0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ The submatrix determinant $\begin{vmatrix}-3&-3\\3&0\end{vmatrix}$ is $9$. $\endgroup$
    – peterwhy
    Jan 7, 2021 at 18:48
  • $\begingroup$ I would have thought the determinant of your second matrix was lots of zeros plus $1 \times (3 \times 1 - (-3)\times 3)=9$ $\endgroup$
    – Henry
    Jan 7, 2021 at 18:49
  • $\begingroup$ 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.. $\endgroup$
    – Pwaol
    Jan 7, 2021 at 18:50

1 Answer 1

2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.