The full question is
Consider the matrix $A=\begin{bmatrix}1&-1&1\\1&1&1\\1&-2&2\end{bmatrix}$.
(a). Find the $LU$ decomposition of $A$.
(b). Find the volume of the box in $R^3$ that is spanned by the columns of $A$.
The $LU$ decomposition I worked out for $A$ is $A=\begin{bmatrix}1&0&0\\1&1&0\\1&-0.5&0.5\end{bmatrix}\begin{bmatrix}1&-1&1\\0&2&0\\0&0&2\end{bmatrix}$, which is different from the solution $A=\begin{bmatrix}1&0&0\\1&1&0\\1&-0.5&1\end{bmatrix}\begin{bmatrix}1&-1&1\\0&2&0\\0&0&1\end{bmatrix}$. I know that a matrix doesn't have a unique $LU$ decomposition but the problem arises in (b) which asks for the volumn of the box in $R^3$ that is spanned by the columns of $A$, which is essentially asking for the determinant of $A$. Since the determinant of a matrix equals to the product of the diagonal entries of an upper/lower triangular matrix or a diagonal matrix. The determinant of $A$ I calculate from my $U$ would be $4$. However, the determinant of $A$ is actually $2$. I'm very sure a matrix could only have one determinant. Where did I do wrong here?
Edit:
$det(A)=det(L)det(U)$