I have started learning linear algebra. One of the exercise problems is as below.
Determine the value(s) of $h$ such that the matrix is the augmented matrix of a consistent linear system.
$$
\left [
\begin{matrix}
1 & 4 &| -2 \\
3 & h &| -6 \\
\end{matrix}
\right ]
$$
A simple row transformation of $R_2 = R_2 – 3R_1$ results in
$$
\left [
\begin{matrix}
1 & 4 &| -2 \\
0 & h-12 &|\ \ \ \ 0 \\
\end{matrix}
\right ]
$$
I reasoned that if $h$ is $12$, though there wouldn’t be any inconsistency, the system would effectively have only one equation to solve for two unknown variables and so $h \neq 12$ if we are to find the solution.
Generally how do we classify such systems, the ones which are consistent but appear unsolvable? On the other hand, is my reasoning wrong?
Best Answer
Your reasoning is correct. At $h = 12$, the system is consistent.
Such a system, in your case, when $h = 12$, or in general, when the augmented coefficient matrix of a system of linear equations has a row of zeros in its row-reduction, we say that the corresponding system has an infinite number of solutions.
Let's look at the system represented by your matrix.
Essentially, we can choose any arbitrary value for $y$: put $y = \alpha$.
Then $x = -2 - 4y = -2(1 + 2y) = -2(1 + 2\alpha)$
We can then represent the infinite "family" of solutions by the vector:
$$\begin{pmatrix}x \\y\end{pmatrix}:=\begin{pmatrix}-2(1 + 2\alpha) \\\alpha \end{pmatrix}$$
For any chosen value (scalar) $\alpha$, $x$ is then determined, so our solution puts some restrictions on the possible solutions, even though there are infinitely many choices for $\alpha$.
So the system having an infinite number of solutions is not equivalent to saying that every vector $\begin{pmatrix} x \\ y \end{pmatrix}$ is a solution to the system, since $x$ is a function of $y$.