determinants

[jfsfroggy]

evaluate the determinate.

0 1 5 4
0 4 5 9
1 9 2 7
1 2 9 8

can anyone please help!!!!

 

[stapel]

What evaluation method are you using? (Minors and cofactors? Doing row operations first? etc) How far have you gotten?

Please reply showing all of your steps so far. Thank you.

Eliz.

 

[tkhunny]

evaluate the determinate.

0 1 5 4
0 4 5 9
1 9 2 7
1 2 9 8

can anyone please help!!!!

Ever hear of "Expansion by Minors"? That first column looks particularly useful.

0*stuff - 0*otherstuff + 1*(40+144+90-40-81-160) - 1*(35+32+405-180-18-140)

 

[jfsfroggy]

i haven't even started i have no idea where to begin with this. i understand how to do the 3x3. my teacher said something about a quick method that gets you down to the 3x3, but i dont know how to get it down to the 3x3.

 

[jfsfroggy]

ok am i on the right track here...
0(-1) 1 5 4 +4(+1) 0 5 4 +5(-1) 0 1 4 +9(1) 0 1 5
9 2 7 1 2 7 1 9 7 1 9 2
2 9 8 1 9 8 1 2 8 1 2 9

 

[soroban]

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!

 

[jfsfroggy]

thank you so much soroban you are a life saver.