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 Post subject: Matrix with moduloPosted: Tue, 8 Nov 2011 08:39:32 UTC
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Joined: Sat, 20 Aug 2011 12:38:23 UTC
Posts: 39
Let p be a prime. For every a ∈ Z, p, a^-1 ∈ Z
p is the unique element such that a × a^-1 mod p = 1. The operation ‘÷ mod p’ is defined as
(b = 1 ÷ a = (1/a) = a^-1 (mod p) ) ⇔ (b ∈ Zp ∧ b × a = a × b = 1 (mod p))

The Gauss-Jordan Algorithm can be adapted to compute the inverse of an invertible
matrix in modulo arithmetic over a prime modulus. Compute the following using

matrix A = ( 5 3 16
10 18 8
3 16 9 )

[(1/3) A^-1] mod 19

Could someone kindly explain to me what does this question exactly mean. I do not really understand it at all

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 Post subject: Re: Matrix with moduloPosted: Tue, 8 Nov 2011 08:43:41 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12075
Location: Austin, TX
aceminer wrote:
Let p be a prime. For every a ∈ Z, p, a^-1 ∈ Z
p is the unique element such that a × a^-1 mod p = 1. The operation ‘÷ mod p’ is defined as
(b = 1 ÷ a = (1/a) = a^-1 (mod p) ) ⇔ (b ∈ Zp ∧ b × a = a × b = 1 (mod p))

The Gauss-Jordan Algorithm can be adapted to compute the inverse of an invertible
matrix in modulo arithmetic over a prime modulus. Compute the following using

matrix A = ( 5 3 16
10 18 8
3 16 9 )

[(1/3) A^-1] mod 19

Could someone kindly explain to me what does this question exactly mean. I do not really understand it at all

You're trying to find a matrix such that

Hint:

will simplify the number of calculations you need to do.

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 Post subject: Re: Matrix with moduloPosted: Tue, 8 Nov 2011 09:04:03 UTC
 Member

Joined: Sat, 20 Aug 2011 12:38:23 UTC
Posts: 39
aceminer wrote:
Let p be a prime. For every a ∈ Z, p, a^-1 ∈ Z
p is the unique element such that a × a^-1 mod p = 1. The operation ‘÷ mod p’ is defined as
(b = 1 ÷ a = (1/a) = a^-1 (mod p) ) ⇔ (b ∈ Zp ∧ b × a = a × b = 1 (mod p))

The Gauss-Jordan Algorithm can be adapted to compute the inverse of an invertible
matrix in modulo arithmetic over a prime modulus. Compute the following using

matrix A = ( 5 3 16
10 18 8
3 16 9 )

[(1/3) A^-1] mod 19

Could someone kindly explain to me what does this question exactly mean. I do not really understand it at all

You're trying to find a matrix such that

Hint:

will simplify the number of calculations you need to do.

But how does the mod 19 come into play.
i am given the mapping of a to a^- 1 mod 19 values.

I am not sure how to even start. Do i reduce the matrix to its inverse form to find A^-1 first? but the inverse encompasses a such that
[(1/3)(A)]^-1.
Not too sure if 1/3 will play a part in this.

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 Post subject: Re: Matrix with moduloPosted: Tue, 8 Nov 2011 09:23:38 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12075
Location: Austin, TX
aceminer wrote:
aceminer wrote:
Let p be a prime. For every a ∈ Z, p, a^-1 ∈ Z
p is the unique element such that a × a^-1 mod p = 1. The operation ‘÷ mod p’ is defined as
(b = 1 ÷ a = (1/a) = a^-1 (mod p) ) ⇔ (b ∈ Zp ∧ b × a = a × b = 1 (mod p))

The Gauss-Jordan Algorithm can be adapted to compute the inverse of an invertible
matrix in modulo arithmetic over a prime modulus. Compute the following using

matrix A = ( 5 3 16
10 18 8
3 16 9 )

[(1/3) A^-1] mod 19

Could someone kindly explain to me what does this question exactly mean. I do not really understand it at all

You're trying to find a matrix such that

Hint:

will simplify the number of calculations you need to do.

But how does the mod 19 come into play.
i am given the mapping of a to a^- 1 mod 19 values.

I am not sure how to even start. Do i reduce the matrix to its inverse form to find A^-1 first? but the inverse encompasses a such that
[(1/3)(A)]^-1.
Not too sure if 1/3 will play a part in this.

The mod 19 is to tell you how to calculate inverses multiplicatively. Just apply the algorithm as usual with this new way of inverting stuff.

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 Post subject: Re: Matrix with moduloPosted: Tue, 8 Nov 2011 09:54:25 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6005
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
Topic moved from Foundations to Matrix Algebra.

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