Solve System of Equations for a×b×c×d

[Albert1]

$a,b,c,d \in R$

$ad - bc =1 -------(1)$

$a^2+b^2+c^2+d^2 - ab +cd =1--------(2)$

$\text {find :} a\times b\times c\times d$

 

[BAdhi]

Re: find: a ᵡ b ᵡ c ᵡ d

$\text{2$\times$(2)-2$\times$(1)}\implies$
$$\begin{align*}
(2a^2+2b^2+2c^2+2d^2-2ab+2cd)-(2ad-2bc)&=2-2\\
(a^2-2ad+d^2)+(a^2-2ab+b^2)+(b^2+2bc+c^2)+(c^2+2cd+d^2)&=0\\
(a-d)^2+(a-b)^2+(b+c)^2+(c+d)^2&=0
\end{align*}$$
to stand the equality,
$$
(a-d)=0,\quad (a-b)=0, \quad (b+c)=0,\quad (c+d)=0\\
d=a, \quad b=a, \quad c=-b=-a
$$

from (1),
$$\begin{align*}
ad-bc&=1\\
a(a)-(a)(-a)&=1\\
2a^2&=1\\
a^2&=\frac 12
\end{align*}$$

$$\therefore a\times b\times c\times d=a\times a \times (-a)\times a= -a^4=-\left(\frac{1}{2}\right)^2=-\frac 14$$

 

[Albert1]

Re: find: a ᵡ b ᵡ c ᵡ d

perfect ! you got it