Solving System of linear equation using LU decomposition

I am working on a simultaneous linear equation problem using LU decomposition and I'm unsure if this is the correct approach/answer to solve a system of simultaneous equations using LU decomposition. I'm looking for a way to check my resulting calculations and to understand if I have done the calculations correctly. This is my first time doing LU decomposition or doing matrix calculations at this level.

$$5x1 + 6x2 + 2.3x3 + 6x4 = 4$$

$$9x1 + 2x2 + 3.5x3 + 7x4 = 5$$

$$3.5x1 + 6x2 + 2x3 + 3x4 = 6.7$$

$$1.5x1 + 2x2 + 1.5x3 + 6x4 = 7.8$$

$$\begin{bmatrix}5&6&2.3&6\\9&2&3.5&7\\3.5&6&2&3\\1.5&2&1.5&6\end{bmatrix}$$

I multiply the top, third and fourth row by 10 and the second by 2 to make it easier to work with.

$$\begin{bmatrix}50&60&23&60\\18&4&7&14\\35&60&20&30\\15&20&15&60\end{bmatrix}$$

I calculated the L matrix as:

$$\begin{bmatrix}1&0&0&0\\9/25&1&0&0\\7/10&45/44&1&0\\3/10&5/44&1.584572&1\end{bmatrix}$$

and U matrix as: $$\begin{bmatrix}50&60&23&60\\0&88/25&-1.28&-38/5\\0&0&5.2&-93/22\\0&0&0&12.6\end{bmatrix}$$

I have y as the following: $$\begin{bmatrix}4\\5\\6.7\\7.8\end{bmatrix}$$

and solved the system with the following values for x $$\begin{bmatrix}4\\3.56\\0.2590\\5.734906\end{bmatrix}$$

• @Moo Thanks for the reply! Was everything up until the missed sign error of -45/44 correct? Would you mind sharing the result you obtained for me to verify my answer as I re-work through it?
– ASH
Mar 25, 2022 at 12:37

1 Answer

Your answer has a few issues, it was close.

You made some sign errors in values of $$L$$, and that messed up the final $$L$$.

For example, you should have $$−45/44$$ and $$−5/44$$ and that looks like it caused $$L_{43}$$ to be wrong. Of course, these issues caused similar errors in $$U$$.

The result should be

$$A = \begin{pmatrix}50&60&23&60\\18&4&7&14\\35&60&20&30\\15&20&15&60\end{pmatrix} = LU = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ \frac{9}{25} & 1 & 0 & 0 \\ \frac{7}{10} & -\frac{45}{44} & 1 & 0 \\ \frac{3}{10} & -\frac{5}{44} & \frac{175}{57} & 1 \\ \end{array} \right)\left( \begin{array}{cccc} 50 & 60 & 23 & 60 \\ 0 & -\frac{88}{5} & -\frac{32}{25} & -\frac{38}{5} \\ 0 & 0 & \frac{57}{22} & -\frac{435}{22} \\ 0 & 0 & 0 & \frac{1935}{19} \\ \end{array} \right)$$

Hopefully, you can take it from here.

Notes:

• $$1.$$ You can always verify your result by substituting the values you got for $$x$$ back into the original system and seeing if the LHS = RHS.

• $$2.$$ You could have also multiplied $$LU$$ and see if that returned $$A$$.

• $$3.$$ Also, please recall that you multiplied your equations by $$10$$ and $$2$$, but forgot to make the corresponding update to $$y$$ for the final steps.

We now want to use back substitution to solve

$$A x = LU x = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ \frac{9}{25} & 1 & 0 & 0 \\ \frac{7}{10} & -\frac{45}{44} & 1 & 0 \\ \frac{3}{10} & -\frac{5}{44} & \frac{175}{57} & 1 \\ \end{array} \right)\left( \begin{array}{cccc} 50 & 60 & 23 & 60 \\ 0 & -\frac{88}{5} & -\frac{32}{25} & -\frac{38}{5} \\ 0 & 0 & \frac{57}{22} & -\frac{435}{22} \\ 0 & 0 & 0 & \frac{1935}{19} \\ \end{array} \right)x = y = \begin{pmatrix}40\\10\\67\\78\end{pmatrix}$$

The final result is

$$x = \begin{pmatrix}-\frac{1976}{645}\\-\frac{281}{860}\\ \frac{1327}{129}\\-\frac{256}{645} \end{pmatrix}$$

• Thank you so so much for the detailed explanation. I appreciate it so much!
– ASH
Mar 25, 2022 at 18:53