantom{-}2 & 0 & \phantom{-}4 \end{matrix} \right] \end{align*}$

Now looking at the elements on or below the main diagonal in column 2, the element with the highest magnitude is in row 3, so we have to interchange rows 2 and 3.

$\displaystyle \begin{align*} \left[ \begin{matrix} \phantom{-}1 & 4 & \phantom{-}1 & 0 & \phantom{-}5 \\ \phantom{-}0 & 6 & -8 & 2 & -4\\ \phantom{-}0 & 2 & -2 & 4 & \phantom{-}2 \\ \phantom{-}0 & 1 & \phantom{-}2 & 0 & \phantom{-}4 \end{matrix} \right] \end{align*}$

Now we must apply R3 - (1/3)R2 to R3 and R4 - (1/6)R2 to R2.

$\displaystyle \begin{align*} \left[ \begin{matrix} \phantom{-}1 & 4 & \phantom{-}1 & \phantom{-}0 & \phantom{-}5 \\ \phantom{-}0 & 6 & -8 & \phantom{-}2 & -4\\ \phantom{-}0 & 0 & \phantom{-}\frac{2}{3} & \phantom{-}\frac{10}{3} & \phantom{-}\frac{10}{3} \\ \phantom{-}0 & 0 & \phantom{-}\frac{10}{3} & -\frac{1}{3} & \phantom{-}\frac{14}{3} \end{matrix} \right] \end{align*}$

Now looking at the elements on or below the main diagonal in column 3, the element with the highest magnitude is in row 4, so we must interchange rows 3 and 4.

$\displaystyle \begin{align*} \left[ \begin{matrix} \phantom{-}1 & 4 & \phantom{-}1 & \phantom{-}0 & \phantom{-}5 \\ \phantom{-}0 & 6 & -8 & \phantom{-}2 & -4\\ \phantom{-}0 & 0 & \phantom{-}\frac{10}{3} & -\frac{1}{3} & \phantom{-}\frac{14}{3}\\ \phantom{-}0 & 0 & \phantom{-}\frac{2}{3} & \phantom{-}\frac{10}{3} & \phantom{-}\frac{10}{3} \end{matrix} \right] \end{align*}$

Now we must apply R4 - (1/5)R3 to R4.

$\displaystyle \begin{align*} \left[ \begin{matrix} \phantom{-}1 & 4 & \phantom{-}1 & \phantom{-}0 & \phantom{-}5 \\ \phantom{-}0 & 6 & -8 & \phantom{-}2 & -4\\ \phantom{-}0 & 0 & \phantom{-}\frac{10}{3} & -\frac{1}{3} & \phantom{-}\frac{14}{3}\\ \phantom{-}0 & 0 & \phantom{-}0 & \phantom{-}\frac{17}{5} & \phantom{-}\frac{12}{5} \end{matrix} \right] \end{align*}$We can now read off $\displaystyle \begin{align*} \mathbf{g} = \left[ \begin{matrix} \phantom{-}5 \\ -4 \\ \phantom{-}\frac{14}{3} \\ \phantom{-}\frac{12}{5} \end{matrix} \right] \end{align*}$ and solving for $\displaystyle \begin{align*} \mathbf{x} \end{align*}$ using back substitution we have

$\displaystyle \begin{align*} \frac{17}{5}\,x_4 &= \frac{12}{5} \\ x_4 &= \frac{12}{17} \\ \\ \frac{10}{3}\,x_3 - \frac{1}{3}\,x_4 &= \frac{14}{3} \\ \frac{10}{3}\,x_3 - \frac{4}{17} &= \frac{14}{3} \\ 10\,x_3 - \frac{12}{17} &= 14 \\ 10\,x_3 &= \frac{250}{17} \\ x_3 &= \frac{25}{17} \\ \\ 6\,x_2 - 8\,x_3 + 2\,x_4 &= -4 \\ 6\,x_2 - \frac{200}{17} + \frac{24}{17} &= -4 \\ 6\,x_2 &= \frac{108}{17} \\ x_2 &= \frac{18}{17} \\ \\ x_1 + 4\,x_2 + x_3 &= 5 \\ x_1 + \frac{72}{17} + \frac{25}{17} &= 5 \\ x_1 &= -\frac{12}{17} \end{align*}$So the solution to the system is $\displaystyle \begin{align*} \mathbf{x} = \frac{1}{17}\,\left[ \begin{matrix} -12 \\ \phantom{-}18 \\ \phantom{-}25 \\ \phantom{-}12 \end{matrix} \right] \end{align*}$.

 

[jvicens]

I would like to point out that solving a system of equations using the augmented matrix method is a commonly used technique in various fields of science, including physics, chemistry, and engineering. This method allows us to efficiently solve systems with multiple variables and equations, which are often encountered in real-world problems.

In addition, the process of upper triangularizing the matrix is an important step in solving the system. This process helps to reduce the complexity of the system and makes it easier to solve using back substitution. It also allows us to easily identify the pivot elements, which are crucial in the process of row operations.

Furthermore, it is important to note that the solution to the system is not unique. This means that there can be multiple sets of values for $\displaystyle \begin{align*} \mathbf{x} \end{align*}$ that satisfy the given equations. This is a common occurrence in scientific problems, and it is important to carefully consider the context of the problem to determine the most appropriate solution.

Overall, the augmented matrix method is a powerful tool in solving systems of equations, and its application is widely used in various scientific fields. It allows us to efficiently and accurately solve complex systems, providing us with valuable insights and solutions to real-world problems.