For what values of $a$, $b$, and $c$ the above system has: One solution. Infinitely many solutions. No solutions.

Question

I am stuck with this now, I tried reducing the matrix to row echelon form, but it gets a bit hard. Is there not a simpler way?

The system is:

\begin{align*} a x + b y − 3 z &= −3\\ −2 x − b y + c z &= −1\\ a x + 3 y − c z &= −3 \end{align*}

Answer 1

HINT: eliminating $y,z$ we get the following equation for $x$: $$2 b c x-7 b c+9 b+9 c-18 x+9=a x (2 b c-3 b-3 c)$$

Answer 2

Compute the determinant of the system, you will get $$ 2abc-3ab-3ac-2bc+18 $$ When is this zero? Well, we can solve for $c$, $$ c = \frac{3(ab-6)}{2ab-3a-2b} $$ as long as $2ab-3a-2b \neq 0$. So when is $2ab-3a-2b = 0$? More algebra gives us $$ b = \frac{3a}{2(a-1)} $$ if $a \neq 1$. What about when $a=1$? Then $2ab-3a-2b = -3 \neq 0$, so we would be in the first case. Otherwise, let's substitute that value of $b$ in to the original equation, we get $$ 0 =3(ab-6) = 3(a\frac{3a}{2(a-1)} - 6) = \frac{a-2}{a-1} $$ which implies $a=2$ which implies $b=3$.

To summarize, the equation will have a unique solution unless $$ 2ab-3a-2b \neq 0 \text{ and } c = \frac{3(ab-6)}{2ab-3a-2b} $$ or $$ a=2, b=3, c = \text{anything} $$

Now for these possibilities, what happens? If $a=2,b=3$ then row 2 is the negative of row 3, but $-1$ is not the negative of $-3$, so there are no solutions. Then try using row echelon form on the matrix.