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Vous préparez un test ? Révisez avec ces 16 leçons sur Les matrices.
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Transcription de la vidéo
Voiceover:In order to get into Battle School cadets have to pass a rigorous entrance exam which includes mathematics. Help Commander Graff grade the next wave of students' tests. The last step of a problem in the matrix multiplication section is the matrix A times B times C where A, B, and C are square matrices. Which of the following candidates have answers that are equivalent to this expression? Select all who are right for any A, B, and C. Select all that apply. I encourage you to pause this video and think about it. Which of these expressions for any square matrices A, B, and C are equivalent to this right over here. I'm assuming you've given a go at it. Let's think through each of them. This one is B, A, C so if they have changed the order and we've already seen that matrix multiplication is not commutative in general and so this will not be true for any square set of matrices A, B, and C, so this is not going to be true. Matrix multiplication is not commutative. Here you have Bernard, who says A times, C times B. We already know that that's going to be the equivalent to A, C, B which once again they've swapped the order between the B and the C, matrix multiplication is not commutative. You can't just swap order and expect to get the same product for any square matrices A, B, and C so we could rule that one out. A times, B, C, so we've already seen matrix multiplication is associative, so this is the same thing as A times B, times C which of course is the same thing as A, B, C. What Caren has right over here, that is right, that is equivalent for any square matrices A, B, and C that is equivalent to A, B, C. Now Ducheval, let's see, now this looks like a bit of a crazy expression but let's think it through a little bit. First of all matrix multiplication, as long as you keep the order right, the distributive property does hold. This first part right over here is equivalent to ... Let me write this down, this one's interesting. We have A times B, C plus A, minus A squared. You can actually distribute this A and I encourage you to prove it for yourself maybe using some two by two matrices for simplicity. This is going to be equal to, this part over here is going to be A, B, C plus A A, A times A which we could write as A squared and then we're going to subtract A squared. These two things are going to cancel out, they're going to end up being the zero matrix and if you take the zero, so these are going to be the zero matrix right over here. If you take the zero matrix and add it to A, B, C you're just going to end up with A, B, C. This one was a little bit tricky, this one actually is equivalent. This one is right and this one is right. Here this is A times B, plus C so this is kind of not-y right over here. They're not even multiplying B and C, so this one's definitely not going to be true for all square matrices A, B, and C.