MatricesWatch

#1
Let be an nxn matrix.

Find using elementary row operations.

I can see by just doing a few that the answer will be

.

However I think I should prove this somehow, induction perhaps. I have written A as

And similarly

and tried to do something with that but I haven't got anywhere. Am I making more of it than is necessary?

Thanks.
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7 years ago
#2
(Original post by Scott3142)
Thanks.
I suggest putting an example into Excel or whatever your favourite matrix multiplying software is; what you have there doesn't work.

Nor do your formulae match the matrices you've quoted at the top.
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#3
I have worked out the first 3 rows from the bottom of and it seems to be upper triangular with alternating 1s and minus 1s. I think this is right. Ahh I see how my formulae are wrong now but how do I generalise the operations to nxn?
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7 years ago
#4
(Original post by Scott3142)
I have worked out the first 3 rows from the bottom of and it seems to be upper triangular with alternating 1s and minus 1s. I think this is right. Ahh I see how my formulae are wrong now but how do I generalise the operations to nxn?
What you are doing is starting from the nth row working up, subtracting the i'th row from the (i-1)'th row, which is generating your alternating pattern in the upper triangle.

If you look at it in terms of the augmented matrix (A|I); these operations reduce A to I, and the I to A^(-1). So just formalise that some way.

Alternatively you could take the approach of considering the elementary row operations as matrices in their own right, and you are generating a string of matrices that perform the reduction, and this allows you to state the inverse, but I'm not familiar with the nomenclature, etc. as it's donkey's years since I looked at this - and have never used it since..
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