Question

Algebra 2 help needed

Answer 1

`\det \left[\begin{array}{ccc}3&3&1\\-2&0&-2\\-2&4&-3\end{array}\right] =\\ 3\cdot0\cdot(-3)+3\cdot(-2)\cdot(-2)+1\cdot(-2)\cdot4-1\cdot0\cdot(-2)-3\cdot(-2)\cdot4-\\ -3\cdot(-2)\cdot(3)= 0+12-8-0+24-18=\boxed{10}`