= Solution
{c}
The centered regressors make the first <normal equation> read $n\hat\beta_0=\sum_iY_i$. Define
$$
A=\sum_i x_{i1}^2,\quad B=\sum_i x_{i1}x_{i2},\quad C=\sum_i x_{i2}^2,\quad
u=\sum_i x_{i1}Y_i,\quad v=\sum_i x_{i2}Y_i.
$$
The remaining equations are $A\hat\beta_1+B\hat\beta_2=u$ and $B\hat\beta_1+C\hat\beta_2=v$. Provided $D=AC-B^2>0$, <matrix inversion> gives
$$
\boxed{\hat\beta_0=\overline Y,\qquad
\hat\beta_1=\frac{Cu-Bv}{D},\qquad \hat\beta_2=\frac{Av-Bu}{D}.}
$$
By the <Cauchy-Schwarz inequality>, $D\geq0$. If $D=0$, the two regressor columns are <linearly dependent>, and their separate slopes are not uniquely identifiable. One still has well-defined <linear least squares> fitted values, but cannot use the displayed divisions by $D$.
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