Linear Algerbra. Inverses and Algerbraic Properties of Matrices

Click For Summary
The discussion focuses on solving the equation C^{T}B^{-1}A^{2}BAC^{-1}DA^{-2}B^{T}C^{-2}=C^{T} for the matrix D, given that all matrices involved are n x n and invertible. Participants suggest a method of "unpeeling" D by systematically applying matrix operations to isolate it. The initial step involves manipulating the equation by left-multiplying with (C^T)^{-1} to simplify the expression. Further guidance includes continuing to apply matrix operations, such as multiplying by B and C^2, to progressively isolate D. The conversation emphasizes the importance of following algebraic properties of matrices to arrive at the solution.
stumpoman
Messages
10
Reaction score
1

Homework Statement


Assuming that all matrices, A, B, C, and D, are n x n and invertible, solve for D.

C^{T}B^{-1}A^{2}BAC^{-1}DA^{-2}B^{T}C^{-2}=C^{T}

Homework Equations



C^{T}B^{-1}A^{2}BAC^{-1}DA^{-2}B^{T}C^{-2}=C^{T}

The Attempt at a Solution



I must have missed something in the reading of this section. All I can think of is

B^{-1}A^{2}BAC^{-1}DA^{-2}B^{T}C^{-2}=I

but I don't know where to go from there or if it is even the right way to start.
 
Physics news on Phys.org
Left-multiply the third equation with B.
 
stumpoman said:

Homework Statement


Assuming that all matrices, A, B, C, and D, are n x n and invertible, solve for D.

C^{T}B^{-1}A^{2}BAC^{-1}DA^{-2}B^{T}C^{-2}=C^{T}
Just "undo" every thing that is done to D, one step at a time. For example if you multiply both sides, on the left, by (C^T)^{-1} you get B^{-1}A^{-2}BAC^{-1}DA^{-2}B^TC^{-2}= (C^T)^{-1}C^T= I
Then multiply on the left by B, then on the right by C^2, or do both together to get A^{=2}BAC^{-1}DA^{-2}B^T= BC^2.
Continue to "unpeel" D.

Homework Equations



C^{T}B^{-1}A^{2}BAC^{-1}DA^{-2}B^{T}C^{-2}=C^{T}

The Attempt at a Solution



I must have missed something in the reading of this section. All I can think of is

B^{-1}A^{2}BAC^{-1}DA^{-2}B^{T}C^{-2}=I

but I don't know where to go from there or if it is even the right way to start.
Yes, that was the first step as I showed. Continue in the same way.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

Replies
3
Views
1K
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
1K
Replies
2
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
6
Views
2K
Replies
0
Views
1K
Replies
6
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K