Hill's Cipher - Known Plaintext Attack I know this question has already been asked few times but I'm struggling a bit on a problem. 

I have a plaintext FRIDAY and its ciphertext PQCFKU, using $M = 2$, with corresponding integers $x = fr id ay = (5, 17),(8,3),(0,24)$ and $y = pq cf ku = (15, 16),(2,5),(10,20)$.

In order to find the key $k$:

$$\pmatrix{y1 \\y2 \\} = \pmatrix{x1 \\x2 \\}k$$
$$\pmatrix{15&amp;16 \\2&amp;5} = \pmatrix{5&amp;17 \\8&amp;3 \\}k$$

Now $x$ determinant, $det(x)$ is given by:

$$det(x)=5*3-8*17 = -121$$ and, using modulo 26:
$$-121\pmod{26} = 9$$
Then, $det(x^{-1})$ is given by:
$$9^{-1}\pmod{26} = 3$$

Now $X^{c}$ is:
$$X^{c}=\pmatrix{3&amp;8\\-17&amp;5}$$
And $(X^{-1})^{T}$ is:
$$(X^{-1})^{T}=\pmatrix{3&amp;-17\\-8&amp;5}$$

So, In order to find $k$:
$$k=(\pmatrix{3&amp;-17\\-8&amp;5}det(x^{-1}))\pmod{26}$$
Which is:
$$k=(\pmatrix{3&amp;-17\\-8&amp;5}3)\pmod{26}$$
$$k=\pmatrix{9&amp;-51\\-24&amp;15}\pmod{26}$$
$$k=\pmatrix{9&amp;1\\2&amp;15}$$


  Now comes the problems, doing the decryption using $k$ and $pq$ I
  can't get $fr$ back.


If I'm not wrong:

$$x1 = (15,16)\pmatrix{9&amp;1\\2&amp;15}=(15*9+16*1, 15*2+16*15)=(151, 270)$$

And now, to obtain the plaintext:

$$(151, 270)\pmod{26} = (21, 10) $$


  and, $$(21, 10) $$ leads to $vm$ which is clearly not $fr$


Is my reasoning wrong? Am I doing errors?

I've been scratching my head for a while but I was unable to find  valid solution. Thanks in advance.
Actually it seems to me that you are using the wrong basis and got the key for the other way around:

Suppose you have the plain text $x_1 = \pmatrix{5\\17}$ and $x_2= \pmatrix{8\\3 \\}$ and the corresponding ciphertexts $y_1=\pmatrix{15 \\16}$ and $y_2=\pmatrix{2\\5}$ and please note how I represent those as vectors and not line matrices, then everything seems to work fine with your method:

Hill's cipher translates into:
$$\pmatrix{k_{11}&amp;k_{12}\\k_{21}&amp;k_{22}}\times x_1 \equiv y_1 \mod 26$$
$$\pmatrix{k_{11}&amp;k_{12}\\k_{21}&amp;k_{22}}\times x_2 \equiv y_2 \mod 26$$

This leads to the system you are solving (but you may also represent it using 4 equations) and you get:
$$\begin{cases}5k_{11}+17k_{12} &amp;= 15\\
5k_{21}+17k_{22} &amp;= 16\\
8k_{11}+3k_{12} &amp;= 2\\
8k_{21}+3k_{22} &amp;= 5\end{cases}$$
Which correspond, once represented using matrices, to solving:
$$\pmatrix{5&amp;17 \\ 8&amp;3} \times \pmatrix{k_{11}\\k_{12}} \equiv \pmatrix{15\\2} \mod 26$$
$$\pmatrix{5&amp;17 \\ 8&amp;3} \times \pmatrix{k_{21}\\k_{22}} \equiv \pmatrix{16\\5} \mod 26$$
Since you have computed it, you know that 
$$X^{-1} \equiv \pmatrix{9&amp;1 \\2&amp;15 \\} \mod 26$$
for $X=\pmatrix{5&amp;17 \\8&amp;3 \\} $. 

But now, this is where it seems to me that you are doing something wrong: to get the key, you can simply solve the above equations by left-multiplying by $X^{-1}$ so that:
$$ \pmatrix{k_{11}\\k_{12}} \equiv X^{-1}\times X\times \pmatrix{k_{11}\\k_{12}} \equiv X^{-1}\times y_1\equiv \pmatrix{9&amp;1 \\ 2&amp;15} \times \pmatrix{15\\2}\mod 26$$ 
and the same holds for $X^{-1}\times y_2$:
$$ \pmatrix{k_{21}\\k_{22}} \equiv X^{-1}\times X\times \pmatrix{k_{21}\\k_{22}} \equiv X^{-1}\times y_2\equiv \pmatrix{9&amp;1 \\ 2&amp;15} \times \pmatrix{16\\5}\mod 26$$ 
 and you end up with the key which seems to be, after computing the latest results:
$$\begin{cases}k_{11}=7\\ k_{12}=8 \\k_{21}=19\\k_{22}=3\end{cases}$$
And which translates into $$K=\pmatrix{7&amp;8\\19&amp;3}$$
And you can check that $ay$ get encrypted into $ku$
$$ K \times \pmatrix{0\\24}\equiv \pmatrix{10\\20} \mod 26$$
But to decrypt anything, you'll need to invert the key to get:
$$K^{-1}=\frac{1}{det(K)}\pmatrix{3&amp;-8\\-19&amp;7}\equiv 25 \pmatrix{3&amp;-8\\-19&amp;7} \equiv \pmatrix{23&amp;8\\19&amp;19} \mod 26$$
And so you can decode the $fr$ from the $pq$ by:
$K^{-1}\times \pmatrix{15\\16} \equiv \pmatrix{5\\17} \mod 26$
which checks out.

This is because decoding Hill's cipher necessitate to left-multiply the cipher by the inverse of the key.