Once again I'm doing a math team practice packet. The problem was: I found the determinant of a 3x3 matrix the way I know how to, which is:
Determine all values of $x$ for which the determinant of $\begin{pmatrix} 1 & x & 2\\\ 3 & 4 & 5\\\ 6 & x & 7 \end{pmatrix}$ equals zero.
I found the determinant of a 3x3 matrix the way I know how to, which is: $$\det\begin{pmatrix} a & b & c\\\ d & e & f\\\ g & h & i \end{pmatrix} = a \times \det\begin{pmatrix}e & f\\\ h & i\end{pmatrix} - b \times \det\begin{pmatrix}a & b\\\ c & d\end{pmatrix} + c \times \det\begin{pmatrix}d & e\\\ g & h\end{pmatrix} $$
I solved the problem using the way I know how, and I got some random fraction when the answer is supposed to be 2. I then looked at the way that they solved the problem and the formula they used was:
$$determinant\ = -b \times \begin{pmatrix}d & f\\\ g & i\end{pmatrix} + e \times \begin{pmatrix}a & c\\\ g & i\end{pmatrix} - h \times \begin{pmatrix}a & c\\\ d & f\end{pmatrix} $$
Using those numbers, it simplified to $0=10x-20$ which means $x=2$
Going back to my original question, is there a different way that you need to find the determinants of matrices based off of what you are given or am I missing something? I'm mostly wondering why they used the formula that they did instead of the standard 3x3 determinant finder.
One approach avoids computing the determinant completely.
The determinant is zero if and only if the column vectors are linearly dependent. Since the left column and right column are independent, this means the middle column, $\begin{pmatrix}x\\4\\x\end{pmatrix}$ must be a linear combination of the left and right column.
This means there must be an $a,b$ such that: $$a+2b=x, 3a+5b=4, 6a+7b=x.$$ Then you have three equations in three variables, and it is easily solved, $a=-2,b=2,x=2.$
This works because all the occurrences of $x$ are in one column. The same would work if all occurrences of $x$ were in a row.