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 $k$:

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

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

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

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

If I'm not wrong:

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

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.

• Not sure if it matters, but on Wikipedia the multiplication is done from the right rather than from the left (first line)
– SEJPM
Jan 22 '17 at 19:42

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: $$\pmatrix{k_{11}&k_{12}\\k_{21}&k_{22}}\times x_1 \equiv y_1 \mod 26$$ $$\pmatrix{k_{11}&k_{12}\\k_{21}&k_{22}}\times x_2 \equiv y_2 \mod 26$$

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

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: $$\pmatrix{k_{11}\\k_{12}} \equiv X^{-1}\times X\times \pmatrix{k_{11}\\k_{12}} \equiv X^{-1}\times y_1\equiv \pmatrix{9&1 \\ 2&15} \times \pmatrix{15\\2}\mod 26$$ and the same holds for $X^{-1}\times y_2$: $$\pmatrix{k_{21}\\k_{22}} \equiv X^{-1}\times X\times \pmatrix{k_{21}\\k_{22}} \equiv X^{-1}\times y_2\equiv \pmatrix{9&1 \\ 2&15} \times \pmatrix{16\\5}\mod 26$$ 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 $$K=\pmatrix{7&8\\19&3}$$ 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: $$K^{-1}=\frac{1}{det(K)}\pmatrix{3&-8\\-19&7}\equiv 25 \pmatrix{3&-8\\-19&7} \equiv \pmatrix{23&8\\19&19} \mod 26$$ 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.

• Thanks for the answer. Tomorrow morning I'll check it out and let you know. Jan 22 '17 at 22:20
• I think you made an error on the four equations system: $5k_{11}+3k_{12}=2$ and $5k_{21}+3k_{22}=5$ should be $8k_{11}+3k_{12}=2$ and $8k_{21}+3k_{22}=5$. Also, I don't understand how you got $25$ from $\frac{1}{det(K)}$ since $det(K) = -131$. I'll be thankful if you explain me how. Jan 23 '17 at 9:36
• My, I should never copy-paste anything. I did, indeed. To obtain $25$ from $\frac{1}{det(K)}$, keep in mind we are inverting it modulo $26$, so basically I have $det(K) = -131 \equiv 25 \mod 26$ and the modular inverse of $25$ is $25$.
– Lery
Jan 23 '17 at 18:34
• Uhm, $-131 \mod26$ gives me $1$, that's why I don't get it. Jan 23 '17 at 18:50
• Well $-131=-26*5-1 = 0-1=-1 \equiv 25 \mod 26$
– Lery
Jan 23 '17 at 20:29