Is there any other way beside the standard one of calculating the determinant of the two following matrices without it getting long and ugly? This was an exam exercise so I think there must be some other way beside expanding by rows or columns like we usually do.
$\det \begin{pmatrix}1-a&b&a&b\\ \:b&1-a&b&a\\ \:a&b&1-a&b\\ \:b&a&b&1-a\end{pmatrix}$
and
$\det \begin{pmatrix}1&1&1&1\\ r&s&t&u\\ r^2&s^2&t^2&u^2\\ stu&rtu&rsu&rst\end{pmatrix}$
Thanks in advance
For the first matrix, we can write the matrix in the form $$ M = I + a\underbrace{\pmatrix{-1 & 0 & 1 & 0\\0 & -1 & 0 & 1\\ 1 & 0 & -1 & 0\\ 0 & 1 & 0 & -1}}_P + b \underbrace{\pmatrix{0&1&0&1\\1&0&1&0\\0&1&0&1\\1&0&1&0}}_Q. $$ One elegant approach is to note that the matrices $P$ and $Q$ can be nicely rearranged with a permutation similarity. In particular, if we take $S$ to be the permutation matrix $$ S = \pmatrix{1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1}, $$ then we find that $$ SPS^T = \pmatrix{-1&1&0&0\\1&-1&0&0\\0&0&-1&1\\0&0&1&-1},\quad SQS^T = \pmatrix{0&0&1&1\\0&0&1&1\\ 1&1&0&0\\1&1&0&0}. $$ Thus, we have $$ \begin{align*} \det(M) &= \det(SMS^T) = \det(I + a(SPS^T) + b(SQS^T)) \\ & = \det \left[ \begin{array}{rr|rr} 1-a & a & b & b\\ a & 1-a & b & b\\ \hline b & b & 1-a & a\\ b & b & a & 1-a \end{array}\right] = \det\pmatrix{A & B\\B & A}. \end{align*} $$ (If your prefer, this all amounted to switching the middle columns and the middle rows). With the block-determinant formula given here, we have $$ \det \pmatrix{A & B\\ B& A} = \det(A - B) \det (A + B) $$ This finally gives us $$ \det(M) = [(1 - a - b)^2 - (a-b)^2][(1 - a + b)^2 - (a+b)^2]\\ = [(1 - 2a)(1 - 2b)][(1 - 2a)(1 + 2b)]\\ = (1 - 2a)^2(1 - 2b)(1 + 2b). $$
For the second matrix, one approach is as follows. Let $M$ denote this matrix. Suppose for the time being that $r,s,t,u$ are all non-zero. Note that the dividing the last row by $rstu$ produces the matrix. $$ M_1 = \pmatrix{ 1 & 1 & 1 & 1\\ r & s & t & u\\ r^2 & s^2 & t^2 & u^2\\ 1/r & 1/s & 1/t & 1/u} $$ Cyclically permuting the rows of this matrix produces the matrix $$ M_2 = \pmatrix{ 1/r & 1/s & 1/t & 1/u\\ 1 & 1 & 1 & 1\\ r & s & t & u\\ r^2 & s^2 & t^2 & u^2 }. $$ Finally, multiplying the columns of this matrix by $r,s,t,u$ respectively produces the Vandemonde matrix $$ M_3 = \pmatrix{1 & 1 & 1 & 1\\ r & s & t & u\\ r^2 & s^2 & t^2 & u^2\\ r^3 & s^3 & t^3 & u^3 }. $$ You do not say what facts would be at your disposal for this exam, but I suspect that one such result is that the determinant of this final matrix is given by $$ \det(M_3) = (s - r)(t - r)(u - r)(t - s)(u - s)(u - t). $$ Some proofs of this result are presented here. From there, we can note that the nature of our steps imply that $$ \det(M_1) = \frac 1{rstu} \det(M)\\ \det(M_2) = -\det(M_1)\\ \det(M_3) = rstu \det(M_2). $$ Putting all this together, we have $\det(M_3) = -\det(M)$, which means that our final result is $$ \det(M) = -(s - r)(t - r)(u - r)(t - s)(u - s)(u - t). $$