simple systems of equations problem

Question

Choose h and k such that the system has 1) no solution, 2) a unique solution, and 3) many solutions. Give separate answers for each part.

x-3y=1, 2x+hy=k

For 1) and 3), isn't that impossible? And for 2) can't h and k be any real number?

Answer 1

You have that $x-3y=1$, $2x+hy=k$. When you multiply the first equation with $2$, you get $2x-6y=2$. Subtrating the two equations, you get, $2x+hy-(2x-6y)=k-2\Rightarrow (h+6)y=k-2$. If $h=-6$ and $k\neq 2$, you get $0y\neq 0$, so a contradiction. Therefore, for $h=-6, k\neq 2$, there are no solutions. If you take $h=-6$ and $k=2$, you get $0y=0$, so $y$ can have any value. If $h\neq -6$, it follows that $y=\frac{k-2}{h+6}$ and $x=1+3\frac{k-2}{h+6}$, therefore you get a unique solution.

You could solve this, using determinants and matrix for systems, if you want a general algorithm.

Answer 2

You wish to analyze the system \begin{array}{rcrcr} x & - & 3\,y & = & 1 \\ 2\,x & + & h\,y & = & k \end{array} To do so, note that $h\neq-6$ implies $$ \DeclareMathOperator{rref}{rref}\rref \begin{bmatrix} 1&-3&1\\ 2&h&k \end{bmatrix} = \begin{bmatrix} 1 & 0 & 3\,\frac{k-2}{h+6}+1 \\ 0 & 1 & \frac{k-2}{h+6} \end{bmatrix} $$ This means that the system has a unique solution when $h\neq-6$. Do you see why?

Can you find the reduced row-echelon form of the system when $h=-6$?