MHB No Real Solutions to System of Equations

AI Thread Summary
The system of equations is shown to have no real solutions through matrix algebra. By defining matrix A and calculating A squared, it results in a determinant of -1, indicating that the eigenvalues of A are imaginary (±i). This implies that the variables a, b, c, and d cannot all be real numbers. The conclusion reinforces the impossibility of finding real solutions for the given equations. The discussion highlights the effectiveness of matrix methods in solving such problems.
anemone
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Show that the following system of equations have no real solutions:

$a^2+bd=0$

$c^2+bd=0$

$ab+bc=1$

$ad+cd=1$
 
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anemone said:
Show that the following system of equations have no real solutions:

$a^2+bd=0$

$c^2+bd=0$

$ab+bc=1$

$ad+cd=1$

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
 
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]
 
Hi kaliprasad and Opalg!

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