Show that if \(\mu_{i j}=\mu+\alpha_{i}+\beta_{j}\), the mean of the \(\mu_{i j}\) (summed on \(j\) ) is equal to \(\mu+\alpha_{i}\), and the mean of \(\mu_{i j}\) (summed on \(i\) and \(j\) ) is equal to \(\mu\), it follows that

\[\sum_{i=1}^{a} \alpha_{i}=\sum_{j=1}^{b} \beta_{j}=0\]