linear algebra - Find the determinant of a $4 times 4$ matrix using row reduction

The original matrix is:

\begin{bmatrix}
1 &2 &0 &1 \\
3 &3 &3 &9 \\

1 & 4 & 1 &4 \\
1 & 1 & 2 &0
\end{bmatrix}

Every time I reduced this to row echelon form, I got $\dfrac{1}{48}$ as the determinant when the actual determinant is $48$. Here are the row operations.

The rows that I have highlighted are the ones that change the determinant since we are changing a row by a factor. All the other operations don't change the determinant and we never switch rows here.

So we get $\det(A) = \dfrac{-1}{3}\times\dfrac{1}{3}\times\dfrac{-3}{16}\times 1 \text{ (entries along diagonal)} = \dfrac{1}{48}$. But the actual determinant is $48$. Where have I went wrong? I did the row operations a couple of times but I am making the same mistake.

Answer

Let's say we start with a matrix $A$ and perform a row operation of multiplying some row $R_i$ by a non-zero scalar $c$ and get a new matrix $A'$. What is the relation between $\det(A)$ and $\det(A')$? Since the determinant is a multilinear functions of the rows of $A$, we have

$$ \det(A') = c \det(A) \iff \det(A) = \frac{1}{c} \det(A'). $$

If we perform various row operations on $A$, the only operations which change the determinant are the multiplication operations. If we have performed multiplication operations by $c_1,\dots,c_k$ (and maybe other row operations) and arrived to the matrix $A'$, we have

$$ \det(A) = \frac{1}{c_1 \dots c_k} \det(A'). $$

In particular, if you reduced $A$ to an upper diagonal matrix $A'$ whose diagonal entries are $1$ then $\det(A') = 1$ and so

$$ \det(A) = \frac{1}{c_1 \dots c_k}. $$