in-exercises-3136-mention-an-appropriate-theorem-in-your-explanation-let-a-and-b-be-square-matrices-

Question

In Exercises 31–36, mention an appropriate theorem in your explanation.
Let $A$ and $B$ be square matrices. Show that even though $A B$ and $B A$ may not be equal, it is always true that det $A B=\operatorname{det} B A$

Chris Trentman · Accepted Answer

were given a statement and were asked to prove this statement using the theorems from the section. So let A and B B square matrices and were asked to show that even though multiplication is not necessarily communicative so that a B is not necessarily equal to be a it is always true that the determinant of a B is equal to the determinant of the matrix B A no, we have since both and beer square matrices It follows that the major sees a bee and be a are also square matrices. And so the determinant is defined and by the&#x27;re, um six we have of the determinant of a B is equal to the determinant of a times the determinant of B which is equal to since scaler zehr communicative determinant of b times the determinant of a and then again by the&#x27;re, um six. This is the determinant of B A. So we see that these two determinants are

James Chok · Answer

okay. In this question, we want to show that determines right here is equal to determine off, eh? So how do we do this? You first separate out the products. Then it will be determines off a as a determinant four p invest. So what we do is we arranged. We rearrange. Hey, turn it off. He even this turn this all day. So pushing in the two peas? Yeah, he pilot, he in this. How&#x27;s my determinant? Okay, now, Pete Pete in verse is the identity matrix Identity matrix has, like, determinants off, eh? Now, determinants off the identity matrix is just simply one. So therefore, this is just a terminus, okay?

Nicholas Barvinok · Answer

We want to verify that for two by two matrices A and B that the determined of a times B it was the determinant of a times the determinant to be alone. Let&#x27;s just fix to arbitrary two by two matrices So a B C D multiplied by, well, normal&#x27;s a B C D. Call that a and be will be F G h k perhaps J. We don&#x27;t want eyes, but Jay&#x27;s okay, actually, I was okay too. Let&#x27;s do I f g h I skip tea. So the determined of well, let&#x27;s let&#x27;s first find the product So a being by matrix multiplication Where we get we get a F plus b h a g plus b i cde f plus de age C g plus d I. And from this we get that the determinant of a B is a f plus. The h times C g plus the I&#x27;m minus a G plus the eye times cf waas the age And on the other hand, we get that the determinant of a times The determinant of B is a D minus. BC, multiplied by f i minus gh and this gets us If we expand everything a loose a D death I minus a d g h minus B c f i plus B c g h. Now let&#x27;s actually expand the determinant of the product on another pain because we are running out of room so we get determinant of a times B is a F plus B H time CG plus D I minus a G plus B I times cf plus the h And if we expand this, we get a F C G plus a D I plus B h c g plus b e h d I minus a g c f minus a g d age minus b i c f minus b i d h. So let&#x27;s see, Does anything cancel? Yes, we have a f c g a f c g here. So these to cancel we have be hd I hear and be hd I hear. And now we have four terms. FBI plus B H C G minus a g D h minus. B I c f. So here we have positive a deify here we have a positive i d. If I Here we have a positive B c g h. Here we have a positive bcg age. Here we have a negative 80 th nated h gh. And here we also negative B i c f negative b i c A. So this formula does hold for two by two matrices and we&#x27;re done.

Ankit Gupta · Answer

en garde A on this bush in and be, uh, white practices. So endless be I was his is endless Gave me us. You did? Oh, it must be. It is. It must be all right. Just give us. No. You know that Majesty, my deep love emerges a Is that that is a It&#x27;s in me at this being just okay. Just be Are deep love eats now adding it all he was hes with this and me as okay, us. Okay. Now comes to party. No, it is no necessary that aimless. We always there is good us. Who? Maybe let&#x27;s be script. It&#x27;s Yeah, this well.

Michael Jacobsen · Answer

in this video, we&#x27;re provided with two Major sees A and B, and we&#x27;re told that they&#x27;re convertible and of the same size and buy in. Our goal now is to show that the product a times B, must also be in vertebral. The way we&#x27;re going to achieve this proof is by the portion of the vertical matrix there. Um, we see down below. Here we&#x27;re told that if an inverse exists, that will happen if and only if this matrix equation has only the trivial solution. So lets us argue to the contrary. Let&#x27;s say assume that a Times B Times X equals zero Vector has a nontrivial solution. Let&#x27;s call that nontrivial solution you. And because we said it&#x27;s not trivial, it&#x27;s not equal to the zero vector. Now let&#x27;s consider what this would mean. We can say. Then eight times be times are solution. You is the zero vector since we said you is a solution to the equation above. But we can re associate the parentheses on this matrix and vector product. Here we can write that eight times the vector B times you equals zero vector. Now, this is where the proof gets exciting because we&#x27;re told a is in vertebral and now we&#x27;re looking at the equation A X equals zero where X is playing the role of B times. You we can say, since a is convertible, we have that b times. You must be equal to the zero vector because eight Times X would only have the trivial solution which is the zero vector here now at the same time be is an in vertebral matrix. So we can say since be is convertible, we have that you must be the zero vector But that&#x27;s contrary to what we said at the beginning. That to you is not the zero vector. So we have a contradiction. We can resolve our contradiction as follows will say. This contradiction shows that a Times B times X equals zero vector. This equation must have on Lee the trivial solution which would be X equals zero vector. The reason we have this contradiction and Louisa reason we can assume we only have the trivial solution is because the moment we assumed we did not have a trivial solution, you not equal to zero. We obtained this contradiction you then equal zero. The contradiction is avoided by assuming that we have on Lee the trivial solution. But now we go back to the vertical matrix theorem again, since there&#x27;s an if and only if right here. Knowing that this equation has only trivial solution allows us to make this following conclusion, I will say next that this shows that a Times B is in vertebral and now the proof is complete.

Geena Pullo · Answer

s so we have A and B which are both two by two matrices on We have a plus B squared. So a plus B squared is equal to a plus B times a plus B And because they&#x27;re both two by two matrices, we know we could technically add them together. Um, but this is equal to a squared plus eight times B plus B times a plus be squared. So, yes, it is equal to that. And this is with those A B is not equal to be a, um, which is true, because in this case, uh, we don&#x27;t know what A and R p r. But this is possible because of the fact that matrix multiplication is not communicative. Um, And so now we have our second part, and we have is a plus B squared, always equal Teoh a squared plus Teoh a B plus B squared. And the answer is no. So I&#x27;ll say no. Um because that would only be true. Um Onley True, when a B is equal to be a um so when a be is equal to be a then a plus, B squared is equal to a plus p times a plus B, and that&#x27;s equal to, um, a squared plus a B plus be, um plus B squared. And then that would be equal. I want to say one here, um, a skirt plus to a be plus B squared. But this isn&#x27;t always true, of course, because, as we said earlier, Major specification is non community of And so because of that, um, we could have our A and R B R A B and R B a. Excuse me not equal to each other. So this would only be true if they were both equal to each other.

Problem

In Exercises 31–36, mention an appropriate theore…

Problem 33 Easy Difficulty

In Exercises 31–36, mention an appropriate theorem in your explanation.
Let $A$ and $B$ be square matrices. Show that even though $A B$ and $B A$ may not be equal, it is always true that det $A B=\operatorname{det} B A$

Answer

see the proof

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Caleb E.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

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see the proof

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Caleb E.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Catherine R.

Anna Marie V.

Caleb E.

Maria G.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications