Hello, jfsfroggy!
Am i on the right track here...
\(\displaystyle \;\;0(-1)\begin{vmatrix}1 & 5 & 4 \\ 9 & 2 & 7 \\ 2 & 9 & 8\end{vmatrix}\;+\;4(+1)\begin{vmatrix}0 & 5 & 4\\1&2&7\\1&9&8\end{vmatrix}\;+\;5(-1)\begin{vmatrix}0&1&4\\1&9&7\\1&2&8\end{vmatrix}\;+\;9(+1)\begin{vmatrix}0&1&5\\1&9&2\\1&2&9\end{vmatrix}\)
Yes, you are . . . Good work!
I see you "used" the second row.
Did you know that you can "use" any
column, too?
The first column is the best (as tknunny pointed out) . . . look at the 0's!
We have: \(\displaystyle \;0(+1)\begin{vmatrix}- - \\ any \\ - - \end{vmatrix} \;+\;0(-1)\begin{vmatrix}- - \\any \\ - -\end{vmatrix} \;+\;1(+1)\begin{vmatrix}1 & 5 & 4\\ 4 & 5 & 9\\ 2&9&8\end{vmatrix}\;+\;1(-1)\begin{vmatrix}1 & 5 & 4\\ 4 & 5 & 9\\ 9 & 2 & 7\end{vmatrix}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
You probably haven't learned about Row Operations yet, but I'll give you a taste.
We can subtract any row from any other row, term by term,
\(\displaystyle \;\;\)and still get the same value.
In your example, if we subtract row 4 from row 3 (and replace row 3),
\(\displaystyle \;\;\)we get: \(\displaystyle \:\begin{vmatrix}0&1&5&4\\0&4&5&9\\ 0&7&-7&-1\\ 1&2&9&8\end{vmatrix}\)
Using the first column, we have: \(\displaystyle \:1(-1)\begin{vmatrix}1&5&4\\4&5&9\\7&-7&-1\end{vmatrix}\;\;\)
. . . that's all!