No Real Solutions to System of Equations

[anemone]

Show that the following system of equations have no real solutions:

$a^2+bd=0$

$c^2+bd=0$

$ab+bc=1$

$ad+cd=1$

 

[kaliprasad]

Show that the following system of equations have no real solutions:

$a^2+bd=0$

$c^2+bd=0$

$ab+bc=1$

$ad+cd=1$

Spoiler

from 1st 2 equations we have

$a^2=c^2$

so a = c or a = -c

now for a = c

we have

2ab = 1 and 2ad = 1 so bd and d both are same

so $a^2 + b^2 = 0$ from 1st equation so a = b = 0

hence a = b = c= d = 0 so we have contradiction in 3rd relation

if a = -c

then

ab + bc = 0 which is contradicton to 3rd relation

hence in both cases no solution

 

[Opalg]

Solution using matrix algebra:
[sp]Let $A = \begin{bmatrix} a&b \\ d&c \end{bmatrix}$. Then $A^2 = \begin{bmatrix} a^2 + bd&ab+bc \\ ad+dc&c^2+bd \end{bmatrix} = \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}$. The determinant of $A^2$ is $-1$, so the determinant of $A$ is $\pm i$. That is not real, so $a,b,c,d$ cannot all be real.[/sp]

 

[anemone]

Hi kaliprasad and Opalg!

Very well done(Yes) and thanks for participating to this challenge!:)