On the determinant of a matrix The matrix 
$$\left[\begin{array}{ccc}
30&20&30\\
40&50&20\\
30&30&20
\end{array}\right]$$
I tried solving it for myself and got $12000$, but math way tells me its $-1000$.
I dont understand how you get a negative, Where did I mess up my calculations?
I did
30 x ((50x20)-(30x20)) = 12000
      20 x ((40x20)-(30x20)) = 4000
      30 x ((40*30)-(30x50)) = -9000  
12000 - 4000 + (-9000) is -1000
oh I see! Sorry guys I messed up ):
Thx <3
 A: Hint: Note that: $$\left|\begin{array}{ccc} 30 & 20 & 30 \\ 40 & 50 &20 \\ 30 &30 &20 \end{array}\right| = 10^3 \left|\begin{array}{ccc} 3 &2 & 3\\ 4 & 5 &2 \\ 3 &3 &2\end{array} \right|$$
Can you do it now? Maybe not worrying with all these zeros makes it easier for you. :)
A: We want to find the determinant of
$$\begin{pmatrix} 30 & 20 & 30 \\
40 &50 &20 \\
30 &30 &20 \end{pmatrix}.$$
There are many ways of writing this down, but let's do it in the expand-by-minors way. First, since everything is divisible by $10$, let's factor out a $10$. Since it's a $3 \times 3$ matrix, this affects the overall determinant by $10^3$. So we look for
$$\begin{align}
10^3 \begin{bmatrix} 3&2&3\\4&5&2\\3&3&2 \end{bmatrix} &= 10^3 \left( 3 \begin{bmatrix} 5&2\\3&2\end{bmatrix} - 2\begin{bmatrix} 4&2\\3&2\end{bmatrix} + 3\begin{bmatrix} 4&5\\3&3 \end{bmatrix}\right) \\
&= 10^3 \left( 3(10 - 6) - 2(8-6) + 3(12-15)\right) \\
&= 10^3 (12 - 4 + -9) \\
&= -1000,
\end{align}$$
which is just as was claimed. $\diamondsuit$
A: $$\begin{vmatrix} 
30 & 20 & 30 \\ 
40 & 50 & 20 \\ 
30 & 30 & 20 \end{vmatrix} = 
10^3 
\begin{vmatrix} 
3 & 2 & 3\\ 
4 & 5 & 2 \\ 
3 & 3 & 2
\end{vmatrix} $$ 
To simplify the calculation of this determinant, we can use elementary row operations:
$\begin{vmatrix}
3 & 2 & 3\\
4 & 5 & 2 \\
3 & 3 & 2
\end{vmatrix}=
\begin{vmatrix}
3 & 2 & 3\\
1 & 2 & 0 \\
3 & 3 & 2
\end{vmatrix}=
\begin{vmatrix}
3 & 2 & 3\\
1 & 2 & 0 \\
0 & 1 &-1
\end{vmatrix}$
Now you can compute this determinant using Sarrus' rule. Since we have two zeroes there, it will be a bit simpler than in original determinant. Or you can continue with elementary row/column operations until you simplify the determinant a bit more. (For instance, if you get to upper triangular matrix, you can simply multiply the elements on the diagonal. Or if you have one row/column which only have one non-zero elements, the problem reduces to calculation of $2\times2$ determinant.)
You can check whether you have correct result, for example, on WolframAlpha.
