Finding the determinent or the identity of the matrix?

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Homework Help Overview

The discussion revolves around determining the values of variables in matrix D such that it equals the inverse of matrix C. The matrices involved are 3x3, and the problem is situated within the context of linear algebra, specifically focusing on matrix inverses and products.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the multiplication of matrices C and D to achieve the identity matrix. Questions arise about the correct identification of variables a and b based on the resulting product. Some participants express confusion about the presence of variable b in the product.

Discussion Status

There is ongoing exploration of the matrix multiplication process, with participants attempting to derive values for a and b that satisfy the conditions for the identity matrix. Some guidance has been offered regarding the correct approach to multiplication and the need for both variables to appear in the product.

Contextual Notes

Participants note the importance of ensuring that the product of matrices C and D results in the identity matrix, and there is mention of various methods for finding matrix inverses, though no specific method is endorsed as the only solution.

carltonblues
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I have the following problem:

Let C= | 1 2 3 |
| 2 5 3 |
| 1 0 8 |

and D = | -40 16 a |
| b -5 -3 |
| 5 -2 -1 |

Determine the values of and and b such that D = C^-1

Do i start off by finding the determinent or the identity or whatever?
 
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If D = C^(-1), then CD = I, where I is the identity matrix. Multiply C and D to produce another 3x3 matrix and determine values for "a" and "b" such that this product has "1" for diagonal elements and "0" elsewhere. Take your time with the product so you don't make careless mistakes. If done correctly, the answer will be quite apparent.


~~
 
Last edited:
I got the product of CD to be:

13 0 a-9
-35 1 2a-18
0 0 a-8

I multiplied that by I and got the same number. What do I do now?

Cheers
 
What happened to "b"?? The variable "b" should appear in your product CD (just like "a" appears in CD). First multiply C & D correctly, then choose values of "a" & "b" such that diagonal elements are "1" and all others are "0". It should be quite apparent!
(Note: Do not multiply by "I". You want CD to BE "I" after selecting proper values for "a" and "b".)


~~
 
Last edited:
Thanks for the help. I get:

2b-25 0 a-9
5b-65 1 2a-18
0 0 a-8

So does that mean b= 13 and a=9 if you want the diagonals to be 1?

Thanks,
 
carltonblues said:
Thanks for the help. I get:

2b-25 0 a-9
5b-65 1 2a-18
0 0 a-8

So does that mean b= 13 and a=9 if you want the diagonals to be 1?

Thanks,
CORRECT. That's exactly how to solve the problem.
 
xanthym said:
CORRECT. That's exactly how to solve the problem.
Muchly appreciated man. Thank you :!)
 
Actually, you could have just found the inverse of the original matrix. Then the value of a and b would also be obvious.

There's at least three ways to find the inverse matrix:

The elimination or http://math.uww.edu/faculty/mcfarlat/inverse.htm method.

The determinant method .

The Cayley-Hamilton method (unfortunately, I can't find a good link that really breaks this one out - I have a book, Quaternions and Rotation Sequences by Jack Kuipers, that breaks this out so well even a beginner can understand it).
Edit: Okay, even though text only, this isn't too bad: http://mathforum.org/library/drmath/view/51978.html

There are also several other methods or variations. This link presents the first two in a different manner (Gaussian elimination and a LaPlace expansion).
http://qucs.sourceforge.net/tech/node15.html

The advantage of the determinant and the Cayley-Hamilton method is that you find out pretty quick whether or not there even is an inverse.
 
Last edited:

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