LU factorization of a matrix

I am trying to computer the LU factorization of the following matrix by hand. This is the first time I have tried a question like this so i am checking to see if my workings are correct, and if they need anymore detail. Thanks!

$$A= \left[ \begin{matrix} 2&1&3\\1&2&1\\1&2&3 \end{matrix} \right]$$

Finding the upper triangular matrix by reduced form

$$\left[ \begin{matrix} 2&1&3\\1&2&1\\1&2&3 \end{matrix} \right] R_2\rightarrow R_1-2R_2 = \left[ \begin{matrix} 2&1&3\\0&-3&1\\1&2&3 \end{matrix} \right]\\ R_3 \rightarrow R_1-R_2 = \left[ \begin{matrix} 2&1&3\\0&-3&1\\0&-3&-3 \end{matrix} \right] \\ R_3\rightarrow R_2-R_3 = \left[ \begin{matrix} 2&1&3\\0&-3&1\\0&0&4 \end{matrix} \right]$$

Then using the $$k$$s we used to modify each row to form the lower triangular matrix we get:

$$\left[ \begin{matrix} 1&0&0\\2&1&0\\2&1&1 \end{matrix} \right]$$

Therefore,

$$\left[ \begin{matrix} 2&1&3\\1&2&1\\1&2&3 \end{matrix} \right] = \left[ \begin{matrix} 1&0&0\\2&1&0\\2&1&1 \end{matrix} \right] \left[ \begin{matrix} 2&1&3\\0&-3&1\\0&0&4 \end{matrix} \right]$$

• no it does not equal A, seems that i have gone wrong – jh123 Oct 2 '18 at 16:41
• why not just L and U? – jh123 Oct 2 '18 at 16:48
• Two of your steps are not elementary row operations. – amd Oct 2 '18 at 18:46

This is from another answer of mine.

Suppose that

$$A = \begin{bmatrix} 2 & 1 & 3 \\ 1 & 2 & 1 \\1 & 2 & 3 \end{bmatrix} \tag{1}$$ $$A = LU \tag{2}$$

$$U =A, L=I \tag{3}$$ $$\ell_{21} = \frac{u_{21}}{u_{11}} = \frac{a_{21}}{a_{11}} = \frac{1}{2} \tag{4}$$

This is where you made the mistake $$R_{2} \to R_{2} - \frac{1}{2} R_{1} \tag{5}$$ Then we're going to subtract $$\frac{1}{2}$$ times the 1st row from the 2nd row $$\begin{bmatrix} 1 & 2 & 1 \end{bmatrix} - \frac{1}{2} \cdot \begin{bmatrix} 2 & 1 & 3 \end{bmatrix} = \begin{bmatrix} 0 & \frac{3}{2} & \frac{-1}{2} \end{bmatrix} \tag{6}$$ Updating each of them $$L= \begin{bmatrix} 1 & 0 & 0 \\ \frac{1}{2} & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \tag{7}$$ $$U = \begin{bmatrix} 2 & 1 & 3 \\ 0 & \frac{3}{2} & \frac{-1}{2} \\1 & 2 & 3 \end{bmatrix} \tag{8}$$ $$\ell_{31} = \frac{u_{31}}{u_{11} } = \frac{1}{2} \tag{9}$$

$$R_{3} \to R_{3} - \frac{1}{2} R_{1} \tag{10}$$ $$R_{3} = \begin{bmatrix} 1 & 2 & 3\end{bmatrix} - \frac{1}{2} \begin{bmatrix} 2 & 1 & 3\end{bmatrix} = \begin{bmatrix} 0 & \frac{3}{2} & \frac{3}{2}\end{bmatrix} \tag{11}$$

Updating each of them $$L= \begin{bmatrix} 1 & 0 & 0 \\ \frac{1}{2} & 1 & 0 \\ \frac{1}{2} & 0 & 1 \end{bmatrix} \tag{12}$$ $$U = \begin{bmatrix} 2 & 1 & 3 \\ 0 & \frac{3}{2} & \frac{-1}{2} \\0 & \frac{3}{2} & \frac{3}{2} \tag{13} \end{bmatrix}$$

$$\ell_{32} = \frac{u_{32}}{u_{22}} = \frac{\frac{3}{2}}{\frac{3}{2}} = 1 \tag{14}$$

$$R_{3} \to R_{3} - R_{2} \tag{15}$$

$$R_{3} = \begin{bmatrix} 0 & \frac{3}{2} & \frac{3}{2} \end{bmatrix} - \begin{bmatrix} 0& \frac{3}{2} & \frac{-1}{2} \end{bmatrix} = \begin{bmatrix} 0 &0 & 2\end{bmatrix} \tag{16}$$

Updating each of them $$L= \begin{bmatrix} 1 & 0 & 0 \\ \frac{1}{2} & 1 & 0 \\ \frac{1}{2} & 1& 1 \end{bmatrix} \tag{17}$$ $$U = \begin{bmatrix} 2 & 1 & 3 \\ 0 & \frac{3}{2} & \frac{-1}{2} \\0 & 0 & 2 \end{bmatrix} \tag{18}$$ It now terminates $$A = LU$$ $$\underbrace{\begin{bmatrix} 2 & 1 & 3 \\ 1 & 2 & 1 \\1 & 2 & 3 \end{bmatrix}}_{A} = \underbrace{\begin{bmatrix} 1 & 0 & 0 \\ \frac{1}{2} & 1 & 0 \\ \frac{1}{2} & 1& 1 \end{bmatrix} }_{L} \underbrace{\begin{bmatrix} 2 & 1 & 3 \\ 0 & \frac{3}{2} & \frac{-1}{2} \\0 & 0 & 2 \end{bmatrix}}_{U} \tag{19}$$

Just to confirm this in python

import scipy.linalg

A = scipy.array([[2 ,1,3],[1, 2, 1 ] ,[1,2,3]])
P,L,U = scipy.linalg.lu(A)

L
Out[6]:
array([[1. , 0. , 0. ],
[0.5, 1. , 0. ],
[0.5, 1. , 1. ]])

U
Out[7]:
array([[ 2. ,  1. ,  3. ],
[ 0. ,  1.5, -0.5],
[ 0. ,  0. ,  2. ]])

• @jh123: Note that this very clean presentation of the solution avoids any syntaxs errors such as writing "(linear combination of rows) = matrix", for which any serious grader will penalize you despite the fact that you will be understood. – Carl Christian Oct 3 '18 at 6:25