What is the inverse of the 3x3 matrix mod 26

[DODGEVIPER13]

Homework Statement

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

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:

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

I have so far without multiplying it by 17:

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

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.

 

[.Scott]

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

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

I have so far without multiplying it by 17:

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

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.

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

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