Linear model with two columns where one column is a linear transformation of another If we have a linear model
$y_t = β_1x_{1t} + β_2x_{2t} + e_t$
where the errors $e_t$ are i.i.d normally distributed: $e_t$ ~ $IIN(0, σ^2)$, t = 1,...,T. The vectors of random regressors $X_1$ and $X_2$ satisfy the relation $X_1 = c + aX_2$ where a and c are constants. Both regressors are uncorrelated with error vector e. All variables in the model (i.e. the dependent variable and the regressors) have mean of zero.
(a) Find the covariance and correlation between $X_1$ and $X_2$
(b) Write the matrix X'X and show that it is non-invertible. 
I think the covariance is
$$Cov(X_1, X_2) = Cov(c + aX_2, X_2) = aVar(X_2)$$
and I think the correlation is
$$Corr(c + aX_2, X_2) = \frac{aVar(X_2)}{a^2Var(X_2)} = a$$.
Is that correct? As for showing X'X is not invertible, I'm lost. Everything I've tried shows that X'X is, in general, invertible but the premise of the question seems to say I am wrong about that. 
 A: $\mathrm{Var}(X_1)=\mathrm{Var}(c+aX_2) = a^2\mathrm{Var}(X_2)$
$\mathrm{Corr}(X_1,X_2)= \frac{\mathrm{Cov}(X_1,X_2)}{\sqrt{\mathrm{Var}(X_1)\mathrm{Var}(X_2)}} \\= \frac{a\mathrm{Var}(X_2)}{\sqrt{a^2\mathrm{Var}(X_2)\mathrm{Var}(X_2)}} = \text{sign}(a)$
$X=(X_1, X_2) =(c+aX_2, X_2)$ 
$X'X = \left(\matrix{(nc^2 + 2ac\sum X_2 +a^2\sum X_2^2 & c\sum X_2+a\sum X_2^2\\ c\sum X_2+a\sum X_2^2& \sum X_2^2}\right) $
Let $X_2 = (1,2,3)'$ and $X_1=1+X_2=(2,3,4)'$. Then $X'X=\left(\matrix{29& 20 \\20 & 14} \right)$
$(X'X)^{-1} =\frac 16\left(\matrix{14& -20\\ -20& 29}\right)$
So that $(X'X)$ is non-invertible is not true.
A: Covariance is indeed equal to
$$Cov(c + aX, X) = aVar(X)$$
As for the correlation, assuming $a \neq 0, Var(X) \neq 0$, there is an important detail
$$Cor(c + aX, X) = \frac{Cov(c + aX, X)}{\sqrt{Var(c + aX)}\sqrt{Var(X)}} = \frac{aVar(X)}{|a|Var(X)} = sgn(a)$$
As for the non-invertibility, model without the bias term, i.e.
$$Y = \beta_1X_1 + \beta_2X_2 + \epsilon$$
indeed has an invertible design matrix $X$, if $c \neq 0$.
However, if you add the bias
$$Y = \beta_0 + \beta_1X_1 + \beta_2X_2 + \epsilon$$
then the design matrix $X \in R^{n \times 3}, n \geq 3$ is no longer invertible, since it does not have rank $3$.
