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James' question about Gaussian Elimination

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
gaussian elimination.jpg

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