Csdrhrt

I have the matrix
$$A=
begin{pmatrix}
-1&1&1\1&-1&1\1&1&-1
end{pmatrix}
$$

It is invertible so it has an LU-factorization (Am I right about that?)

I tried to solve it, I first reached that the upper matrix is equal to
$$U=P(2,3)E_1A=
begin{pmatrix}
-1&1&1\0&2&0 \0&0&2
end{pmatrix}
$$

Where
$$E_1=
begin{pmatrix}
1&0&0\1&1&0 \1&0&1
end{pmatrix}
$$
And $P(2,3)$ is the matrix that switches the 2nd row with the $3$rd row.

But I continued as I usually do, the matrix $L$ usually turns out to be $L=[P(2,3)E_1]^{-1}$, but this wasn't a lower matrix !!

I got that
$$L=
begin{pmatrix}
1&0&0\-1&0&1 \-1&1&0
end{pmatrix}
$$

I checked my calculations $3$ times I don't think I have calculation mistakes...

Can anyone help please by giving a correct method and tell me what I did wrong in the method I used?

You have to perform Gaussian elimination without row swaps.
begin{align}
begin{pmatrix}
-1&1&1\
1&-1&1\
1&1&-1
end{pmatrix}
&to
begin{pmatrix}
1&-1&-1\
1&-1&1\
1&1&-1
end{pmatrix}
&& R_1gets -R_1
\[6px] &to
begin{pmatrix}
1&-1&-1\
0&0&2\
0&2&0
end{pmatrix}
&&begin{aligned} R_2&gets R_2-R_1 \ R_3&gets R_3-R_1 end{aligned}
end{align}
No, the matrix does not admit an $LU$ decomposition, with $L$ lower triangular and $U$ upper triangular.

If you swap rows to begin with,
begin{align}
P(2,3)A=begin{pmatrix}
-1&1&1\
1&1&-1\
1&-1&1\
end{pmatrix}
&to
begin{pmatrix}
1&-1&-1\
1&1&-1\
1&-1&1\
end{pmatrix}
&& R_1gets -R_1
\[6px] &to
begin{pmatrix}
1&-1&-1\
0&2&0\
0&0&2\
end{pmatrix}
&&begin{aligned} R_2&gets R_2-R_1 \ R_3&gets R_3-R_1 end{aligned}
\[6px] &to
begin{pmatrix}
1&-1&-1\
0&1&0\
0&0&1\
end{pmatrix}
&&begin{aligned} R_2&gets tfrac{1}{2}R_2 \ R_3&gets tfrac{1}{2}R_3 end{aligned}
end{align}
Thus you get
$$
U=begin{pmatrix}
1&-1&-1\
0&1&0\
0&0&1\
end{pmatrix}
qquad
L=begin{bmatrix}
-1 & 0 & 0 \
1 & 2 & 0\
1 & 0 & 2
end{bmatrix}
$$
so that
$$
A=P(2,3)LU
$$
If you don't do pivot reduction, the idea is essentially the same.

Since you are swapping the second and third rows, you must multiply by the corresponding elemenetary matrix:
$$A=
begin{pmatrix}
-1&1&1\1&-1&1\1&1&-1
end{pmatrix}=
begin{pmatrix}
1&0&0\0&0&1\0&1&0
end{pmatrix}
begin{pmatrix}
1&0&0\-1&1&0\-1&0&1
end{pmatrix}
begin{pmatrix}
-1&1&1\0&2&0\0&0&2
end{pmatrix}=PLU.$$
See WA answer.

You know that $L$ will be a lower triangular matrix with ones in the diagional. Now look at the pivots in $U$, and conclude all columns (and rows) contain pivots. Now you go back to your $A$ matrix, and divide the columns by the diagonal pivot. In your first column, $-1$ is on the diagonal, thus you divide the entire column by $-1$. In your second column, $-1$ is on the diagional, thus all the entries below the diagonal (in this case only the third row) will be divided by $-1$. The same applies for the third, getting

$ L=begin{bmatrix}
1 & 0 & 0 \
-1 & 1 & 0 \
-1 & -1 & 1
end{bmatrix} ,$ does this make sense?

If $A$ is invertible, then it admits an LU factorization if and only if all its leading principal minors are non-zero.

So $A$ doesn't have an LU factorization since a leading principal minor is zero, which means that,
$$D_2=
bigg|
begin{matrix}
-1&1\1&-1
end{matrix}
bigg| =0
$$

But matrix $A$ has a PLU decomposition