1. ## solve system

Hi again, I think I have the correct answer to part a of this question but I I am not sure about part b, so would someone be kind enough to check my working please.

$\begin{bmatrix}6 & -3 \\2 & 6 \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \end{bmatrix} = \begin{bmatrix}3 \\1 \end{bmatrix}$

a) solve in $F_p$ for $p=5,11,17$
b) find the number of solutions for $p=7$

Working:
a) $det \begin{bmatrix}6 & -3 \\2 & 6 \end{bmatrix} =42=2 \cdot 3 \cdot 7$
so it is not invertible for $p=2,3,7$
so I get
$\begin{bmatrix}x_1 \\x_2 \end{bmatrix} = \frac{1}{42} \begin{bmatrix}6 & 3 \\-2 & 6 \end{bmatrix}\begin{bmatrix}3 \\1 \end{bmatrix}= \begin{bmatrix}2^{-1} \\0 \end{bmatrix}$

For $p=5$: $2^{-1}=3$ and $0=0$ so solution for $p=5$ is $\begin{bmatrix}3 \\0 \end{bmatrix}$

For $p=11$, solution is $\begin{bmatrix}6 \\0 \end{bmatrix}$

For $p=17$, solution is $\begin{bmatrix}9 \\0 \end{bmatrix}$

b) For p=7 the matrix in not invertible so for the two linear equation one is a scalar of the other.

$3 \cdot (2,6)=(6,18)=(6,-3)$

therefore
$2x_1+6x_2=1=6$
$x_1+3x_2=3$

Thus, for each choice of $x_2 \in F_7$, $x_1=3-3x_2$.
So there are 7 possible solutions:
$\begin{bmatrix} 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 \\ 5 \end{bmatrix}, \begin{bmatrix} 3 \\ 0 \end{bmatrix}, \begin{bmatrix} 4 \\ 2 \end{bmatrix}, \begin{bmatrix} 5 \\ 4 \end{bmatrix}, \begin{bmatrix} 6 \\ 6 \end{bmatrix}$

2. ## Re: solve system

1 does not equal 6 mod 7.

$2x_1 + 6x_2 = 1 \implies x_1 + 3x_2 = 4 \implies x_1 = 4 + 4x_2$

3. ## Re: solve system

oh yes, stupid mistake. Thanks Deveno. Got it now.

you mean $x_1=4-3x_2 \implies x_1 = 4 +(7-3)x_2=4+4x_2$