Determinant of a 4 x 4 Matrix Using Row Operations
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- เผยแพร่เมื่อ 16 ก.ย. 2024
- Linear Algebra: Is the 4 x 4 matrix A = [ 1 2 1 0 \ 2 1 1 1 \ -1 2 1 -1 \ 1 1 1 2] invertible? We test invertibility by checking the determinant. We compute the determinant by performing row operations before using cofactors.
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Easily the best video on this topic you can find on youtube.
He seems so tough, hardcore, and clear. Very helpful. Thanks!
So simple, but yet perfectly detailed. Nice job and thnx, this helped a lot.
you make great videos for linear algebra!
Thanks! It's my favorite subject, for a broad definition of linear algebra.
were u in the army?
Why can't everyone just teach like this? Thank you in advance if this question is on tomorrow's final! :) Happy New Year!
Explaining math well is very hard. Explaining math well to a camera will harm your sanity. :)
The one time I lost it, I was about 30 takes into a 3 minute segment - had it nailed and UPS rang in the last 30 seconds. Boom. - Bob
MathDoctorBob if we do row operations then the detarminant value might be changing. please explain to me.how the subtractions of row does not affect the detarminant??
MathDoctorBob is the best one on youtube !
You have done well
thank you dr. Bob .I realy enjoyed watching this .
You're welcome, and thanks for the kind words! I have a Linear Algebra playlist which is mostly solutions to old exam problems. The links are nicely organized at the website listed at the channel page.
It depends on the matrix and whether you have technology to use. Cramer's Rule is usually not optimal when the size is large. If determinant = 0, you'll need row echelon form to solve (many solutions or none).
I don't think I have one specifically. It only works for a 3x3 matrices. Set the matrix next to itself. Multiply down the first three diagonals to the right and add. Then multiply down the next three diagonals, but to the left, and subtract.
You can check that it agrees with the general formula or other methods.
@Diabianeya You're welcome. Thanks for the comment. - Bob
@DJAnthonyillWill Calculators are a mixed blessing. They save time, but you need to calculate to develop intuition. And then there's the professor thing. - Bob
@timugin Thank you for the high praise! - Bob
Just did it in my TI-83 and saved 4 minutes of time lol. Too bad my professor didn't allow us to use our calculators on our last exam involving matrices
It can be either. It will depend on the given matrix. Of course whichever has the most zeros will require less work. - Bob
@robmunene33 You're welcome, and thanks for the comment. I'm glad to be of help. Please let me know if you have any requests. - Bob
You're welcome!
We need to get those A and Ainv in the determinant to make sense.
What can we assume? If A and B are 2x2, brute force cofactor expanding is manageable to get both = det(A)det(B). If you have det(XY)=det(X)det(Y) to use, we can factor the second matrix.
Thanks! That's just cruel - finding it for a 3x3 is enough to check understanding. I'd avoid row reducing, but make sure you know the diagonal trick for 3x3 matrices.
You have five coefficients. The lead is always 1. The next one is always
minus Trace(A). The last one is always (-1)^4 det(A) = det(A). Trace is easy, det probably not.
Some checks: det is 0 means you have eigenvalue 0 at least once. If all rows (or columns) sum to the same number, you have an eigenvalue of that sum.
@husamaldean You're welcome! - Bob
Thanks
@rockyelz You're welcome! - Bob
finally i understand it thamk you :)mr.Bob....
thanks, do you anymore of this your a so good.
@Tigeress8482 You're welcome, and good luck on exams! Happy 2012! - Bob
You're welcome. Please let me know if you have any requests.
thank you!
where is the diagonal trick video to find out 3 by 3 matrix determinant sir.
Great videos, I'm gonna have to find the characteristic polynomial det(A-lambda*I) of a 4x4 matrix in my exam.
The process takes forever, and row/column reductions are not obvious because of lambda being in the way.
Any tricks would be appreciated.
Oh, I think I got it. Since A and B are 2x2 matrices, the operations are done twice, so we end up having A*A*-A^-1*-A^-1*det() = -I*-I*det=I*det(), where I=1
thank you very much :)
This video saved my butt on my first test. Do you have a video that explains vector spaces and axioms and the stuff from chapter 4 of the howard anton 11th ed. linear book?
Thanks! There are two ways to teach linear algebra: for engineers, or for mathematicians/phyicists. Most of the first semester linear algebra videos are shot for engineering style, so not much abstract nonsense.
If we are working with a block matrix (let's say C = [[A,0][-I,B]]
we want to show that det C = det [[-I,B][0,AB]]
we apply the following row operations : (1) R2 --> AR2 + R1; (2) R1 --> R1-R2; (3) R1--> -A^(-1)*R1; to get [[-I,B][0,AB]].
Can we apply the properties on block matrices? So we would get the following modifications to the det : A*-A^-1*det([[-I,B][0,AB]]), which leads to -I*det([[-I,B][0,AB]]), which is the negative...
or when dealing with block matrices, det stays the same?
To solve a linear system of 5 equations and 5 unknowns, I assume you would use a 5x5 matrix and Cramer´s Rule? Is my assumption correct MathDr.?
where A and B are 2x2 matrices