= Solution
For $v_1=(1,1,1,0)^T$, direct multiplication by the <sample covariance matrix> gives
$$
\Sigma v_1=(2250,2250,2250,0)^T=2250v_1.
$$
It is therefore an <eigenvector>, with <eigenvalue> $2250$. Normalization gives $u_1=v_1/\sqrt3$, so the corresponding <sample principal component> is
$$
\boxed{Y_1=\frac{(X_1-\bar X_1)+(X_2-\bar X_2)+(X_3-\bar X_3)}{\sqrt3},\qquad\widehat{\operatorname{Var}}(Y_1)=2250.}
$$
Its coefficients are proportional to the required vector. The remaining <eigenvalues> calculated in (ii) are smaller, confirming that this is the first <principal component>.
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