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I'm doing some work in linear algebra and these came up and I realized I don't know how to solve them as they have quadratics in them. I'm sure I've done this before but if someone could give me a crash course on how to find the values of the variables, it would be most appreciated. For the following questions, let $a,b,c,d \in \mathbb{R}$

1)

$\begin{cases} a^2+b^2=a\\ (a+d)b=b\\ b^2+d^2=d \end{cases}$

2)

$\begin{cases} a^2+bc=a\\ (a+d)b=b\\ (a+d)c=c\\ bc+d^2=d \end{cases}$

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  • $\begingroup$ $a,b,c,d$ are reals? Or integers? $\endgroup$ – Inceptio May 29 '13 at 4:50
  • $\begingroup$ @Inceptio They are all real numbers. $\endgroup$ – Sujaan Kunalan May 29 '13 at 4:52
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1)

$\begin{cases} a^2+b^2=a\\ (a+d)b=b\\ b^2+d^2=d \end{cases}$

in equation $2$ divide both side by b $\begin{cases} a^2+b^2=a\\ a+d=1\\ b^2+d^2=d \end{cases}$

from eqn $2$ and $3$

$$a=1-d$$ $$b^2=d-d^2$$

put these value in eqn $1$

$$(1-d)^2+d-d^2=1-d\implies d=0$$ it will give $a=1,b=0$

2) $\begin{cases} a^2+bc=a\\ (a+d)b=b\\ (a+d)c=c\\ bc+d^2=d \end{cases}$

in eqn $2,3$ divide both side by b and c

$ a^2+bc=a\\ a+d=1\\ a+d=1\\ bc+d^2=d $

$ a^2+bc=a\\ a+d=1\\ bc+d^2=d $

from eqn $1$ and $2$ $$bc=a-a^2$$ $$d=1-a$$

put these value in $3$ $$a-a^2+(1-a)^2=1-a\implies a=0$$ that will give $d=1$ and either of b and c will 0. $$$$

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  • $\begingroup$ Many solutions are missing. $\endgroup$ – André Nicolas May 29 '13 at 5:51
  • $\begingroup$ yes it is a one set of solution.If I will substitute different values from different eqn then it will give another set.Should I mention in answer? $\endgroup$ – iostream007 May 29 '13 at 6:48
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I tried to solve 1) in your question. $$\begin{cases} a^2+b^2=a ----(1)\\ (a+d)b=b ----(2)\\ b^2+d^2=d ----(3) \end{cases}$$ when $b=0$, the equations (1) and (3) boil down to $a^2=a$ and $d^2=d$. In this case, we have the solutions below $$\{a=0,b=0,d=0\},\\ \{a=0,b=0,d=1\},\\ \{a=1,b=0,d=0\},\\ \{a=1,b=0,d=1 \}.$$
If $b\ne0$, then equations (1), (2) and (3) becomes \begin{cases} a^2-a +b^2=0 ----(4)\\ a+d =1 -------(5)\\ d^2-d +b^2= 0----(6) \end{cases} From (4),(5)and (6), we can see $a$ and $d$ are the solutions of the equation $x^2-x+b^2=0$, where $b$ can be any real number. So in this case, the solutions could be $$\left\{a=\frac{1}{2} \left(1-\sqrt{1-4 b^2}\right),d=\frac{1}{2} \left(\sqrt{1-4 b^2}+1\right)\right\},\\ \left\{a=\frac{1}{2} \left(\sqrt{1-4 b^2}+1\right),d=\frac{1}{2} \left(1-\sqrt{1-4 b^2}\right)\right\}. $$ Hint: your question 2) can be solved in a similar way.

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