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Question

Suppose $A$ and $B$ are $n 	imes n, B$ is invertible, and $A B$ is invertible. Show that $A$ is invertible. [Hint: Let $C=A B,$ and solve this equation for $A .$ ]

Angelo Rendina · Accepted Answer

So let&#x27;s go see a times B and seizing veritable So C minus one exists. So what we do here is we multiply both sides from the right by the inverse of C So we have C see in verse equal a B see embers. But now see, see in verse is by definition of inverse get entity metrics And on the inside we have a me CM birth Now about the associative property or matrix multiplication. We can put the brackets here and this formula is telling us that BC in verse he&#x27;s the inverse away so easy in vegetable.

Joshua Eastwood · Answer

this question tells us that we have three matrices A, b and C, and we&#x27;re trying to If if a is in vertebral, prove that a times B was he means that b is equal to see. Okay, So what we&#x27;re gonna do is we&#x27;re gonna multiply by a inverse on both of these because we know that a is invulnerable and only told me that. And then what happens is a inverse times a that becomes the identity matrix which then leads me to i a Times B equals I time see? And if you multiply I times be the eid any major times bu get B and if you multiply the identity matrix time see you get See So therefore, if this is true, if a B is equal a see and a vertical and B is equal to see

Angelo Rendina · Answer

so we have be minus C that multiplies the equal to zero. So since the is in vertebral, we can multiply from the right by the inverse of the so right B minus c the and then the inverse. And that&#x27;s equal to zero times Demers. Now the product the D in verse is just it. Entity metrics. So this becomes B minus c that multiplies the again. Did he is equal to now Any metrics multiplied by 00 So on the right hand side, we just have zero and now must be bribed by the identity matrix does nothing. So on the left hand side we have B minus c and on the right inside we have zero Now we some ah si on both sides. So that&#x27;s right, B minus c and then we add plus C on the right. Inside we have zero Blasi Well now minus C plus C is just zero. So are we on the left inside? We just have B and on the right inside something zero does nothing. So we&#x27;re just left with C, which is what we wanted

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.

Runpeng Li · Answer

So in this problem, we&#x27;re going to show if a b a Times B is in veritable Hubert. Oh, then he is in vertical. Okay, So in order to show this by our convertible matrix B room, we know that Bye. I am t. There exists a matrix c Such that. See, time&#x27;s a b, his identity. So that is equivalent to guess the matrix multiplication, he said. So associative associative. So we have see, Time&#x27;s a time speedy his identity. So Dad means See, Time&#x27;s a is the inverse. Uh, so we conclude that be is convertible and we are done.

Michael Jacobsen · Answer

in this video, we&#x27;re assuming that the Matri, CES and B are both convertible and of size and buy em Merkel. Here&#x27;s to show that its product, a times B, must also be in vertebral. First, let&#x27;s write the following Since a is convertible, we can write a as the following factories ation. They will be equal to E one times e to up to E. K, where each of these matrices e i our elementary matrices. The reason we could do this is a being convertible will happen if and only if we can do this decomposition here for elementary matrices. Likewise, we could do the same for B. Well, right? Also B, it&#x27;s in vertical can be written as Let&#x27;s Use d one d two through d. C. Api, where each of these major sees d i. R also elementary matrices. Now let&#x27;s look at a Times B will say then the product a Times B can be written as first take these major cities, he won times e to multiplying all the way through Ikea, and this multiplies these matrices here coming from B So take d one d two all the way through diese P now Let&#x27;s look at what we have here. This is still a product of elementary matrices. So then vertical being tricks there and tells us that then a B is in vertebral since I could do a better see their. Since A B is a product off elementary matrices, this completes this proof.

Problem

Solve the equation $A B=B C$ for $A,$ assuming th…

Problem 16 Easy Difficulty

Suppose $A$ and $B$ are $n \times n, B$ is invertible, and $A B$ is invertible. Show that $A$ is invertible. [Hint: Let $C=A B,$ and solve this equation for $A .$ ]

Answer

$A$ is invertible

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

Linear Algebra and Its Applications

Alayna H.

Kayleah T.

Maria G.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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$A$ is invertible

Linear Algebra and Its Applications

Alayna H.

Kayleah T.

Maria G.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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

Alayna H.

Kayleah T.

Maria G.

Kristen K.

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

Linear Algebra and Its Applications