linear algebra - Find the triangular matrix and determinant.

I have a 4x4 matrix and I want to find the triangular matrix (lower half entries are zero).

$$A=
\begin{bmatrix}

2 & -8 & 6 & 8\\
3 & -9 & 5 & 10\\
-3 & 0 & 1 & -2\\
1 & -4 & 0 & 6
\end{bmatrix}
$$

Here are the elementary row operations I performed to get it into triangular form.

row swap rows 1 and row 4

$r_2 - 3\cdot r_1$ replacing $r_2$

$r_3 + 3\cdot r_1$ replacing $r_3$

$r_4 - 2\cdot r_1$ replacing $r_4$

I get this matrix

$$A= -

\begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\
0 & -12 & 1 & 16\\
0 & 0 & 6 & -4
\end{bmatrix}
$$

I then did $4\cdot r_2 + r_3$ to replace $r_3$ and got

$$A= -
\begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\
0 & 0 & 21 & -16\\
0 & 0 & 6 & -4
\end{bmatrix}
$$

I then did $-21\cdot r_4 + 6\cdot r_3$ to replace $r_4$ and got

$$A= -
\begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\
0 & 0 & 21 & -16\\
0 & 0 & 0 & -12
\end{bmatrix}
$$

I am not sure if I did this correctly but the determinant of the matrix should be -36. When I multiply the diagonal entries it isn't -36. I can't figure out what I am doing wrong.

Answer

"I then did -21*row 4 + 6*row 3 to replace row 4 and got"

This is a determinant altering operation and not an elementary operation.

Don't write that $A$ equals something which isn't $A$.

Picking up where you errored and using the same idea you had one gets:

$$\begin{align} \begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\
0 & 0 & 21 & -16\\
0 & 0 & 6 & -4
\end{bmatrix}&\leadsto \begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\

0 & 0 & 6\cdot 21 & -6\cdot 16\\
0 & 0 & -21\cdot 6 & (-21)\cdot (-4)
\end{bmatrix}\\
&\leadsto \begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\
0 & 0 & 6\cdot 21 & -16\\
0 & 0 & 0 & -12
\end{bmatrix}_.\end{align}$$

Making the proper compensation yields
$$\det(A)=-\dfrac{1\cdot 3\cdot (6\cdot 21)\cdot (-12)}{-21\cdot 6}=-36.$$