# What is the inverse of the 3x3 matrix mod 26

1. Jul 13, 2014

### DODGEVIPER13

1. The problem statement, all variables and given/known data
What is the inverse of the 3x3 matrix mod 26?
K = $$\begin{pmatrix} 17 & 17 & 5\\ 21 & 18 & 21\\ 2 & 2 & 19 \end{pmatrix}$$

2. Relevant equations

3. The attempt at a solution
So I found all the cofactors and then took the transpose of the matrix. I then divided new matrix, by the determinate -939. After which I would multiply this by 17 because 23-1 mod 26 = 17 to get the inverse. I found 17 by using the euclidean algorithm. This was all UPLOADED. However I am confused because even if I do this I do not get the answer in the book. They get:

$$\begin{pmatrix} 4 & 9 & 15\\ 15 & 17 & 6\\ 24 & 0 & 17 \end{pmatrix}$$

I have so far without multiplying it by 17:

$$\begin{pmatrix} 300/-939 & -313/-939 & 267/-939\\ -357/-939 & 313/-939 & -252/-939\\ 6/-939 & 0 & -51/-939 \end{pmatrix}$$

I realize that even if I go ahead I will not reach what the book has, what have I done wrong? All of my work has been UPLOADED.

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2. Jul 14, 2014

### .Scott

I'll do the center column - because it's easy:
$$\begin{pmatrix} k_{11} & k_{12} & k_{13}\\ k_{21} & k_{22} & k_{23}\\ k_{31} & k_{32} & k_{33} \end{pmatrix} \begin{pmatrix} 17 & 17 & 5\\ 21 & 18 & 21\\ 2 & 2 & 19 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} \ \\ \ \\ \ \\ A)\ 17K_{11} + 21K_{12} + 2K_{13} = 1 \\ B)\ 17K_{11} + 18K_{12} + 2K_{13} = 0 \\ C)\ 5K_{11} + 21K_{12} + 19K_{13} = 0 \\ D)\ 17K_{21} + 21K_{22} + 2K_{23} = 0 \\ E)\ 17K_{21} + 18K_{22} + 2K_{23} = 1 \\ F)\ 5K_{21} + 21K_{22} + 19K_{23} = 0 \\ G)\ 17K_{31} + 21K_{32} + 2K_{33} = 0 \\ H)\ 17K_{31} + 18K_{32} + 2K_{33} = 0 \\ I)\ 5K_{31} + 21K_{32} + 19K_{33} = 1 \\ \ \\ \ \\ Modulo 26\ table\ for\ Y=3X: 0, 3, 6, 9, 12, 15, 18, 21, 24, 1, 4, 7, 10, 13, 16, 19, 22, 25, ...\\ A-B)\ \ \ 3K_{12} = 1 \\ A-B/3)\ K_{12} = 9 \\ D-E)\ \ \ 3K_{22} = 25 \\ D-E/3)\ K_{22} = 17 \\ G-H)\ \ \ 3K_{32} = 0 \\ G-H/3)\ 3K_{32} = 0 \\ \ \\ \ \\ \begin{pmatrix} k_{11} & 9 & k_{13}\\ k_{21} & 17 & k_{23}\\ k_{31} & 0 & k_{33} \end{pmatrix} \ \\$$
That book answer is looking good to me.

3. Jul 14, 2014

### .Scott

Mod 26 (-939) is 23.
Mod 26 (1/23) is 17.
So 17 is the right number to use.

$$\begin{pmatrix} 17*300 & 17*-313 & 17*267\\ 17*-357 & 17*313 & 17*-252\\ 17*6 & 0 & 17*-51 \end{pmatrix} = \begin{pmatrix} 17*14 & 17*25 & 17*7\\ 17*7 & 17*1 & 17*8\\ 17*6 & 0 & 17*1 \end{pmatrix} = \begin{pmatrix} 4 & 9 & 15\\ 15 & 17 & 6\\ 24 & 0 & 17 \end{pmatrix}$$