Solution
= Solution
Direct substitution into the three equations gives
$$
\frac{4\cdot13+13+12}{11}=7,\qquad
\frac{13+9\cdot13+2\cdot12}{11}=14,\qquad
\frac{13+2\cdot13+16\cdot12}{11}=21.
$$
Since the <covariance matrix> is invertible, this identifies its unique solution:
$$
\boxed{z=\frac1{11}(13,13,12)^T.}
$$
The original PDF has all three components; the converted TeX's damaged binomial notation should not be used.