How can I find an LU factorisation of this $3 \times 3$ matrix?

Question

$$A=\begin{bmatrix}1&2&-3\\-2&-4&8\\-3&-4&14\end{bmatrix}$$

This is what I found:

$$U=\begin{bmatrix}1&2&-3\\0&0&2\\0&0&8\end{bmatrix}$$

$$L=\begin{bmatrix}1&0&0\\-2&1&0\\-3&1&1\end{bmatrix}$$

There's something wrong about $U$, but I don't know what it is. Can someone help me please?

Answer 1

You claim that something is wrong about $U$.

Assuming that $L$ is correct, then we will have $U=L^{-1}A$. Let's compute that.

octave:7> A = [1,2,-3; -2, -4, 8; -3, -4, 14]
A =

    1    2   -3
   -2   -4    8
   -3   -4   14

octave:8> L
L =

   1   0   0
  -2   1   0
  -3   1   1

octave:9> inv(L)*A
ans =

   1   2  -3
   0   0   2
   0   2   3

Yikes, $L$ is not correct.

Also,

octave:11> rank(A)
ans =  3

$U$ certainly can't have $(2,2)$-th entry being $0$.

The thing is there is no such decomposition for this matrix.

Suppose on the contrary that such decomposition exists.

Then we have

$$A=\begin{bmatrix} l_{11} & 0 & 0 \\ -2l_{11} & l_{22} & 0 \\ l_{31} & l_{32} & l_{33} \end{bmatrix} \begin{bmatrix} \frac1{l_{11}} & \frac2{l_{11}} & \frac{-3}{l_{11}} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{bmatrix} $$

From the $(2,2)$-th entry, we have $l_{22}u_{22}=0$, this would contradict the rank of matrix $A$.

You might like to consider an $LUP$ decomposition instead.