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 $$b- 2c +2 = 0$$
Best Answer
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}