My optimization problem:
$$ Max U(C_1,C_2) = C_1C_2 $$
subject to:
$$ C_1 + \frac {C_2}{1 + r} = I_1 $$
Where $ C_1 = Consumption\ in \ Period \ 1, C_2 = Consumption\ in\ period\ 2\ and\ I_1\ is\ income\ in\ period\ 1 $
I formed my lagrange equation as: $$L = C_1C_2 + \lambda (I_1 - C_1 + \frac {C_2}{1 + r}) $$
The solutions at FOC is: 
The bordered hessian I'm getting is:
\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{matrix}
While the solution given in my reference is:
To find the bordered hessian, I first differentiate the constraint equation with respect to C1 and and C2 to get the border elements of the matrix, and find the second order differentials to get the remaining elements. Where am I going wrong?
Firstly take care of the signs. The lagrange function is $$L = C_1C_2 + \lambda (I_1 - C_1 \color{red} - \frac {C_2}{1 + r}) $$
The bordered Hessian is defined as $$\tilde H=\left( \begin{array}{} 0 & \frac{\partial^2 \mathcal L}{\partial \lambda\partial C_1}& \frac{\partial^2 \mathcal L}{\partial \lambda\partial C_2} \\ \frac{\partial^2 \mathcal L}{\partial \lambda\partial C_1} & \frac{\partial^2 \mathcal L}{\partial C_1\partial C_1} & \frac{\partial^2 \mathcal L}{\partial C_1\partial C_2} \\ \frac{\partial^2 \mathcal L}{\partial \lambda\partial C_2} & \frac{\partial^2 \mathcal L}{\partial C_1\partial C_2} & \frac{\partial^2 \mathcal L}{\partial C_2\partial C_2} \end{array}\right)$$
And the first derivatives are
$\frac{\partial \mathcal L }{\partial C_1}= C_2-\lambda$, $\frac{\partial \mathcal L }{\partial C_2}= C_1-\frac{\lambda}{1+r}$,$\frac{\partial \mathcal L }{\partial \lambda }= I_1-C_1-\frac{C_2}{1+r}$
Consequently the second derivatives are:
$\frac{\partial^2 \mathcal L }{\partial C_1^2}=0$
$\frac{\partial^2 \mathcal L }{\partial C_2^2}=0$
$\frac{\partial^2 \mathcal L }{\partial C_1C_2}=1$
$\frac{\partial^2 \mathcal L }{\partial \lambda C_1}=-1$
$\frac{\partial^2 \mathcal L }{\partial \lambda C_2}=-\frac{1}{1+r}$
Thus the bordered hessian is
$$\tilde H=\left( \begin{array}{ccc} 0 & -1& -\frac{1}{1+r} \\ -1 & 0 & 1 \\ -\frac{1}{1+r} & 1 & 0 \end{array}\right)$$
The solution is not right with respect to the order of the second derivatives.