# Determining the secret using the Blakley's scheme?

I am solving a example with Blakley's scheme problem. The issue that I am having is using the row operation to determine the secret $x_{0}$. So what I have so far is:

Given three planes: $z = 7x + 84y + 41$, $z = 17x + 5y + 3$, and $z = 37x + 46y + 66$. Then I set up the matrix: $$$$\begin{pmatrix} 7 & 84 & -1 \\ 17 & 5 & -1 \\ 37 & 46 & -1 \end{pmatrix} \begin{pmatrix} x_{0} \\ y_{0} \\ z_{0} \end{pmatrix} \equiv \begin{pmatrix} -41 \\ -3 \\ -66 \end{pmatrix} (mod \ 87)$$$$ I used the row operations calculator to get the matrix, however it seems way off and incorrect. Am I suppose to do a row operations first and once I do that what do I need to do after to get $x_{0}$,$y_{0}$, and $z_{0}?$ Any help is appreciated.

I don't see any problem with your matrix, it is invertible.

Once you've setup the matrix, you can work out the solution of your system. Just don't forget you are working modulo 87, so:

\begin{aligned}& \begin{pmatrix} 7 & 84 & -1 \\ 17 & 5 & -1 \\ 37 & 46 & -1 \end{pmatrix} \begin{pmatrix} x_{0} \\ y_{0} \\ z_{0} \end{pmatrix} \equiv \begin{pmatrix} -41 \\ -3 \\ -66 \end{pmatrix} \pmod{87} \\ \equiv& \begin{pmatrix} 7 & 84 & 86 \\ 17 & 5 & 86 \\ 37 & 46 & 86 \end{pmatrix} \begin{pmatrix} x_{0} \\ y_{0} \\ z_{0} \end{pmatrix} \equiv \begin{pmatrix} 46 \\ 84 \\ 21\end{pmatrix} \pmod{87} \end{aligned}

And for example the inverse of $5 \bmod{87}$ is $35$, so I would multiply $\bmod{87}$ the second row with $35$ to get a $1$ in the second column: $$\begin{pmatrix} 7 & 84 & 86 \\ \color{red}{73} & \color{red}{1} & \color{red}{52} \\ 37 & 46 & 86 \end{pmatrix} \begin{pmatrix} x_{0} \\ y_{0} \\ z_{0} \end{pmatrix} \equiv \begin{pmatrix} 46 \\ \color{red}{69} \\ 21\end{pmatrix} \pmod{87}$$

As mentioned, since the matrix has an invertible determinant $\bmod{87}$, the matrix is invertible, so if there is a solution (don't worry, there is one), you can find it using Gaussian Elimination, I think you should try it too, until you obtain in the end a system of the form: $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x_{0} \\ y_{0} \\ z_{0} \end{pmatrix} \equiv \begin{pmatrix} ?? \\ ?? \\ ??\end{pmatrix} \pmod{87}$$

PS: Note that the rank's theorem for a matrix does not exist in a commutative ring.

• Thanks for the explanation, I shall try and see what I get. – KonoDDa Nov 22 '17 at 19:56