Navigating Linear Algebra Ep. 7: Determinant

[@LetsSolveMathProblems]

Thank you for those of you still keeping up with this series! In case you are curious, I plan to end this series (at least for this summer) with the release of Episode 10. I will give a list of some important topics *not* covered in this series in the video of Episode 10 or its description. The problems for this video can be found in this link: github.com/LetsSolveMathProblems/Navigating-Linear-Algebra/blob/main/Episode%207.pdf. Of course, please free to post solutions to these problems in the comment section, as well as peer-review other people's proofs. :)

[@cantcommute]

oh noo it's gonna end ;

[@particleonazock2246]

I am from Liaoning Province, China, and this really helped.

[@axemenace6637]

I may have missed something, but in 25:20, how do you know that the determinant of elementary matrices is multiplicative? Also, would the following be a valid (and similar) proof that A is invertible -> det A = 0? You can verify that multiplying a row by a constant m will multiply the determinant by m (consider the transpose, and the result follows from multilinearity) Adding one row to another does not change the determinant due to alternating property. Swapping two rows multiplies the determinant by -1. So, if the determinant of a matrix is nonzero, then applying elementary row operations keeps the determinant nonzero. Since the rref of A is just I (because A is invertible), and det I != 0, we conclude det A != 0.

[@leonilsonnunes3755]

I was anxiously waiting for Maths videos from your channel. Please, if you can produces more videos of Linear Algebra, I'll be really grateful. :)

[@LetsSolveMathProblems]

Thank you! I will almost definitely be making a few more linear algebra videos (up till Episode 10). :)

[@cantcommute]

For problem 2: We take the transpose of B as the hint suggested and expand over the first row. We do this again with the new determinant and find the det(B)=-90. The only elementary row operations that we should consider are scaling and swapping as they can change the determinant. Going backwards from (v) we see that we have to change the determinant in the following way: scale by 1/15, nothing, multiply by -1,nothing, scale by 1/2. Hence detA=3.

[@LetsSolveMathProblems]

For problem 2, this is correct.

[@axemenace6637]

​@@LetsSolveMathProblems I think that another interesting (and potentially more elegant) way of solving this is through the permutation formula. Note that the only permutation with nonzero coefficients is going to be the identity permutation (since the permutation must map e1 -> e1, e2 -> e2, and so on). So the answer is just the product of the diagonals.

[@cantcommute]

For problem 1: (a) We can apply a vector (a,b) to it and see that it sends it the equivalent vector if we just rotated it by 𝜃. This does not change the scale or orientation hence the determinant is 1. (b) Via multilinarity we can factor out the 10 three times and get 1000det(I) which is 1000. (c) Via the alternating property, it is zero. (d) Need 3 swaps to make it the identity matrix, hence its determinant is -1.

[@LetsSolveMathProblems]

For problem 1, this is correct.

[@cantcommute]

For problem 4: First we will shift this parallelogram up by three to get (-2,1),(-3,6),(-1,5),(0,0). Hence we can take the determinant of [(-2,1),(-1,5)] and take the absolute value (we know this is a correct choice between the three because they're linearly independent.) So the area is 9.

[@LetsSolveMathProblems]

For problem 4, this is correct.

[@MathwithMing]

This is how determinants should be introduced

[@cantcommute]

10:56 i was like 'he's gonna show us a universal property!!' but turns out it's a problem sheet problem lol

[@cantcommute]

For problem 3: Suppose at least one of A and B are invertible (wlog assume it's A), then there exists a series elementary row operations M1,...Mn such that M1M2...MnA=I. They are all invertible, and hence A=Mn'...M2'M1' where every Mi' is the inverse of Mi. The inverses are also elementary row operations but going backward. Hence the determinant of the product can be split and we have: det(AB)=det(Mn'...M2'M1'B)=det(Mn'...M2'M1')det(B)=det(A)det(B). If both A and B are not invertible, then their product isn't. As if it was, then at least one of A or B would be surjective. Hence via rank-nullity theorem, they're invertible which gives a contradiction. So det(AB)=0=det(A)det(B).

[@LetsSolveMathProblems]

For problem 3, this is correct.