I was given the following two equations and told to find $a, b$ and $c$:
$$\left\{\begin{array}{c}
a+b-c=1\\
a^2+b^2-c^2=-1\end{array}\right.$$
I tried to form a matrix to solve it and I made this:
$$\left[\begin{array}{ccc|c}
1 & 1 & -1 & 1\\
a & b & -c & -1
\end{array}
\right]$$
Then I performed elimination and got this:
$$\left[\begin{array}{ccc|c}
1&0&\frac{-b+c}{-a+b}&\frac{b+1}{-a+b}\\
0&1&\frac{a-c}{-a+b}&-\frac{a+1}{-a+b}
\end{array}\right]$$
After this I tried to find a specific solution when $c=0$ and a made this:
$$\left[\begin{matrix}
\frac{1\pm i\sqrt 3}{2}\\
\frac{1\mp i\sqrt 3}{2}\\
0
\end{matrix}\right]$$
Then, to find all the solutions I tried to find the null space while making $c=1$, then I found that all values are of the form:
$$\left[\begin{matrix}
\frac{1\pm i\sqrt 3}{2}\\
\frac{1\mp i\sqrt 3}{2}\\
0
\end{matrix}\right]
%
+C\left[\begin{matrix}
0/1\\
1/0\\
1
\end{matrix}\right]$$
Note: ‘/’ denotes $“or”$ and $C$ can be any integer
But when tried to plug this back into my question taking $C=1$ it didn’t work, but when $C=0$, it works, please help me.
Best Answer
There are only $2$ equations for $3$ variables so you can try to find two of them in function of the third one.
Since $a,b$ play a symmetrical role, we can choose between $b$ or $c$ as free variable. The choice of $c$ leads to solve quadratic equations, so we will choose $b$ instead.
First let $b=1$ then the system becomes $\begin{cases}a=c\\a^2-c^2=-2\iff 0=-2\end{cases}\quad$ the system has no solution.
So now we can exploit $a^2-c^2=\underbrace{(a-c)}_{1-b}(a+c)=-1-b^2\quad$ and solve $\begin{cases}a=\frac b{b-1}\\c=\frac{b^2-b+1}{b-1}\end{cases}$