Choose $h$ and $k$ so that the system has no solution, one solution andinfinitely many

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Choose $h$ and $k$ so that the system has no solution, one solution and infinitely many

$x_1 + 3x_2 =2$

$3x_1 + hx_2 = k$

So I put it into a augmented matrix and row-reduced to get it in row echelon form:

$\begin{bmatrix}3&9&|&6\\0&9-h&|&6-k\end{bmatrix}$

I'm not sure what to do next.

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3 Answers

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There is no solution when $9-h = 0$ and $6 - k \ne 0$ i.e. $h = 9, k \ne 6$

There is a unique solition when $9-h \ne 0$ i.e; when $h \ne 9$

There are infinitely many solutions when $9-h = 6-k = 0 $ or when $h =9, k = 6$

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Hint:

What happens if $9-h=0$, and $6-k\neq 0$?

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Hint:

The simpler way is to find $x_1$ from the first equation: $x_1=2-3x_2$ and substitute in the second equation that becomes: $$ x(h-9)=k-6 $$

Now You can see for what values of $h$ and $k$ this equation has no solutions or one solution or many solutions.

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