James' question about Gaussian Elimination

[Prove It]

View attachment 5489

We start by writing the system as an augmented matrix

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

Once we have used Gaussian Elimination to upper triangularise the system, the right hand column ends up becoming the elements of $\displaystyle \begin{align*} \mathbf{g} \end{align*}$.

As we are told to do this without pivoting, we will apply R2 - 2R1 to R2 and R3 - 2R1 to R3 giving

$\displaystyle \begin{align*} \left[ \begin{matrix} 2 & -2 & 1 & \phantom{-}1 & \phantom{-}4 \\ 0 & \phantom{-}3 & 0 & \phantom{-}1 & -\frac{21}{2} \\ 0 & \phantom{-}0 & 3 & -1 & \phantom{-}0 \\ 0 & \phantom{-}2 & 3 & \phantom{-}4 & -5 \end{matrix} \right] \end{align*}$

We will now apply R4 - (2/3)R2 to R4 giving

$\displaystyle \begin{align*} \left[ \begin{matrix} 2 & -2 & 1 & \phantom{-}1 & \phantom{-}4 \\ 0 & \phantom{-}3 & 0 & \phantom{-}1 & -\frac{21}{2} \\ 0 & \phantom{-}0 & 3 & -1 & \phantom{-}0 \\ 0 & \phantom{-}0 & 3 & \phantom{-}\frac{10}{3} & \phantom{-}2 \end{matrix} \right] \end{align*}$

We will now apply R4 - R3 to R4 giving

$\displaystyle \begin{align*} \left[ \begin{matrix} 2 & -2 & 1 & \phantom{-}1 & \phantom{-}4 \\ 0 & \phantom{-}3 & 0 & \phantom{-}1 & -\frac{21}{2} \\ 0 & \phantom{-}0 & 3 & -1 & \phantom{-}0 \\ 0 & \phantom{-}0 & 0 & \phantom{-}\frac{13}{3} & \phantom{-}2 \end{matrix} \right] \end{align*}$

So we can now read off $\displaystyle \begin{align*} \mathbf{g} = \left[ \begin{matrix} \phantom{-}4 \\ -\frac{21}{2} \\ \phantom{-}0 \\ \phantom{-}2 \end{matrix} \right] \end{align*}$, and solving for $\displaystyle \begin{align*} \mathbf{x} \end{align*}$ we have

$\displaystyle \begin{align*} \frac{13}{3}\,x_4 &= 2 \\ x_4 &= \frac{6}{13} \\ \\ 3\,x_3 - x_4 &= 0 \\ 3\,x_3 - \frac{6}{13} &= 0 \\ 3\,x_3 &= \frac{6}{13} \\ x_3 &= \frac{2}{13} \\ \\ 3\,x_2 + x_4 &= -\frac{21}{2} \\ 3\,x_2 + \frac{6}{13} &= -\frac{21}{2} \\ 3\,x_2 &= -\frac{285}{26} \\ x_2 &= -\frac{95}{26} \\ \\ 2\,x_1 - 2\,x_2 + x_3 + x_4 &= 4 \\ 2\,x_1 + \frac{95}{13} + \frac{2}{13} + \frac{6}{13} &= 4 \\ 2\,x_1 &= -\frac{51}{13} \\ x_1 &= -\frac{51}{26} \end{align*}$

So the solution to the system is $\displaystyle \begin{align*} \mathbf{x} = \frac{1}{26}\,\left[ \begin{matrix} -51 \\ -95 \\ \phantom{-}4 \\ \phantom{-}12 \end{matrix} \right] \end{align*}$.

 

[jvicens]

I would like to add a few comments about this solution process. First of all, Gaussian Elimination is a powerful tool for solving systems of linear equations, but it is important to note that it is not always the most efficient method. In some cases, it may be more efficient to use other techniques such as LU decomposition or matrix inversion. It is always important to consider the specific characteristics of the system at hand when choosing a solution method.

Additionally, it is important to note that in practice, pivoting is often necessary when using Gaussian Elimination to avoid division by zero or to improve the accuracy of the solution. Therefore, it is important to understand the limitations and potential issues with this method.

Finally, it is worth mentioning that this solution process assumes that the system is consistent, meaning that there is at least one solution. If the system is inconsistent, meaning that there is no solution, then Gaussian Elimination will lead to a row of all zeros in the augmented matrix, and the solution process will not be able to continue. It is important to check for consistency before using Gaussian Elimination to solve a system of linear equations.