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!:)

 

[CellCoree]

I understand the importance of finding solutions to systems of equations in order to better understand and solve real-world problems. However, in this particular system of equations, it can be shown that there are no real solutions.

Firstly, let's examine the first two equations: $a^2+bd=0$ and $c^2+bd=0$. We can see that both equations have the same left-hand side, which means that $a^2=c^2$. Since the square of a number is always positive, this can only be true if both $a$ and $c$ are equal to 0. This means that the first two equations can be rewritten as $bd=0$ and $bd=0$, respectively.

Next, let's look at the third and fourth equations: $ab+bc=1$ and $ad+cd=1$. By rearranging the terms, we can see that both equations have the same right-hand side, which means that $ab+bc=ad+cd$. By factoring out $b$ and $d$, we get $b(a+c)=d(a+c)$. Since $a=c=0$ from the previous equations, this becomes $0=0$. This means that the third and fourth equations are actually the same equation and do not provide any additional information.

Putting all of this together, we can see that the system of equations reduces to $bd=0$ and $0=0$. This means that either $b$ or $d$ must equal 0 in order for the first equation to be satisfied. However, this would contradict the fact that $ad+cd=1$ in the fourth equation.

Therefore, there are no real solutions to this system of equations. It is possible that there may be complex solutions, but in the context of real numbers, this system has no solutions. This highlights the importance of carefully examining and analyzing systems of equations in order to determine their solutions.