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Question

If $A, B,$ and $C$ are $n 	imes n$ invertible matrices, does the equation $C^{-1}(A+X) B^{-1}=I_{n}$ have a solution, $X ?$ If so, find it.

Michael Jacobsen · Accepted Answer

for this exercise were provided with three matrices A, B and C, and we&#x27;re told that these matrices are in vertebral and that their size is and buy in. Well, what we&#x27;re going to do next is consider a matrix equation where the unknown is also a matrix. The equation is see in verse times a plus. X times be in verse equals the end by N identity matrix, and the unknown is X. But here X is going to be considered a matrix. The reason it must be a matrix is because of the addition here. That means X must be a matrix of the same size and buy in now to solve an equation like this. One thing we can do is pre multiply by a matrix C. Then it will multiply the left side of the equation by sea in verse, times a plus X times be in verse and the right side of the equation will be see times the identity matrix. Now the next step here we know that there is a cancellation with sea by sea inverse. It results in the identity matrix, but there is no difference when we multiply here with identity we still get a plus X times be in verse for the left side, on the right hand site we get See by the same logic. Now recall. We&#x27;re trying to isolate X here, so the next problem is that be inverse matrix. But we post multiply by. Be on the same side of that both sides of the equation we&#x27;ll have in another elimination. Just as before, this turns into the identity matrix and eight times x times. The identity matrix is still a plus X, so the right hand side is see. Time&#x27;s be left hand side is a plus X, the less operation we can take to solve for the Matrix A is subtract a from both sides, so we next have that X is equal to see times B minus A, and this is the solution to our matrix equation. Let&#x27;s indicate what we&#x27;ve shown. We&#x27;ve shown that if this matrix equation has a solution than the solution must be see time&#x27;s B minus A to really prove without shadow of a doubt that we know that this is the solution. What you need to do is take this solution substituted here and show that we get a true statement that the left hand side would be thy Danny Matrix

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 example, we&#x27;re starting out with the square matrix of size A buy in for a and then two more matrices B and C, which are of size and by P. Next, let&#x27;s consider the Matrix equation that a Times B is equal to a time seat. This is an interesting equation to look at because if we were dealing with ordinary real numbers, we would be able to divide by a and conclude B is equal to see. But for matrices, that is not always the case. We need extra assumptions. So let&#x27;s assume that a inverse exists or a different way of saying this. We&#x27;re assuming that a is an inverted matrix. If that&#x27;s the case, what this allows us to do is to take a inverse here and let&#x27;s left multiply by a inverse life. Have a inverse times A B equals a inverse times a C. Now we know, since A is in vertebral a times, a inverse is equal to the identity matrix. So we&#x27;ll have identity matrix here on the left side and an identity matrix here on the right side. But that results in B Times C. Since any matrix times identity gives back that matrix. So if we have that a inverse exists, then this equation does indeed imply that be equal C. However, if a inverse does not exist, then be is not necessarily equal to see. For example, it&#x27;s possible to come up with the Matrix B so that this results in zero. But the right hand side does not, and so we can&#x27;t conclude that B is equal to see unless we know that a inverse exists.

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 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.

Adhish Rele · Answer

in this example we have been given three atr isis each off order in their old square matrices. So it&#x27;s an engrossing metrics and these that it&#x27;s fair to conditions. The 1st 1 being be prepared by a gives you the under to meet tricks and see multiplied. But it also gives us the admitting me tricks and we need to prove that b equals c. So first mooted noticed that since be military but he gives us the identity matrix. This just in place a modifier baby will also give us the energy matrix because envy border in crossing the traces on since be multiple body is the energy matrix. This means in Watts equals B and B in must also equals e. And this weekend I don&#x27;t this equation that is be multiplied by a is equal identity matrix. Similarly, since see modifiable e gives you the energy matrix is also in place E multiple Basie equal to the energy matrix in order to start with B no beacon bitter Dennis be into the under the matrix. I am No, I m. That is the energy matrix of order in converting as be in he murdered better by sea by using the creation marked by green. He&#x27;s equals. He was in the secretion. Now we can use associative property to see that be re first, multiply be into a and that multiplayer by sea Now be more prepared by a is the under the metrics we have by the situation in green again So being please are the matrix Haven&#x27;t you matrix into sea? And that unlimited embassy is again Just see the metrics e So we have shown that b equals see and there is a solution to this problem.

Problem

Suppose $A, B,$ and $X$ are $n \times n$ matrices…

Problem 19 Medium Difficulty

If $A, B,$ and $C$ are $n \times n$ invertible matrices, does the equation $C^{-1}(A+X) B^{-1}=I_{n}$ have a solution, $X ?$ If so, find it.

Answer

$X = C B - A$

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

Linear Algebra and Its Applications

Catherine R.

Alayna H.

Maria G.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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$X = C B - A$

Linear Algebra and Its Applications

Catherine R.

Alayna H.

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

Catherine R.

Alayna H.

Maria G.

Kristen K.

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

Linear Algebra and Its Applications