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Calculate $\det(A)$ and $\det(A)^5$: $$A= \begin{bmatrix}a&a&a&a\\a&b&b&b\\a&b&c&c\\a&b&c&d\end{bmatrix}$$
I found $\det A$ with Laplace expansion: $$a(-abc+b^2c+a c^2-b c^2+a b d-b^2 d-a c d+b c d) .$$ But how can I determine $\det A^5$ easily/fast? I know that there is the possibility to use $\det(A^5)=\det(A)^5)$, but this is too long for given time to resolve the problem.
Of course $\det (A^5) = (\det A)^5$ is just the fifth power of the expression you produced, but notice that we can also factor the given expression as $$\det A = a (b - a) (c - b) (d - c) ,$$ which one can also produce quickly by row reducing $A$ to an upper triangular matrix.
You can much easier get your expression by subtracting the third line from the fourth, then the second from the third and finally, the first from the second: $$ \det A= \det \begin{pmatrix} a&a&a&a\\ a&b&b&b\\ a&b&c&c\\ 0&0&0&d-c \end{pmatrix}= \det \begin{pmatrix} a&a&a&a\\ a&b&b&b\\ 0&0&c-b&c-b\\ 0&0&0&d-c \end{pmatrix}= \det \begin{pmatrix} a&a&a&a\\ 0&b-a&b-a&b-a\\ 0&0&c-b&c-b\\ 0&0&0&d-c \end{pmatrix}=a(b-a)(c-b)(d-c) $$ You can easily find $\det (A^5)$ now and even say that $$ \det A^n = \det \begin{pmatrix} a^n&*&*&*\\ 0&(b-a)^n&*&*\\ 0&0&(c-b)^n&*\\ 0&0&0&(d-c)^n \end{pmatrix} $$
If calculations are done mentally but not jotted down, I think the quickest way to calculate $\det A$ is to note that \begin{align} A&=a(e_1+e_2+e_3+e_4)(e_1+e_2+e_3+e_4)^T\\ &+(b-a)(e_2+e_3+e_4)(e_2+e_3+e_4)^T\\ &+(c-b)(e_3+e_4)(e_3+e_4)^T\\ &+(d-c)e_4e_4^T. \end{align} Therefore $A$ has the $LDL^T$ decomposition $$ A=\pmatrix{1&0&0&0\\ 1&1&0&0\\ 1&1&1&0\\ 1&1&1&1} \pmatrix{a\\ &b-a\\ &&c-b\\ &&&d-c} \pmatrix{1&1&1&1\\ 0&1&1&1\\ 0&0&1&1\\ 0&0&0&1} $$ and $\det A=a(b-a)(c-b)(d-c)$. And as noted by the others, $\det A^5$ is just $(\det A)^5$.
The above decomposition has the additional merit that it gives also the signature of $A$. In particular, $A$ and $A^5$ are positive semidefinite if and only if $d\ge c\ge b\ge a\ge0$.
@Michael Freimann
What follows is not an answer. It is just an aside comment, that might interest some readers.
The kind of matrices considered in this question can be described by the following formula for their coefficients
$$A_{ij}=a_{min(i,j)}$$ for a given sequence $a_1,a_2,...a_n$, this sequence being here {a,b,c,d}.
Their inverses have a tridiagonal form with simple coefficients. For example, here:
$A^{-1}=\begin{pmatrix} \frac{1}{a} + \frac{1}{b-a}&\frac{1}{a - b}&0&0\\ \frac{1}{a - b}&\frac{1}{b-a} + \frac{1}{c-b}&\frac{1}{b - c}&0\\ 0&\frac{1}{b - c}&\frac{1}{c-b} + \frac{1}{d-c }&\frac{1}{c - d}\\ 0&0&\frac{1}{c - d}&\frac{1}{d-c}\end{pmatrix}$
In particular, when $a=b=c=d=1$,
$$\text{The inverse of} \ \ \begin{pmatrix} 1&1&1&1\\ 1&2&2&2\\ 1&2&3&3\\ 1&2&3&4 \end{pmatrix} \ \ \text{is} \ \ \frac{1}{2}\begin{pmatrix} 2&-1&0&0\\ -1&2&-1&0\\ 0&-1&2&-1\\ 0&0&-1&1\end{pmatrix} $$ One can recognized in the last matrix an approximation of the opposite of the second derivative, a very important matrix in numerical analysis: see this (slides 10, 29, 35, 37, 42...).
This is in clear connection with the decomposition given by @user1551, which, for $a=c=d=1$, gives:
$$A=\pmatrix{1&0&0&0\\ 1&1&0&0\\ 1&1&1&0\\ 1&1&1&1} \pmatrix{1&1&1&1\\ 0&1&1&1\\ 0&0&1&1\\ 0&0&0&1}$$
which is equivalent to a double discrete integration operator.
Of course, this property extends naturally to any dimension. Moreover, it has a nice correspondence with the "continuous world": see this.
