Inverse Matrices

MIT 18.06SC Linear Algebra, Fall 2011
Instructor: Ana Rita Pires

A teaching assistant works through a problem on inverse matrices.

License: Creative Commons BY-NC-SA

[inverse_functor]

It takes courage to make a mistake and admit it before your peers. Bravo.

[tanmaybhayani]

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.

[ixine-fxwd]

pretty neat way of deducing the conditions

[angfeng]

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;

[rddegreeyt]

I would straight write that det(A) should not equal to 0 instead of the whole fuss of inequalities

[captainheretic]

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)?

[vardhabhai]

When did the professor talk about easy tests to spot if a matrix is not invertible?

[SpeaksYourWord]

But getting rid of those (a-b) would mess up the identity on the left

[moayadyaghi]

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?

[sohrabgh]

5:09 if sum row2 and row3 with row 1 and divide row1 by 'a' after this... there is no computationally heavy thing!

[delongzhai]

at 5:04, can you just assume a to be 1 and b to be 0?

[JulianAC]

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.

[clashroyalekrypton]

a, a-b, a-b common liya lkein kaise bhai koi btayega

[MrSazid]

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