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We row reduce the corresponding augmented matrix to RREF.

$$ \left[\begin{array}{rrr|r}2&3&3&-4\\-1&3&-6&2\\3&5&4&-6\end{array}\right] \begin{array}{c}R_1+R_2\\\\R_3+3R_2\end{array}\sim \left[\begin{array}{rrr|r}1&6&-3&-2\\-1&3&-6&2\\0&14&-14&0\end{array}\right] \begin{array}{c}R_2+R_2\end{array}\sim \left[\begin{array}{rrr|r}1&6&-3&-2\\0&9&-9&0\\0&14&-14&0\end{array}\right] \begin{array}{c}\frac19R_2\end{array}\sim \left[\begin{array}{rrr|r}1&6&-3&-2\\0&1&-1&0\\0&14&-14&0\end{array}\right] \begin{array}{c}R_1-6R_2\\\\R_3-14R_2\end{array}\sim \left[\begin{array}{rrr|r}1&0&3&-2\\0&1&-1&0\\0&0&0&0\end{array}\right] $$

Thus, we can see the system if consistent and contains $1$ free variable. Thus, the system is consistent with infinitely many solutions. Therefore, the answer is (b).

I was looking at my Linear Algebra quiz solutions and I saw the following:

Thus from RREF, we can see the system if consistent and contains 1 free variable. Thus, the system is consistent with infinitely many solutions.

  1. How can one understand if a system is consistent from RREF (that it has at least one solution). Is this about the free variable?

  2. Which theorem does the answer use to conclude "Thus, the system is consistent with infinitely many solutions"?

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A system is CONSISTENT if it has a unique solution, or infinitely many solutions.


When you look at the RREF of a square matrix augmented with the RHS, every complete row of zeros (last column included) corresponds to a free variable.

A row of zeros also indicates infinitely many solutions. Therefore, it has to be "consistent" in this context.


I'm not sure how your book presents the material and don't know if this result is presented as a separate theorem in your book.

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  • $\begingroup$ Thanks! I understand it now. $\endgroup$ – user3605771 Aug 4 '15 at 1:41

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