Finding system with infinitely many solutions The question asks to find equation for which the system has infinitely many solutions.
The system is:
\begin{cases}
       -cx + 3y + 2z = 8\\
         x      + z  = 2\\
        3x + 3y + az = b
\end{cases}
How should I approach questions like this?
I tried taking it to row reduced echelon form but it got kind of messy.
The answer is supposed to be:
 $$a - c -5 = 0$$  and $$b- 2c +2 = 0$$
 A: You can do row reduction; consider the matrix
\begin{align}
\left[\begin{array}{ccc|c}
-c & 3 & 2 & 8 \\
1 & 0 & 1 & 2 \\
3 & 3 & a & b
\end{array}\right]
&\to
\left[\begin{array}{ccc|c}
1 & 0 & 1 & 2 \\
-c & 3 & 2 & 8 \\
3 & 3 & a & b
\end{array}\right]\quad\text{swap 1 and 2}\\
&\to
\left[\begin{array}{ccc|c}
1 & 0 & 1 & 2 \\
0 & 3 & 2+c & 8+2c \\
0 & 3 & a-3 & b-6
\end{array}\right]\quad\text{reduce first column}\\
&\to
\left[\begin{array}{ccc|c}
1 & 0 & 1 & 2 \\
0 & 3 & 2+c & 8+2c \\
0 & 0 & a-c-5 & b-2c-14
\end{array}\right]\quad\text{reduce second column}\\
\end{align}
The system has infinitely many solutions if and only if the last row is zero, that is
\begin{cases}
a-c=5\\
b-2c=14
\end{cases}
A: Use Cramer's rule. If the determinant:
$$\begin{vmatrix}
-c && 3 && 2 \\
1 && 0 && 1 \\
3 && 3 && a 
\end{vmatrix} = 0$$
then the system has either no solution or infinite amount of solutions. If all three determinants:
$$\begin{vmatrix}
8 && 3 && 2 \\
2 && 0 && 1 \\
b && 3 && a 
\end{vmatrix} = \begin{vmatrix}
-c && 8 && 2 \\
1 && 2 && 1 \\
3 && b && a 
\end{vmatrix} = \begin{vmatrix}
-c && 8 && 2 \\
1 && 2 && 1 \\
3 && b && a 
\end{vmatrix} = 0$$
Then the system has infinite amount of solution. Can you solve it on your own now?
