Finding the determinent or the identity of the matrix?

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To determine the values of "a" and "b" such that matrix D equals the inverse of matrix C, one must first multiply C and D to yield a product matrix that has 1s on the diagonal and 0s elsewhere. The product was calculated, revealing expressions involving "a" and "b." After substituting values, it was concluded that setting b=13 and a=9 would satisfy the conditions for the identity matrix. Additionally, alternative methods for finding the inverse of matrix C were mentioned, such as the elimination method and the determinant method. The discussion emphasizes the importance of correctly performing matrix multiplication to achieve the desired identity matrix outcome.
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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.


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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".)


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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.
 
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