# Solving systems of equation with unknown

How do i solve this system?

$$\left\{ \begin{array}{c} ax+y+z=1 \\ x+ay+z=a \\ x+y+az=a^2 \end{array} \right.$$

Ive reduced to this form. How should i continue to get infinitely many solutions, no solution and unique?

$$\left[ \begin{array}{ccc|c} 1&a&1&a\\ 0&(a^2)-1&a-1&(a^2)-1\\ 0&(a-1)(a-2)&0&a-1 \end{array} \right]$$

• Consider the last line. What happens for $a=1$, $a=2$ and $a\notin\{1,2\}$? – Mundron Schmidt Oct 10 '17 at 18:24

hint:$$\left\{ \begin{array}{c} ax+y+z=1 \\ x+ay+z=a \\ x+y+az=a^2 \end{array} \right.$$sum of the equations : $$x(a+2)+y(a+2)+z(a+2)=1+a+a2\\x+y+z=\frac{1+a+a^2}{a+2}\\$$now $$ax+y+z=1-(x+y+z=\frac{1+a+a^2}{a+2})\\(a-1)x=1-\frac{1+a+a^2}{a+2}\\(a-1)=\frac{a+2-1-a-a^2}{a+2}\\x=\frac{1-a^2}{(a+2)(a-1)}$$

If $A$ is the matrix of the coefficients, then for Cramer's theorem the system has one unique solution if $\det A\ne 0$ $\det A=\left| \begin{array}{lll} a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & a \\ \end{array} \right|=a^3-3 a+2$

$\det A=0 \to (a-1)^2 (a+2)=0\to a_1=1;\;a_2=-2$

Then for $a\ne 1;\;a\ne 1$ the system has one and only one solution

$$x=-\frac{a+1}{a+2},\;y=\frac{1}{a+2},\;z=\frac{(a+1)^2}{a+2}$$

$$.$$ If $a=1$ the completed matrix $A|B$ of the system becomes

$A|B=\left( \begin{array}{lll|l} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ \end{array} \right)$

As $\text{rank } A= \text{rank } A|B=1$

the system has $\infty^{3-1}=\infty^2$ solutions given by the equation

$x+y+z=1$ whose solutions are $(t,u,1-t-u)$

they have two parameters which is linked to the $\infty^2$ solutions $$.$$

If $a=-2$

$A|B=\left( \begin{array}{rrr|r} -2 & 1 & 1 & 1 \\ 1 & -2 & 1 & -2 \\ 1 & 1 & -2 & 4 \\ \end{array} \right)$

$\text{rank } A|B=3$ while $\text{rank }A=2$

they are different so the system is impossible, has no solutions.

Hope this helps

We have to obtain the reduced row echelon form of the augmented matri, starting from $$\begin{bmatrix}\!\!\begin{array}{@{}ccc|c} 1&a&1&a\\0&a^2-1&a-1&a^2-1\\0&(a-1)(a-2)&0&a-1 \end{array}\!\!\end{bmatrix}$$

1. If $a=1$, this matrix becomes $\;\smash[b]{\begin{bmatrix}\!\!\begin{array}{@{}ccc|c} 1&1&1&1\\0&0&0&0\\0&0&0&0 \end{array}\!\!\end{bmatrix}}$, which corresponds to the plane with equation: $$\color{red}{x+y+z=1}.$$
2. If $a\ne 1$, we can factor out $x-1$, and thus obtaining the matrix $$\begin{bmatrix}\!\!\begin{array}{@{}ccc|c} 1&a&1&a\\0&a+1&1&a+1\\0&a-2&0&1 \end{array}\!\!\end{bmatrix},\enspace\text{and swapping columns 2 and 3}:\quad \begin{bmatrix}\!\!\begin{array}{@{}ccc|c} 1&1&a&a\\0&1&a+1&a+1\\0&0&a-2&1 \end{array}\!\!\end{bmatrix}$$ (Swapping these columns amounts to exchange unknowns $y$ and $z$).
3. If $a=2$, the linear system is inconsistent.
4. If $a\ne2$, we can proceed with row reduction: \begin{align} \begin{bmatrix}\!\!\begin{array}{@{}ccc|c} 1&1&a&a\\0&1&a+1&a+1\\0&0&a-2&1 \end{array}\!\!\end{bmatrix}&\rightsquigarrow\begin{bmatrix}\!\!\begin{array}{@{}ccc|c} 1&1&a&a\\0&1&a+1&a+1\\0&0&1&\frac1{a-2} \end{array}\!\!\end{bmatrix}\rightsquigarrow\begin{bmatrix}\!\!\begin{array}{@{}ccc|c} 1&1&0&\frac{a(a-3)}{a-2}\\0&1&0&\frac{(a+1)(a-3)}{a-2}\\0&0&1&\frac1{a-2} \end{array}\!\!\end{bmatrix}\\ &\rightsquigarrow \begin{bmatrix}\!\!\begin{array}{@{}ccc|c} 1&0&0&-\frac{(a-3)}{a-2}\\0&1&0&\\0&0&1&\frac1{a-2} \end{array}\!\!\end{bmatrix} \end{align} So the solutions are $$\color{red}{x=-\frac{(a-3)}{a-2}},\quad \color{red}{y=\frac1{a-2}},\quad \color{red}{z=\frac{(a+1)(a-3)}{a-2}}.$$