Indeed $k_1$ must be a matrix, say $\begin{bmatrix} a & b \\ c & d\end{bmatrix}$ and $k_2$ a vector $\begin{bmatrix} p \\ q \end{bmatrix}$. Then looking at the first row of the matrix and $p$: these are the elements that determine the first component of every $(c,p)$ pair. So the plaintext $\begin{pmatrix} 5\\4\end{pmatrix}$ mapping to the ciphertext $\begin{pmatrix}1\\8 \end{pmatrix}$ gives us the equation $$5a + 4b + p = 1$$ the second pair yields $$8a + 10b + p = 8$$ while the third pair gives $$7a + b + p = 8$$ where all computations are done modulo $11$ of course. So in schematic form we get the system $$\left[
\begin{array}{ccc|c} 5 & 4 & 1 & 1 \\ 8 & 10 & 1 & 8 \\ 7 & 1 & 1 & 8 \end{array}\right]$$
Similarly for $c,d,q$ we get a system with the same coefficients but a different right hand side:
$$\left[
\begin{array}{ccc|c}
5 & 4 & 1 & 8\\
8 & 10 & 1 & 5\\
7 & 1 & 1 & 7
\end{array}
\right]
$$
So we can do elimation on two systems at the same time:
$$\left[
\begin{array}{ccc|cc}
5 & 4 & 1 & 1 & 8\\
8 & 10 & 1 & 8 & 5\\
7 & 1 & 1 & 8 & 7
\end{array}
\right]
$$
where the first column right of the bar is for the $a,b,p$ system, and the other for the $c,d,q$ system.
If we had reals we would subtract the first row $\times \frac{8}{5}$ from the second to eliminate the first variable. Modulo $11$ we note that $5^{-1} = 9$ because $5\times 9 = 45 \equiv 1 \pmod{11}$. So we multiply the first equation (both left and right hand side) with $8 \times 5^{-1} = 8 \times 9 = 72 \equiv 6 \pmod{11}$, which gives
$$\left[
\begin{array}{ccc|cc} 30 & 24 & 6 &6 & 48\end{array}\right]$$ which is mod $11$ equivalent to
$$\left[
\begin{array}{ccc|cc} 8 & 2 & 6 &6 & 4
\end{array}
\right]$$
We substract this from the second equation an get a new equation:
$$\left[
\begin{array}{ccc|cc} 0 & 8 & -5 &2 & 1\end{array}\right]$$
which mod $11$ is
$$\left[
\begin{array}{ccc|cc} 0 & 8 & 6 &2 & 1\end{array}\right]$$
Also we want to eliminate the first variable from the third equation.
So we substract $\frac{7}{5} =7 \cdot 5^{-1}= 63 \equiv 8 \pmod{1}$ times the first equation (the original one) from the third.
First compute $8$ times the first:
$$\left[
\begin{array}{ccc|cc}
40 & 32 & 8 & 8 & 64
\end{array}
\right]
$$
which modulo $11$ becomes
$$\left[
\begin{array}{ccc|cc}
7 & 10 & 8 & 8 & 9
\end{array}
\right]
$$
So we get a new third equation:
$$\left[
\begin{array}{ccc|cc}
0 & -9 & -7 & 0 & -2
\end{array}
\right]
$$
or
$$\left[
\begin{array}{ccc|cc}
0 & 2 & 4 & 0 & 9
\end{array}
\right]
$$
So in total we get the equations after elimination step 1:
$$\left[
\begin{array}{ccc|cc}
5 & 4 & 1 & 1 & 8\\
0 & 8 & 6 & 2 & 1\\
0 & 2 & 4 & 0 & 9
\end{array}
\right]
$$
The next phase: eliminate the second variable from eq. 2 and 3.
To this end we do "$\text{eq. } 3 - \frac{2}{8} \text{eq. } 2$"
and we note that $8^{-1} \pmod{11} = 7$ so $\frac{2}{8}$ is $14 \equiv 3$.
$3$ times the new eq. 2 becomes:
$$\left[
\begin{array}{ccc|cc}
0 & 24 & 18 & 6 3
\end{array}
\right]
$$
which is the same (mod $11$) as
$$\left[
\begin{array}{ccc|cc}
0 & 2 & 7 & 6 & 3
\end{array}
\right]
$$
So new eq.3 becomes:
$$\left[
\begin{array}{ccc|cc}
0 & 0 & -3 & -6 &6
\end{array}
\right]
$$
Now we can back-substitute the two equation systems (recall the first column was for the system in terms of $a,b,p$): $-3p = -6$, so $p=2$.
Then $8b + 6p = 2$ from the updated second equation. So $8b + 12 = 2 \rightarrow 8b = -10 = 1$, so $b = 8^{-1} \equiv 7 \pmod{11}$.
Finally $5a + 4b + p = 1 \rightarrow 5a + 28 + 2 = 1 \rightarrow 5a = 4 \pmod{11}$, which gives $a=3$ from $5^{-1} = 9$ again.
The same can now be done for $-3q = 6$, so $q = -2 = 9$.
Then $8d + 6q = 1 \rightarrow 8d + 54 = 1 \rightarrow 8d = 2 \pmod{11}$ hence $d = 2 \cdot 7 \equiv 14 \equiv 3$.
Finally $5c + 4d + q = 7 \rightarrow 5c + 12 + 9 = 7 \rightarrow 5c = 8$ so that $c = 4$.
So the cipher was given by $$\begin{bmatrix} c_1 \\ c_2\end{bmatrix} = \begin{bmatrix} 3 & 7 \\ 4 & 3\\ \end{bmatrix}\begin{bmatrix} p_1 \\ p_2\end{bmatrix} + \begin{bmatrix} 2 \\ 9\end{bmatrix}$$