Solution RREF.M45 Chris Black

Let $l, m, p$ denote the number of lizards, mice and peacocks. Then the statements from the problem yield the equations: \begin{align*} 4l+4m+2p&=108\\ l+m+p &= 30\\ 2l-m&=0 \end{align*} We form the augmented matrix for this system and row-reduce \begin{align*} \begin{bmatrix} 4 & 4 & 2 & 108\\ 1 & 1 & 1 & 30\\ 2 & -1 & 0 & 0 \end{bmatrix} &\rref \begin{bmatrix} \leading{1} & 0 & 0 & 8\\ 0 & \leading{1} & 0 & 16\\ 0 & 0 & \leading{1} & 6 \end{bmatrix} \end{align*} From the row-reduced matrix, we see that we have an equivalent system $l=8$, $m=16$, and $p=6$, which means that you have 8 lizards, 16 mice and 6 peacocks.