<s>[INST] Determinant of a $4 \times 4$ matrix $A$ and $(\det(A))^5$ 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.
 [/INST] 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
$$
\det A^n = \det \begin{pmatrix}
a^n&*&*&*\\
0&(b-a)^n&*&*\\
0&0&(c-b)^n&*\\
0&0&0&(d-c)^n
\end{pmatrix}
$$
</s>