- Regula lui Sarrus: se pun dedesubt primele 2 linii ale determinantului,apoi se calculează prin suma produselor numerelor de pe diagonalele descendente minus suma produselor numerelor de pe diagonalele ascendente, dupa cum urmeaza:
b.
[tex]\left|\begin{array}{ccc}1&1&1\\-x&-y&-z\\x^2&y^2&z^2\end{array}\right|\\\left\begin{array}{ccc}\ \ 1&\ \ \11&\ \ \ 1\\-x&-y&-z\\\end{array}\right=[/tex]
-yz²-xy²-zx²-(-yx²-zy²-xz²)=
-yz²-xy²-zx²+yx²+zy²+xz²=
yx²+zy²+xz²-yz²-xy²-zx²=
Le grupa 2 cate 2 convenabil, astfel
x²y-xy²-zx²+zy²+z²x-z²y=
xy(x-y)-z(x²-y²) + z²(x-y)=
xy(x-y)-z(x-y)(x+y) + z²(x-y)=
(x-y)(xy-zx-zy+z²)
d.
[tex]\left|\begin{array}{ccc}1&2&1\\a&2&b\\-1&0&-1\end{array}\right|\\\left\begin{array}{ccc}\ \ 1&\ \ 2&\ \ 1\\ \ a&\ 2&b\\\end{array}\right[/tex]
-2+0-2b-(-2+0-2a)=
-2-2b+2+2a=
2a-2b=2(a-b)
e.
[tex]\left|\begin{array}{ccc}3&-i&0\\i&3&0\\0&0&\alpha\end{array}\right|\\\left\begin{array}{ccc}\ 3&-i&0\\ \ i&\ 3&0\\\end{array}\right[/tex]
=9α+0+0-(0+0-i²α)=9α-α=8α
