find all the values of a and b so that the system has a) no solution b) 1 solution c) exactly 3 solutions and 4) infinitely many solutions

Question

$$\begin{cases} &x &- &y &+ &2z &= 4 \\ &3x &- &2y &+ &9z &= 14 \\ &2x &- &4y &+ &az &= b \end{cases} $$ I know that $a$ and $b$ has to either equal to something or not in order to satisfy the $4$ conditions stated above.

my matrices looked like $$ \begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & a+12 \\ \end{bmatrix} \begin{bmatrix} x & \\ y &\\ z & \\ \end{bmatrix}= \begin{bmatrix} 6 & \\ 2 & \\ b+8 & \\ \end{bmatrix} $$

Answer 1

Apply Gaussian Elimination to the reduced matrix. $$ \pmatrix{1 & -1 & 2 & 4\\3 & -2 & 9 & 14\\2 & -4 & a & b} \to \pmatrix{1 & -1 & 2 & 4\\0 & 1 & 3 & 2\\0 & -2 & a-4 & b-8} \to \pmatrix{1 & -1 & 2 & 4\\0 & 1 & 3 & 2\\0 & 0 & a+2 & b+4} $$ So

• if you want a unique solution, $\frac{b+4}{a+2}$ must be defined, i.e. $a \neq -2$.
• if you want infinite number of solutions, $a=-2$ and $b = -4$ removes the last equation.
• if you want no solutions, $a=-2$ and $b \neq -4$.