On the determinant of a matrix

Question

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

Answer 1

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. :)

Answer 2

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. $\diamondsuit$

Answer 3

$$\begin{vmatrix} 30 & 20 & 30 \\ 40 & 50 & 20 \\ 30 & 30 & 20 \end{vmatrix} = 10^3 \begin{vmatrix} 3 & 2 & 3\\ 4 & 5 & 2 \\ 3 & 3 & 2 \end{vmatrix} $$

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.