# Problem finding determinant using elementary row or column operations.

I'm trying to find this determinant using row and column operations, but I got $$-9$$ as an answer and the right answer is $$9$$ and I couldn't figure out my mistake. $$\begin{vmatrix} &{1}&&{3}&&{4}&\\ \\ &{3}&&{6}&&{9}&\\ \\ &{1}&&{6}&&{4}&\\ \end{vmatrix}$$ So what I did was I removed $$3$$ times column $$1$$ from column $$2$$, and after that I removed $$4$$ times column $$1$$ from column $$3$$. and I got: $$\begin{vmatrix} &{1}&&{0}&&{0}&\\ \\ &{3}&&{-3}&&{-3}&\\ \\ &{1}&&{3}&&{0}&\\ \end{vmatrix}$$ but now, I tried to calculate the determinant according to first row and got $$(-1)^2 *1 *(-9)=-9$$
Am I missing something? I would be happy if someone can tell me where my mistake is. Thanks in advance.

• The submatrix determinant $\begin{vmatrix}-3&-3\\3&0\end{vmatrix}$ is $9$. Jan 7, 2021 at 18:48
• I would have thought the determinant of your second matrix was lots of zeros plus $1 \times (3 \times 1 - (-3)\times 3)=9$ Jan 7, 2021 at 18:49
• Thanks everyone, I messed up with that .. can't believe I have been just trying to find where my mistake is for like 30 mins.. Jan 7, 2021 at 18:50

$$\begin{vmatrix} 1&0&0\\ 3&-3&-3\\ 1&3&0 \end{vmatrix}=1\begin{vmatrix} -3&-3\\ 3&0 \end{vmatrix}=-3(0)-3(-3)=+9$$