Finding value of unknown coefficient in a matrix I'm beginning to learn linear algebra, and I can't, for the life of me, figure out how to find the value of h in following matrix, so that it satisfies the system.
System is :
\begin{cases}
x + hy=4\\
3x+6y=8
\end{cases}
With the augmented matrix being :
\begin{pmatrix}1 & h &4 \\3 & 6 & 8\end{pmatrix}
I tried substracting 3L1 to L3, but it doesn't help...
My textbook states that h=2, how to get to the result ?
I've looked at topic related : Finding a coefficient of a unknown to have a unique solution in a system , but I need a step by step explanation.
Thank you
 A: The steps after what you have done-
$$\begin{pmatrix}1 & h &4 \\3 & 6 & 8\end{pmatrix} \implies \begin{pmatrix}1 & h &4 \\0 & 6-3h & -4\end{pmatrix} [R_{2}^{'} \leftarrow R_2-3R_1]\implies \begin{pmatrix}1 & h &4 \\0 & 1 & -\frac{4}{(6-3h)}\end{pmatrix} [R_{2}^{'} \leftarrow R_2\times \frac{1}{(6-3h)} \text{ if } 6-3h \neq 0] \implies \begin{pmatrix}1 & 0 & \frac{24-8h}{6-3h} \\0 & 1 & -\frac{4}{(6-3h)}\end{pmatrix} [R_{1}^{'} \leftarrow R_1 - hR_2]$$
So, if $(6-3h) \neq 0 \implies h \neq 2$, then rank of augment matrix is $2$ and the rank of the coefficient matrix is also $2$. So the system of equation has a solution in that case.
If $h = 2$, then rank of augment matrix is $2$, but the rank of the coefficient matrix is $1$. So the system of equation does not have a solution in that case.
Interpretation from Linear Algebra: If $h=2$, then the range space of the coefficient matrix will be in the form of $\lambda \begin{pmatrix}1 \\3\end{pmatrix}$, where $\lambda \in \mathbb{R}$. So, clearly $\begin{pmatrix}4 \\8\end{pmatrix}$ will not be in the range space of that matrix. So, no solution in that case.
Interpretation from Geometry: If $h=2$, then you can see both the equations will represent two line which are parallel to each other. So, they can not cut each other, so no solution.
