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I am new to this so please don't make fun of me. The question:

Suppose that the linear system $$ \begin{align*} 2x + 4y &= f \\ cx + dy &= g \end{align*} $$ is consistent for all possible values of $f$ and $g$. What can you say about the coefficients $c$ and $d$? (Hint: What does row reduction tell you?)

I learned that this would be the augmented matrix: $$ \begin{bmatrix} 2 & 4 & f \\ c & d & g \end{bmatrix} $$ So with row reduction: Each row has to have a leading coefficient of $1$ and where the leading coefficient is $1$ then the the rest in the column has to be zeros if the leading coefficient is $1$.

Am I on the right track?

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We should be able to find a better title, but I can't think of a clear one right now. Also, I'd like to think that we don't make fun of people trying to learn in this community. You shouldn't worry about that. – Dylan Moreland Jan 21 '12 at 20:47
Okay, thanks for the heads up. Is it possible that someone could help me get on the right track so I can answer this question correctly. – Janine Jan 21 '12 at 20:50
I would give it a little more time. I would bet money that someone will answer this within two hours; I will write something later if no one else has, certainly. – Dylan Moreland Jan 21 '12 at 20:51
If the system was not consistent for all values of $f$ and $g$, what would the row reduction end up looking like? – Michael Joyce Jan 21 '12 at 20:54
(It only took three minutes!) – Dylan Moreland Jan 21 '12 at 21:13

If I'm following you correctly, the answer is yes. You can determine when your system always has a solution by examining the row reduction procedure.

You need to figure out when row reduction will give you the proper form in the augmented matrix (the conditions that you specified in your last paragraph).

If $c$ and $d$ are both zero, the system does not always have a solution; in particular, there would be no solution when $g\ne0$.

Otherwise, let's row reduce: $$ \left[\matrix{2&4 \ \cr c & d }\Biggl| \matrix{\ f\cr\ g}\right]\rightarrow\quad \left[\matrix{1&2 \cr c\ & d \ }\Biggl| \matrix{\ f/2\cr\ g}\right] \rightarrow\quad \left[\matrix{1&2\ \cr 0 & d-2c \ }\Biggl| \matrix{\ f/2\cr\ {g }-{cf/2 }}\right] $$ The row reduction can be completed if and only if $d-2c\ne0$.

If $d-2c=0$, then by selecting certain values of $f$ and $g$, we would have a system with no solution (it would have an equation of the form $0= g-cf/2\ne0$)

If $d-2c\ne0$, then we can complete the reduction: $$ \rightarrow\quad \left[\matrix{1&2& \ \cr 0 & 1\vphantom{1\over2}\ } \Biggl| \matrix{\ f/2\cr {g }-{cf/2 }\over d-2c }\right] $$ and we can solve regardless of the values of $f$ and $g$.

So, the system always has a solution if and only if $d-2c\ne0$ (note this condition includes the case we previously considered: $c=d=0$).

As it turns out, the condition is that the first two entries of the second row do not both become 0 in the reduction. This happens if and only if the row $[ c\ d]$ is not a multiple of the row $[2\ 4]$.

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Lol, it did only take three minutes! I was on the right track, thanks! YES!! – Janine Jan 21 '12 at 22:00

A less ad-hoc way of arriving at the same result is this: If your equation system has solutions for all choices of $f$ and $g$, this means that the column space of the matrix $\pmatrix{2&4\\c&d}$ is all of $\mathbb R^2$. This is the same as saying that the matrix has rank 2, and the rank of a square matrix equals the dimension exactly when it is invertible, which is again the same as saying that its determinant is nonzero. So the condition is $$\left|\begin{array}{cc}2&4\\c&d\end{array}\right| = 2c-4d \ne 0$$

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"the rank of a square matrix equals the dimension exactly when it is invertible, which is again the same as saying that its determinant is nonzero. So the condition is" ... or simply, rank $= 2 \Leftrightarrow$ there exists a $2 \times 2$ non-vanishing minor. – user2468 Jan 22 '12 at 19:45

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