Определите число решений следующей системы в зависимости от значений параметров a и b

$
\, \,
\begin{cases}
ax + 6y + 4z = -1
\\
y + 3z = 3
\\
-3x + bz = 3
\end{cases}
\\\\\\
\left(
\begin{array}{ccc|c}
a & 6 & 4 & -1
\\
0 & 1 & 3 & 3
\\
-3 & 0 & b & 3
\end{array}
\right)
=>
\left(
\begin{array}{ccc|c}
a & 0 & -14 & -19
\\
0 & 1 & 3 & 3
\\
-3 & 0 & b & 3
\end{array}
\right)
\\\\\\
1) \, a = 0 \quad
\left(
\begin{array}{ccc|c}
0 & 0 & -14 & -19
\\
0 & 1 & 3 & 3
\\
-3 & 0 & b & 3
\end{array}
\right)
(1) <--> (3)
\left(
\begin{array}{ccc|c}
-3 & 0 & b & 3
\\
0 & 1 & 3 & 3
\\
0 & 0 & -14 & -19
\end{array}
\right)
(1) = (1) / (-3)
\left(
\begin{array}{ccc|c}
1 & 0 & \frac{-b}{3} & -1
\\
0 & 1 & 3 & 3
\\
0 & 0 & -14 & -19
\end{array}
\right)
\\\\\\
(3) = (3) / (-14)
\left(
\begin{array}{ccc|c}
1 & 0 & \frac{-b}{3} & -1
\\
0 & 1 & 3 & 3
\\
0 & 0 & 1 & \frac{19}{14}
\end{array}
\right)
(2) = (2) - 3 * (3)
\left(
\begin{array}{ccc|c}
1 & 0 & \frac{-b}{3} & -1
\\
0 & 1 & 0 & 3 - \frac{57}{14}
\\
0 & 0 & 1 & \frac{19}{14}
\end{array}
\right)
(1) = (1) + \frac{-b}{3} * (3)
\left(
\begin{array}{ccc|c}
1 & 0 & 0 & \frac{-42 + 19b}{42}
\\
\\
0 & 1 & 0 & \frac{-15}{14}
\\
\\
0 & 0 & 1 & \frac{19}{14}
\end{array}
\right)
\\\\\\
\left(
\begin{array}{ccc|c}
x
\\
\\
y
\\
\\
z
\end{array}
\right)
=
\left(
\begin{array}{ccc|c}
\frac{-42 + 19b}{42}
\\
\\
\frac{-15}{14}
\\
\\
\frac{19}{14}
\end{array}
\right)
\\\\\\
2) \, a \neq 0
\\\\\\
\left(
\begin{array}{ccc|c}
a & 0 & -14 & -19
\\
0 & 1 & 3 & 3
\\
-3 & 0 & b & 3
\end{array}
\right)
(1) = (1)/a
\left(
\begin{array}{ccc|c}
1 & 0 & \frac{-14}{a} & \frac{-19}{a}
\\
0 & 1 & 3 & 3
\\
-3 & 0 & b & 3
\end{array}
\right)
(3) = (3) + 3 * (1)
\left(
\begin{array}{ccc|c}
1 & 0 & \frac{-14}{a} & \frac{-19}{a}
\\
0 & 1 & 3 & 3
\\
0 & 0 & b - \frac{42}{a} & 3 - \frac{57}{a}
\end{array}
\right)
\\\\\\
2.1) \, b - \frac{42}{a} = 0
\\\\\\
\left(
\begin{array}{ccc|c}
1 & 0 & \frac{-14}{a} & \frac{-19}{a}
\\
0 & 1 & 3 & 3
\\
0 & 0 & 0 & 3 - \frac{57}{a}
\end{array}
\right)
\\\\\\
2.1.1) \, 3 - \frac{57}{a} \neq 0
\quad
a \neq 19 \quad \text{Нет решений}
\\\\\\
2.1.2) \, 3 - \frac{57}{a} = 0
\quad
a = 19 \quad => b = \frac{42}{19}
\\\\\\
\left(
\begin{array}{ccc|c}
1 & 0 & \frac{-14}{19} & -1
\\
0 & 1 & 3 & 3
\\
0 & 0 & 0 & 0
\end{array}
\right)
=>
\left(
\begin{array}{ccc|c}
x
\\
y
\\
z
\end{array}
\right)
=
\left(
\begin{array}{ccc|c}
-1 + \frac{14}{19}z
\\
3 - 3z
\\
z
\end{array}
\right),
z \in \mathbb{R}
\\\\\\
2.2) \, b - \frac{42}{a} \neq 0
\\\\\\
\left(
\begin{array}{ccc|c}
1 & 0 & \frac{-14}{a} & \frac{-19}{a}
\\
0 & 1 & 3 & 3
\\
0 & 0 & b - \frac{42}{a} & 3 - \frac{57}{a}
\end{array}
\right)
=> (3) = \frac{(3)}{b - \frac{42}{a}}
\left(
\begin{array}{ccc|c}
1 & 0 & \frac{-14}{a} & \frac{-19}{a}
\\
0 & 1 & 3 & 3
\\
0 & 0 & 1 & \frac{3a - 57}{ab - 42}
\end{array}
\right)
=> (2) = (2) - 3 * (3)
\left(
\begin{array}{ccc|c}
1 & 0 & \frac{-14}{a} & \frac{-19}{a}
\\
\\
0 & 1 & 0 & \frac{3ab - 9a + 45}{ab - 42}
\\
\\
0 & 0 & 1 & \frac{3a - 57}{ab - 42}
\end{array}
\right)
\\\\\\
=> (1) = (1) + \frac{14}{a} * (3)
\left(
\begin{array}{ccc|c}
1 & 0 & 0 & \frac{-19ab + 42a - 1596}{a(ab - 42)}
\\
\\
0 & 1 & 0 & \frac{3ab - 9a + 45}{ab - 42}
\\
\\
0 & 0 & 1 & \frac{3a - 57}{ab - 42}
\end{array}
\right)
\\\\\\
\left(
\begin{array}{ccc}
x
\\
y
\\
z
\end{array}
\right)
=
\left(
\begin{array}{ccc|c}
\frac{-19ab + 42a - 1596}{a(ab - 42)}
\\
\\
\frac{3ab - 9a + 45}{ab - 42}
\\
\\
\frac{3a - 57}{ab - 42}
\end{array}
\right)
\\\\\\
\text{Ответ: } \\\\
\text{При а = 0: } \\\\\\
\left(
\begin{array}{ccc|c}
x
\\
\\
y
\\
\\
z
\end{array}
\right)
=
\left(
\begin{array}{ccc|c}
\frac{-42 + 19b}{42}
\\
\\
\frac{-15}{14}
\\
\\
\frac{19}{14}
\end{array}
\right)
\\\\\\
\text{При а} \neq 0 \text{ и } b - \frac{42}{a} = 0 \text{ и } a \neq 19 \text{ нет решений.} \\\\\\
\text{При а} \neq 0 \text{ и } b - \frac{42}{a} = 0 \text{ и } a = 19 \\\\\\
\left(
\begin{array}{ccc|c}
x
\\
y
\\
z
\end{array}
\right)
=
\left(
\begin{array}{ccc|c}
-1 + \frac{14}{19}z
\\
3 - 3z
\\
z
\end{array}
\right),
z \in \mathbb{R}
\\\\\\
\text{При } b - \frac{42}{a} \neq 0:
\left(
\begin{array}{ccc}
x
\\
y
\\
z
\end{array}
\right)
=
\left(
\begin{array}{ccc|c}
\frac{-19ab + 42a - 1596}{a(ab - 42)}
\\
\\
\frac{3ab - 9a + 45}{ab - 42}
\\
\\
\frac{3a - 57}{ab - 42}
\end{array}
\right)
\\\\\\
$