2
$\begingroup$

I know there is no specific way, but what is, in your opinion, the best way to discuss a system of linear equations like: $ \left\{ \begin{array}{l} ax+by+z = 1\\ x+aby+z=b\\ x+by+bz=1 \end{array} \right. $ $(A|B) = \left(\begin{array}{ccc|c} a & b & 1 & 1\\ 1 & ab & 1 & b\\ 1 & b & b & 1 \end{array} \right) $ If $A$ is the coefficient matrix, $\det A = b(a-1)(-2+b+a b)$, so it is a little bit hard to discuss the system using determinants, right?

  • 0
    See my answer below in which I discuss the case $-2+b+ab=0$.2012-12-02

2 Answers 2

1

If $b=0$ the system leads to also $a=2$ and then the solution is $x=1,z=1$ with $y$ arbitrary.

If $a=1$ it leads to also $b=1$ and in this case the system becomes the single relation $x+y+z=1$.

Now if we assume that $b$ is not zero (since the case $b=0$ already dealt with), but that $-2+b+ab=0$, then we have $a=(2-b)/b$ [OK since $b$ nonzero]. In this case the system is inconsistent unless it happens that $b^2+b-2=0$, i.e. $b=1,-2$. The case $b=1$ leads again to simply $x+y+z=1$, while the case $b=-2$ has also $a=-2$ and the solution may be written $x=-z$, $y=-1/2-1/2z$, and $z$ arbitrary.

In the remaining case where the determinant is nonzero, Cramer's rule can be used. It leads to $x = \frac{ (a-b)(b-1)}{(a-1)(ab+b-2)},$ $y=\frac{(ab+a-2)(b-1)}{(a-1)(ab+b-2)b},$ $z=\frac{a-b}{ab+b-2}.$

1

Whenever you have a parameterized system of some sort, it is often useful to think in terms of the parameter space.

In this particular case, there are specific regions in the parameter space where the determinant is zero, and as such the system is singular. It may be useful to visualize these regions, and depending on what your system represents, these regions might separate your parameter space into sub-regions that lead to different intuitive and/or mathematical properties (for a great example, think Hopf bifurcations).

The equation for your determinant gives you an easy way to describe this.