Question 831410: i got this matrix :
3 0 0 -1
0 2 -2 1
-3 3 -2 2
6 1 0 1
i want to find the determinant, i got this :
|2 -2 1| | 0 2 -2|
3 |3 -2 2|- 0 + 0 + |-3 3 -2|
|1 0 1| | 6 1 0|
and got 3((2x-2)+(2x1)+2)+(42+42) = 84 ?
where do i got wrong ? Found 2 solutions by Elomeht, AnlytcPhil:Answer by Elomeht(22) (Show Source):
You can put this solution on YOUR website! The correct answer is 18.
It is very cumbersome to write in this forum.
Send me a "thank you" note and I will get it out to you in a PDF format.
Sorry, but it takes me forever to do it by long hand in this forum.
Best regards.
Don't listen to him. You can't stop and go "PDF-ing" everytime you need
to do a small problem.
First get a 0 in the upper left corner by multiplying the 4th column by 3
and adding it to the first column.
| 3 0 0 -1| | 0 0 0 -1| | 3 2 -2| | 3 2 -2|
| 0 2 -2 1| 3C4+C1->C1 | 3 2 -2 1| = -(-1)| 3 3 -2| = | 3 3 -2|
|-3 3 -2 2| | 3 3 -2 2| | 9 1 0| | 9 1 0|
| 6 1 0 1| | 9 1 0 1|
When you get down to a 3×3 determinant, it's often too tedious to
always expand it by minors. I'm sure your error was a sign or
something. Expand by minors until you get down to a 3×3, then
do the 3×3 this following way. You're less likely to make a mistake.
Copy the first two columns over to the right of the determinant
| 3 2 -2| 3 2
| 3 3 -2| 3 3
| 9 1 0| 9 1
Then multiply the three diagonals that slant this way \ and add them:
(3)(3)(0) + (2)(-2)(9) + (-2)(3)(1) = 0-36-6 = -42
Then multiply the three diagonals that slant this way / and add them
(-2)(3)(9) + (3)(-2)(1) + (-2)(3)(0) = -54 - 6 - 0 = -60
Then subtract (-42)-(-60) = -42 + 60 = 18
Edwin