MHB Question from Jesse about Gaussian Elimination and LU factorisation.

Prove It
Gold Member
MHB
Messages
1,434
Reaction score
20
View attachment 5370

This system can be written as a matrix equation $\displaystyle \begin{align*} A\,\mathbf{x} = \mathbf{b} \end{align*}$ as

$\displaystyle \begin{align*} \left[ \begin{matrix} \phantom{-}2 & 4 & \phantom{-}0 & 1 \\ \phantom{-}2 & 8 & -2 & 7 \\ -2 & 2 & -2 & 7 \\ \phantom{-}0 & 8 & -5 & 11 \end{matrix} \right] \left[ \begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{matrix} \right] = \left[ \begin{matrix} \phantom{-}7 \\ \phantom{-}3 \\ -7 \\ -12 \end{matrix} \right] \end{align*}$

To get the LU factorisation we need to use Gaussian Elimination on the coefficient matrix...

$\displaystyle \begin{align*} A = \left[ \begin{matrix} \phantom{-}2 & 4 & \phantom{-}0 & 1 \\ \phantom{-}2 & 8 & -2 & 7 \\ -2 & 2 & -2 & 7 \\ \phantom{-}0 & 8 & -5 & 11 \end{matrix} \right] \end{align*}$

To eliminate the terms under the main diagonal in the first column, we will apply Row 2 - Row 1 to Row 3 and Row 3 - (-1)Row 1 to Row 3. As we are eliminating the elements $\displaystyle \begin{align*} a_{2,1} \end{align*}$ and $\displaystyle \begin{align*} a_{3, 1} \end{align*}$, that means in our lower triangular matrix, which has 1 as all the elements on the main diagonal and the multipliers of the rows as its coefficients under the main diagonal, will have $\displaystyle \begin{align*} \mathcal{l}_{2,1} = 1 \end{align*}$ and $\displaystyle \begin{align*} \mathcal{l}_{3,1} = -1 \end{align*}$. Since we didn't have to do anything with element $\displaystyle \begin{align*} a_{4,1} \end{align*}$ that means the multiplier is 0, and thus $\displaystyle \begin{align*} \mathcal{l}_{4,1} = 0 \end{align*}$. A has now become

$\displaystyle \begin{align*} \left[ \begin{matrix} 2 & 4 & \phantom{-}0 & 1 \\ 0 & 4 & -2 & 6 \\ 0 & 6 & -2 & 8 \\ 0 & 8 & -5 & 11 \end{matrix} \right] \end{align*}$

To eliminate the terms under the main diagonal in the second column, we will apply Row 3 - (3/2)Row 2 to Row 3 and Row 4 - 2 Row 2 to Row 4. As we are eliminating the elements $\displaystyle \begin{align*} a_{3,2} \end{align*}$ and $\displaystyle \begin{align*} a_{4,2} \end{align*}$, this means that $\displaystyle \begin{align*} \mathcal{l}_{3,2} = \frac{3}{2} \end{align*}$ and $\displaystyle \begin{align*} \mathcal{l}_{4,2} = 2 \end{align*}$. A has now become

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

To eliminate the term under the main diagonal in the third column, we will apply Row 4 - (-1)Row 3 to Row 4. As we are eliminating the element $\displaystyle \begin{align*} a_{4,3} \end{align*}$ that means that $\displaystyle \begin{align*} \mathcal{l}_{4,3} = -1 \end{align*}$. A has now become

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

So we have found that our lower triangular matrix $\displaystyle \begin{align*} L = \left[ \begin{matrix} \phantom{-}1 & 0 & \phantom{-}0 & 0 \\ \phantom{-}1 & 1 & \phantom{-}0 & 0 \\ -1 & \frac{3}{2} & \phantom{-}1 & 0 \\ \phantom{-}0 & 2 & -1 & 1 \end{matrix} \right] \end{align*}$ and our upper triangular matrix $\displaystyle \begin{align*} U = \left[ \begin{matrix} 2 & 4 & \phantom{-}0 & \phantom{-}1 \\ 0 & 4 & -2 & \phantom{-}6 \\ 0 & 0 & \phantom{-}1 & -1 \\ 0 & 0 & \phantom{-}0 & -2 \end{matrix} \right] \end{align*}$.

This reduces the system to $\displaystyle \begin{align*} L\,U\,\mathbf{x} = \mathbf{b} \end{align*}$. Notice that $\displaystyle \begin{align*} U\,\mathbf{x} \end{align*}$ is a column matrix when multiplied out, which we can call $\displaystyle \begin{align*} \mathbf{g} \end{align*}$. This now reduces the system to two simpler matrix equations, as the coefficient matrices are diagonal. They are $\displaystyle \begin{align*} L\,\mathbf{g} = \mathbf{b} \end{align*}$ and $\displaystyle \begin{align*} U\,\mathbf{x} = \mathbf{g} \end{align*}$.

Solving for $\displaystyle \begin{align*} \mathbf{g} \end{align*}$ we have

$\displaystyle \begin{align*} \left[ \begin{matrix} \phantom{-}1 & 0 & \phantom{-}0 & 0 \\ \phantom{-}1 & 1 & \phantom{-}0 & 0 \\ -1 & \frac{3}{2} & \phantom{-}1 & 0 \\ \phantom{-}0 & 2 & -1 & 1 \end{matrix} \right] \left[ \begin{matrix} g_1 \\ g_2 \\ g_3 \\ g_4 \end{matrix} \right] = \left[ \begin{matrix} \phantom{-}7 \\ \phantom{-}3 \\ -7 \\ -12 \end{matrix} \right] \end{align*}$

We can see $\displaystyle \begin{align*} g_1 = 7 \end{align*}$. Forward substituting gives

$\displaystyle \begin{align*} g_1 + g_2 &= 3 \\ 7 + g_2 &= 3 \\ g_2 &= -4 \end{align*}$

Forward substituting gives

$\displaystyle \begin{align*} -g_1 + \frac{3}{2}\,g_2 + g_3 &= -7 \\ -7 - 6 + g_3 &= -7 \\ g_3 &= 6 \end{align*}$

Forward substituting gives

$\displaystyle \begin{align*} 2\,g_2 - g_3 + g_4 &= -12 \\ -8 - 6 + g_4 &= -12 \\ -14 + g_4 &= -12 \\ g_4 &= 2 \end{align*}$

Now solving $\displaystyle \begin{align*} U\,\mathbf{x} = \mathbf{g} \end{align*}$ for $\displaystyle \begin{align*} \mathbf{x} \end{align*}$ gives

$\displaystyle \begin{align*} \left[ \begin{matrix} 2 & 4 & \phantom{-}0 & \phantom{-}1 \\ 0 & 4 & -2 & \phantom{-}6 \\ 0 & 0 & \phantom{-}1 & -1 \\ 0 & 0 & \phantom{-}0 & -2 \end{matrix} \right] \left[ \begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{matrix} \right] = \left[ \begin{matrix} \phantom{-}7 \\ -4 \\ \phantom{-}6 \\ \phantom{-}2 \end{matrix} \right] \end{align*}$

We can see that

$\displaystyle \begin{align*} -2x_4 &= 2 \\ x_4 &= -1 \end{align*}$

Back substituting gives

$\displaystyle \begin{align*} x_3 - x_4 &= 6 \\ x_3 + 1 &= 6 \\ x_3 &= 5 \end{align*}$

Back substituting gives

$\displaystyle \begin{align*} 4\,x_2 - 2\,x_3 + 6\,x_4 &= -4 \\ 4\,x_2 - 10 - 6 &= -4 \\ 4\,x_2 &= 12 \\ x_2 &= 3 \end{align*}$

Back substituting gives

$\displaystyle \begin{align*} 2\,x_1 + 4\,x_2 + x_4 &= 7 \\ 2\,x_1 + 12 - 1 &= 7 \\ 2\,x_1 &= -4 \\ x_1 &= -2 \end{align*}$

So to answer your question we have $\displaystyle \begin{align*} g_1 = 7 , \, g_2 = -4 , \, g_3 = 6 , \, g_4 = 2, \, x_1 = -2 , \, x_2 = 3 , \, x_3 = 5 \end{align*}$ and $\displaystyle \begin{align*} x_4 = -1 \end{align*}$.
 

Attachments

  • gelu.jpg
    gelu.jpg
    67 KB · Views: 98
Mathematics news on Phys.org


Thank you for the detailed explanation, it was very helpful! It's interesting to see how the Gaussian Elimination process can be broken down into smaller, simpler equations.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.

Similar threads

Back
Top