Inverse Matrices
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- Опубликовано: 18 окт 2024
- MIT 18.06SC Linear Algebra, Fall 2011
View the complete course: ocw.mit.edu/18...
Instructor: Ana Rita Pires
A teaching assistant works through a problem on inverse matrices.
License: Creative Commons BY-NC-SA
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A more general solution to this question(Takes into account complex numbers too) is: when a^3 -2(a^2)b+a(b^2)!=0 . (i.e. not equal to zero). You can check, this also satisfies the conditions that she stated above.
It takes courage to make a mistake and admit it before your peers. Bravo.
Peers?
But she was a masters' student at the recording of this video.
pretty neat way of deducing the conditions
love this course
A is convertible is equivalent to Ax=0 has only one solution [0; 0; 0]. Let x = [x1;x2;x3], Ax=0 => ax1+b(x2+x3)=0 and a(x1+x2)+bx3=0 and a(x1+x2+x3)=0, which reduces to (b-a)x2 =0 and a(x1+x2+x3)=0. Thus, if only [0; 0; 0] solves Ax = 0, then we have a/=0, and a /=b;
4:11, thats what she said
4:01 augmented matrix row reduction
best q so far i think
At 7:24 she computes the top left component of the augmented matrix to be: 1 / (a-b)
She says that row 1 is: row1 - b/a(row2 + row3)
So the computation starts with: 1/a - (b/a * (-1/(a-b) + 0))
What are the steps to end up with 1/(a-b)?
1/a -(b/a)(-1/(a-b)) -> (a-b) / (a(a-b)) + b / (a(a-b)) -> (a - b + b) / (a(a-b)) -> a / (a(a - b)) -> 1 / (a-b)
Combine fractions with common denominator in what you did and it resolves to 1/(b-a).
When did the professor talk about easy tests to spot if a matrix is not invertible?
Lecture 3
great recite!
But getting rid of those (a-b) would mess up the identity on the left
0:46 make 3 by 3 matrix invertible
7:58 didnt determine an amd b
I tried to invert this matrix and I used a bit different steps without breaking any rules
but I got another inverse.
Does a matrix have more than an inverse? Or it has a unique one?
unique
for every matrix there is only one inverse.
Thank you. :)
5:09 if sum row2 and row3 with row 1 and divide row1 by 'a' after this... there is no computationally heavy thing!
Correct me if I am wrong, but I think the first row would be made of zeros and there would be no non zero entry in (1,1) in that case
6:15 why would (2,2) automatically equal to 1? I think there wasn't a mistake at the first time. if you come up with 1/a-b and if you want it to equal to 1 then a-b should equal to 1. can someone please explain this to me thank you.
The thing she is doing in that step is to make he entries of the diagonal equal 1. To do so she makes the elementary operation of dividing a row by a scalar, and that scalar being each row's pivot. That second row is being divided by a-b, so the first term is going to be 1. Hope it helps.
at 5:04, can you just assume a to be 1 and b to be 0?
which makes it to be identity matrix
no, the whole idea of the question is to get the generic form of the inverse and not to assume anything. I guess by now you have already passed the course ! enjoy :)
a , a-b, a-b common liya lkein kaise bhai koi btayega
tak a common from A and do same in identity matrix. Perform this operation for a-b too.
When you take out 1 / a -b out of the Matrix in shouldn't it be cubed power. Since you were taking out from each row
yea, good question
@@Abhi-qi6wm I figured it out it's actually in the case of determinant when you pull a scalar out you have to do it row wise but when it comes to a matrix if you'll pull a scalar out you pull it out of the whole Matrix
@@MrSazid1 Yess, I remember it now. Thanks for the reply 😄
That back bench student in the class