Use the method of elimination to evaluate the determinants 
So this is my process, but I used calculator to check if I got them right or not. and it seems like i got both of them wrong. number 6 supposed to be 36 and 1 supposed to be 135. can anyone please explain where did i do wrong?
 A: The operation $\frac{1}{3}R_3$ changes the determinant, dividing it by $3$; similarly, the operation $R_2+2R_3\to R_3$ multiplies the determinant by $2$.
Further, the last operation leaves $1$ in position $(3,3)$. As a consequence, the determinant is
$$
1\cdot(-2)\cdot1\cdot 3\cdot\frac{1}{2}=-3
$$
which the computer confirms:
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             i386 running darwin (x86-64/GMP-6.1.2 kernel) 64-bit version
          compiled: Jul 19 2018, Apple LLVM version 9.0.0 (clang-900.0.39.2)
                               threading engine: single
                    (readline v7.0 enabled, extended help enabled)

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parisize = 8000000, primelimit = 500000
? a=[1,-3,-3;-1,1,2;2,-3,-3]
%1 = 
[ 1 -3 -3]

[-1  1  2]

[ 2 -3 -3]

? matdet(a)
%2 = -3

The same program confirms that the second determinant is $135$; indeed, the last operation you do multiplies the determinant by $-29$; the others don't modify it.
A: For the first determinant, there is a final error: the last determinant should be
$$\begin{vmatrix}
1&-3&-3 \\
0&-2&-1 \\
0&\phantom{-}0&\phantom{-}\color{red}1
\end{vmatrix}, $$
so the final determinant is $-2$.
There are also two conceptual errors:


*

*The step $\frac13 R_3\to R_3$ multiplies the determinant by $\frac13$;

*the step $R_2+2R_3\to R_3$ multiplies the determinant by $2$.


So to get the original determinant, you have to multiply the final determinant by $\frac32$, which yields $\color{red}{-3}$ as the sought determinant.
