For which values of $a,b\in\mathbb{R}$ does the matrix have no solutions, one solution and infinite solutions.

Question

For which values of $a,b\in\mathbb{R}$ does the matrix have no solutions, one solution and infinite solutions.

I have the following functions: $$ax+4y +az= 0 \\ x + ay + 3z = b \\ x(a+1) + y(a+4) + z(a-b^2) = b-2 $$ I have reduced this to: $$\left(\begin{array}{ccc|c} 1& a& 3 & b\\ 0 & 1 & \frac{1}{2a} & \frac{b}{4-a^2} \\ 0 & 0 & 1 & \frac{2}{b^2+3} \end{array}\right) $$ (can't figure out how to make a line in the matrix).

Using this I get to the point where I have $4xb^2+12x-a^2b^2-4a^2 = 8b^3+24b-b^3a^2+3ba^2$ what can I conclude from this, and have I done anything wrong here?

Answer 1

First swap the first and second row. Then by subtracting the first and the second row from the third one and then subtracting $a$ times the first row from the second we get:

$$\left(\begin{array}{ccc|c} 1& a& 3 & b\\ 0 & 4-a^2 & -2a & -ab \\ 0 & 0 & -b^2-3 & -2 \end{array}\right) $$

If $a \not = \pm 2$ we have a unique solution, as the coefficient matrix would be invertible, considering that $-b^2 - 3 \le -3 < 0$

if $a=\pm2$ by substituting the matrix will reduce to:

$$\left(\begin{array}{ccc|c} 1& \pm 2& 3 & b\\ 0 & 0 & 1 & \frac b2 \\ 0 & 0 & 0 & -2 + \frac{b^3 + 3b}{2} \end{array}\right) $$

The last equation has infinitely many solutions if $-2 + \frac{b^3 + 3b}{2} = 0$ and none otherwise.

Answer 2

Hint

The system has solution if an only if $$\rm{ran}\begin{pmatrix} a & 4 & a\\ 1 & a & 3 \\ a+1 & a+4& a-b^2\end{pmatrix}=\rm{ran}\begin{pmatrix} a & 4 & a& 0\\ 1 & a & 3 & b\\ a+1 & a+4& a-b^2& b-2\end{pmatrix}.$$ Moreover, it has only a solution if and only if both ranks are $3$ and infinitely many solutions if and only if both ranks are less than $3$ (and equal, of course).

(See Rouché-Capelli theorem https://en.wikipedia.org/wiki/Rouch%C3%A9%E2%80%93Capelli_theorem)

So, we have study the rank of both matrices in terms of $a,b$ and conclude. Note that

$$\rm{ran}\begin{pmatrix} a & 4 & a\\ 1 & a & 3 \\ a+1 & a+4& a-b^2\end{pmatrix}=\rm{ran}\begin{pmatrix} a & 4 & a\\ 1 & a & 3 \\ 0 & 0 & -b^2-3\end{pmatrix}.$$ Now

$$\begin{vmatrix} a & 4 & a\\ 1 & a & 3 \\ 0 & 0 & -b^2-3\end{vmatrix}=-(b^2+3)(a^2-4).$$ So, if $a\ne \pm 2$ then both ranks are $3$ and the system has only one solution. Study the cases $a=\pm 2.$