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Describe how the number of solutions to the system: \begin{eqnarray*} x+y+z &=&1 \\ 2x+y+a^2z &=&a \\ x+3y+3z &=&3 \end{eqnarray*} is dependant on the value of $a$.

My attempt at a solution:

First we write the the equation system in matrix form:

$$\left[\begin{array}{ccc|c}1 & 1 & 1 & 1\\2 & 1 & a^2 & a\\1 & 3 & 3 & 3\end{array}\right]$$

Now, after subtracting the two times first row from the second we get:

$$\left[\begin{array}{ccc|c}1 & 1 & 1 & 1\\0 & -1 & a^2-2 & a-2\\1 & 3 & 3 & 3\end{array}\right]$$

And subtract one time the first row from the third:

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

Then we subtract one half the third row from the first:

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

We continue with adding two times the second row to the third:

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

We divide the third row by $2$ and simplify the expressions.

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

We subtract one time the third row from the second row:

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

This result gives me the following conclusion:

If $a = 1$ the system will have infinitely many solutions.

If $a = -1$ the system will not have any solutions,

For all other values of $a$ the system will have a unique solution.

Am I thinking correctly?

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  • $\begingroup$ Assuming your computations are right, your answer is correct. $\endgroup$
    – mfl
    Nov 3, 2014 at 15:49
  • $\begingroup$ Can't find any flaws $\endgroup$
    – konewka
    Nov 3, 2014 at 15:51
  • $\begingroup$ Thank you very much for your input! Feel free to add that as an asnwer and I will accept that. $\endgroup$ Nov 3, 2014 at 16:01

1 Answer 1

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If $A$ denotes the $3\times 3$-matrix of your equation $Ax=b$, then $\det(A)=-2(a-1)(a+1)$. Hence if the field has not characteristic two, and $(a-1)(a+1)\neq 0$, then we have a unique solution. You have already solved this case correctly. Note that you have divided by $2$ here. For $char(K)=2$ we always have $\det(A)=0$. The system then is given by \begin{align*} x+y+z& =1, \\ y+a^2z & = a. \\ \end{align*} We can solve this by setting $x=1-y-z$ and $y=a(1-az)$. If $K$ has infinitely many elements, then we have infinitely many solutions, independent of the value of $a$.

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  • $\begingroup$ Thank you very much for your answer! Much appreciated! $\endgroup$ Nov 3, 2014 at 19:14
  • $\begingroup$ You are welcome ! The case with characteristic two is probably not relevant for you. It is needed in coding theory, for example. $\endgroup$ Nov 3, 2014 at 20:25

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