Gaussian elimination involving parameters

Question

The problem is :Solve the given system of equations involving the parameter a :

$$x+y+az=1\\ x+ay+z=a\\ ax+y+z=a^2\\ ax+ay+az=a^3 .$$

I tried to solve this using the Gaussian method but I'm stuck because this is $4\times3$ matrix, and the Gaussian process is used for square matrix ? Please help.....

Answer 1

You can solve the system formed by the first three equations, which will let you express $x,y,z$ as a function of $a$.

Then plugging these expressions in the fourth equation, you will get a condition on $a$ for the system to be resolvable.

Answer 2

recommend rewriting the first three in order $$ ax+y+z=a^2\\ x+ay+z=a\\ x+y+az=1 $$

since the eigenvalues, determinant, and inverse (when there is one) of $$ \left( \begin{array}{ccc} a & 1 & 1\\ 1 & a & 1 \\ 1 & 1 & a \end{array} \right) $$ are pretty easy to find. It is of the form $ T +(a-1)I,$ where $T$ is the 3 by 3 matrix consisting of all entries equal to $1.$

Now that i think of it, the matrix has no inverse when $a=1$ or $a=-2,$ you might as well do those two cases using the actual numbers involved, no symbols.

Answer 3

Subsitute $y=1-x-az$ from the first equation and then $z= - (a - x + 1)$ from the second equation, assuming $a\neq 1$. Then the third equation gives $x=(a^2+2a+1)/(a+2)$, the case $a=-2$ being impossible. Then the fourth equation gives $$ a(a+1)(a-1)=0. $$ For $a=0$ we have a $3\times 3$ system with unique solution $(x,y,z)=(\frac{1}{2},-\frac{1}{2},\frac{1}{2})$. For $a=1$ we obtain $(x,y,z)=(1-y-z,y,z)$ with two free parameters $x,y$. Otherwise it follows that $a=-1$ and $(x,y,z)=(0,1,0)$.