# Finding the A=LDU factorization for matrices

I am trying to factorize the matrix on the picture, but I am not sure how can I get the upper triangular matrix. Do I have to reverse every step of the matrix operation that I performed? Any help will be greatly appreciated!

I usually work out row reduction in Gauss-Jordan form: \begin{align} A=\begin{bmatrix} 1&0&1\\ 2&2&2\\ 3&4&5 \end{bmatrix} &\xrightarrow{\substack{E_{31}(-3)\\E_{21}(-2)}} \begin{bmatrix} 1&0&1\\ 0&2&0\\ 0&4&2 \end{bmatrix} \\[6px]&\xrightarrow{E_2(1/2)} \begin{bmatrix} 1&0&1\\ 0&1&0\\ 0&4&2 \end{bmatrix} \\[6px]&\xrightarrow{E_{32}(-4)} \begin{bmatrix} 1&0&1\\ 0&1&0\\ 0&0&2 \end{bmatrix} \\[6px]&\xrightarrow{E_{3}(1/2)} \begin{bmatrix} 1&0&1\\ 0&1&0\\ 0&0&1 \end{bmatrix}=U \end{align} The notation used is

• $$E_i(c)$$ means multiplying the $$i$$th row by $$c\ne0$$
• $$E_{ij}(d)$$ means summing to the $$i$$th row the $$j$$th row multiplied by $$d$$

These can be seen as elementary matrices that successively multiply on the left, so I get $$L_0=E_{21}(2)E_{31}(3)E_2(2)E_{31}(4)E_3(2)= \begin{bmatrix} 1 & 0 & 0 \\ 2 & 2 & 0 \\ 3 & 4 & 2 \end{bmatrix}$$ the product of the inverses in the reverse order.

It can be shown that the entries are exactly as dictated by the elementary matrices, with $$0$$ added off-diagonal and missing entries on the diagonal are $$1$$.

Now writing $$L_0=LD$$ is easy: $$L= \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix} \qquad D=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$ (divide each column by the corresponding entry in the diagonal of $$D$$).

I am including the $$L$$ component of the answer as well. But $$U$$ is explained clearly too:

The goal of row reduction is to reduce a matrix $$A$$ to an upper-triangular matrix $$A^{'}$$, which you have as a step in your picture. $$A=\begin{bmatrix}1&0&1\\2&2&2\\3&4&5\end{bmatrix}\to \, \, A^{'} = \begin{bmatrix}1&0&1\\0&2&0\\0&0&2\end{bmatrix}$$

But each row operation can be represented as multiplying $$A$$on the left by some matrix $$E$$. In your picture, you use 3 row operations to get to $$A^{'}$$, so $$E = E_1 E_2 E_3$$ and we have $$EA = A^{'}$$

Now we can solve for $$A$$ and see $$A = E^{-1} A^{'}$$

Note that $$A^{'}$$ can be written as the product of a diagonal matrix and an upper triangular matrix with 1 on the diagonals, $$A^{'} = DU = \begin{bmatrix}1&0&0\\0&2&0\\0&0&2\end{bmatrix} \begin{bmatrix}1&0&1\\0&1&0\\0&0&1\end{bmatrix}$$.

It turns out that $$E^{-1}$$ is already lower triangular, so we have

$$A = LDU = E^{-1} \begin{bmatrix}1&0&0\\0&2&0\\0&0&2\end{bmatrix} \begin{bmatrix}1&0&1\\0&1&0\\0&0&1\end{bmatrix}$$