What is the equation of the hypersurface in a different base. We had a test in Linear algebra where a problem was as follows:
In 4 dimentional real number space to a hypersurface belong those vectors who's standartbase coordinates $(x_1;x_2;x_3;x_4)$ make the equation $x_1^2+x_2^2-x_3^2-x_4^2 = 1$ true. Find this hypersurface's equation in a base f coordinates $(x_1^\prime;x_2^\prime;x_3^\prime;x_4^\prime)$. Base f: $f_1=(1;1;1;1)$, $f_2=(1;1;-1;-1)$, $f_3=(1;-1;1;-1)$, $f_4=(1;-1;-1;1)$.
My answer is $8x_1^\prime x_2^\prime+8x_3^\prime x_4^\prime = 1$ while many disagree and say $x_1^\prime x_2^\prime+x_3^\prime x_4^\prime=2$ is the right answer. 
My vector of choice was (5;1;3;4) and after transformation my equation was true.
Thank you in advance to whomever helps with this problem 
 A: It’s very easy to get these transformations backwards (as I did in my initial comment). Suppose that $\mathbf x = M\mathbf x'$. Then, writing the equation in matrix form as $\mathbf x^TQ\mathbf x=1$, we have $$\mathbf x^TQ\mathbf x = (M\mathbf x')^TQ(M\mathbf x') = \mathbf x'^T(M^TQM)\mathbf x' = 1,$$ that is, the matrix $Q'$ of this equation in the primed basis is $M^TQM$. Here, since the original basis is the standard basis, $M$ is just the matrix with the primed basis vectors as its columns, so $$Q' = \begin{bmatrix}1&1&1&1\\1&1&-1&-1\\1&-1&1&-1\\1&-1&-1&1\end{bmatrix} \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix} \begin{bmatrix}1&1&1&1\\1&1&-1&-1\\1&-1&1&-1\\1&-1&-1&1\end{bmatrix} = \begin{bmatrix}0&4&0&0\\4&0&0&0\\0&0&0&4\\0&0&4&0\end{bmatrix},$$ which corresponds to the equation $8x_1'x_2'+8x_3'x_4'=1$, just as you’ve computed.  
The key thing to remember is that quadratic forms transform covariantly: in order to transform the matrix of a quadratic form into a new basis, you use the inverse of the matrix that transforms a vector from the old basis to the new. Indeed, if you were to use the inverse of $M$ in the above computation, you’d get the equation that others came up with.
