A Hill Cipher over the above $27$-symbol alphabet: $A=0$, $B=1,\ldots,Z=25,$ Space$=26$, was used to produce the ciphertext:
KVBDA RYFIBDOAATBDJCZRBDGFGDASSUDEGFSJNOCPXQ PVKCVUMSBZSDFCNUMTKWUIAEUVXYOGQYIFSVYVKWOFFYIWNRWRGCDCCQKSKHWOZFSSBZSVXOMITFFPOTZCCQKJUHDLKUMHSXZAVKVDOCDGXSKHDXZNOPRMZQTXO TSUCDVKOZWYCCQKJUMZFKVXIJTZEEGDSCVXYOIPBCWNSKHWOZFSCCQKHDWOOZIYTZXYVKHSBQBGASBDBOCDVKZSUMFPJXQKXWSKBDMZOQTTHOHGSUWYPPBDGFPFNOXZXYVKCVY CUVLV
Please decipher the text.
Step 1 is to convert the text to numbers in $\mathbb{Z}_{27}$. Call the result $(c_0, c_1, c_2, c_3, \ldots)$
Call the decryption matrix (which you need)
$$D = \begin{bmatrix}x &y \\ u & v\ \end{bmatrix}$$
for definitess. For every possibility of $(x,y) \in \mathbb{Z}_{27}^2$, compute the supposed partial plain text (we get the even indexed plain text, because we only do one row at the time) $P(x,y) = (p_0, p_2 ,p_4, \ldots)$ where $p_{2i} = x\cdot c_{2i} + y \cdot c_{2i+1} \bmod 27$. Assign a point score $+2$ or more to spaces and 'e's that occur in $P(x,y)$, and minus points to q's and x's etc. Or do a $\chi^2$-matching score if you prefer. Probably one $(x,y)$ row will jump out, and be correct.
Then do the same for $(u,v)$ and the odd-places plaintext.
Alternatively, if you know the text is English and not to strange, you can for each position $i$ supose that $c_{i} c_{i+1} c_{i+2} c_{i+3}$ corresponds to $(19, 7, 4, 26)$ (or "the "), which gives 4 assumed equations in the 4 unknowns $(x,y,u,v)$, and which you can solve modulo $27$, and if solvable test it on another part of the text. If it fails try another position. Etc.
Both solutions will involve some programming of course, unless you're very patient.