What is the inverse of the 3x3 matrix mod 26

[DODGEVIPER13]

Homework Statement

What is the inverse of the 3x3 matrix mod 26?
K = <br /> \begin{pmatrix}<br /> 17 &amp; 17 &amp; 5\\<br /> 21 &amp; 18 &amp; 21\\<br /> 2 &amp; 2 &amp; 19 <br /> \end{pmatrix}<br />

Homework Equations

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:

<br /> \begin{pmatrix}<br /> 4 &amp; 9 &amp; 15\\<br /> 15 &amp; 17 &amp; 6\\<br /> 24 &amp; 0 &amp; 17 <br /> \end{pmatrix}<br />

I have so far without multiplying it by 17:

<br /> \begin{pmatrix}<br /> 300/-939 &amp; -313/-939 &amp; 267/-939\\<br /> -357/-939 &amp; 313/-939 &amp; -252/-939\\<br /> 6/-939 &amp; 0 &amp; -51/-939<br /> \end{pmatrix}<br />

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.

 

[.Scott]

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

 

[.Scott]

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:

<br /> \begin{pmatrix}<br /> 4 &amp; 9 &amp; 15\\<br /> 15 &amp; 17 &amp; 6\\<br /> 24 &amp; 0 &amp; 17 <br /> \end{pmatrix}<br />

I have so far without multiplying it by 17:

<br /> \begin{pmatrix}<br /> 300/-939 &amp; -313/-939 &amp; 267/-939\\<br /> -357/-939 &amp; 313/-939 &amp; -252/-939\\<br /> 6/-939 &amp; 0 &amp; -51/-939<br /> \end{pmatrix}<br />

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.

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

<br /> \begin{pmatrix}<br /> 17*300 &amp; 17*-313 &amp; 17*267\\<br /> 17*-357 &amp; 17*313 &amp; 17*-252\\<br /> 17*6 &amp; 0 &amp; 17*-51<br /> \end{pmatrix} =<br /> \begin{pmatrix}<br /> 17*14 &amp; 17*25 &amp; 17*7\\<br /> 17*7 &amp; 17*1 &amp; 17*8\\<br /> 17*6 &amp; 0 &amp; 17*1<br /> \end{pmatrix} =<br /> \begin{pmatrix}<br /> 4 &amp; 9 &amp; 15\\<br /> 15 &amp; 17 &amp; 6\\<br /> 24 &amp; 0 &amp; 17<br /> \end{pmatrix}<br />
So you already had the solution.
If you're using a recent Windows operating system, you have a calculator with the Mod function.

 

[DODGEVIPER13]

Ok cool so I did get the answer.