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?


1 Answer 1


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 $y$.

  • $\begingroup$ Ah, makes sense. The text talked about infinite number of solutions for a system, but had presented no example and I was wondering how such a system would look like. Specifically, in this case, the infinite number of solutions is only when $h = 12$ right? The system has a unique solution if $h \neq 12$. $\endgroup$ Mar 16, 2013 at 2:56
  • $\begingroup$ Thanks. I think your edit provided the necessary clarification. $\endgroup$ Mar 16, 2013 at 2:59
  • $\begingroup$ Yes, exactly! $y$ would equal $0$, and $x$ would then be $-2$ $\endgroup$
    – amWhy
    Mar 16, 2013 at 2:59

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