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Mrs Brown has pet cats, pet parrots, pet fish and pet rocks. Each pet rock has three eyes, no legs and no tail. The other pets have as many of these features as you would expect. Let the number of cats, parrots, fish and rocks be given by c, p, f and r respectively.

(a) Suppose N = [ c p f r ]^T. Find the matrix M so that MN is the matrix whose entries are: the total number of eyes, legs and tails of Mrs Brown’s pets.

(b) Together, Mrs Brown’s pets have 25 eyes, 14 legs and 8 tails. How many pets of each type might she have? Explain your answer carefully.

I'm not sure if part b needs me to state just the general expressions or find actual solutions because it seems to me that the system has infinitely many solutions, i.e. total number of eyes ( 2c + 2p + 2f + 3r = 25 ), legs ( 4c + 2p = 14 ) and tails ( c + p + f = 8 ).

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There is an implicit side condition that the numbers of the various kinds of pet are all non-negative integers.

Since there are $25$ eyes, there cannot be more than $12$ pets. In particular, there cannot be infinitely many solutions.

Added: As you mentioned, we arrive at the equations $2c+2p+2f+3r=25$, $4c+2p=14$, and $c+p+f=8$.

It seems reasonable to start with the simplest equation, $4c+2p=14$, which we rewrite as $2c+p=7$. Clearly $p$ must be odd. We have the possibilities (i) $p=1$, $c=3$; (ii) $p=3$, $c=2$; (iii) $p=5$, $c=1$; (iv) $p=7$, $c=0$.

In case (i), from the equation $c+p+f=8$, we get $f=4$, and then from $2c+2p+2f+3r=25$ we get $r=3$. Indeed in all cases we must have $r=3$. For since $c+p+f=8$, we have $2c+2p+2f=16$, and therefore $3r=25-(2c+2p+2f)=9$.

Cases (ii), (iii), and (iv) are dealt with similarly.

We could also use matrix row reduction procedures to get at the solutions.

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    The explanation was very helpful. My apologies, I hadn't refreshed the page so didn't see it prior to replying. Thank you so much for all your help!2012-09-18