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For the 2x2 case being considered, write $$\mathbf{B}=\left[\begin{array}{cc} b_{1,1} & b_{1,2}\\ b_{2,1} & b_{2,2} \end{array}\right].\quad$$ It follows that the element (1,1) in $B^{-1}$ is given by $\frac{b_{2,2}}{b_{1,1}b_{2,2}-b_{1,2}b_{2,1}}$. Notice that $$\frac{\partial q_1(p_1,p_2)}{\partial p_1}=(\frac{\partial p_1(q_1,q_2)}{\partial q_1 }... 5 Let matrix A = \begin{bmatrix} p_1 & p_2 & \ldots & p_n \end{bmatrix}. Let \mathbf{x}^* be a fixed solution to A \mathbf{x} = c. Then for any vector \mathbf{u} that belongs to the null space of A, we have A \mathbf{u} = 0 hence \mathbf{x} = \mathbf{x}^* + \mathbf{u} is also a solution (furthermore, all solutions \mathbf{x} can be ... 3 This concerns the partial derivatives of the utility function with respect to goods, and not the partial derivative of the Lagrangian of the maximization problem. So a zero derivative, and moreover at the optimum, would imply a threshold quantity after which utility diminishes. In the real world, we all know that consuming excessively may result in ... 2 I am not sure exactly what you divided by what, but suppose your input-output table looks like this:$$ A = \left[ \begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right] If you proceed to divide the first column by \sum\limits_i a_{i,1} the second by \sum\limits_i ... 2 Let the n states of the finite-state markov chain be denoted by \{x_1,...,x_n\} and let \vec e = [e(x_1), ..., e(x_n)]'. Now, first note that because X_{t+1} \mid X_t is independent of W_{t+1}, we can write \begin{align*} \exp(\eta) e(x) &= E[\exp(D'x + x' F W_{t+1})] E[ e(X_{t+1}) \mid X_t = x] \\ &= \exp(D'x + x' F F' x)\, E[ e(X_{t+1}) ... 2 This kind of projections in econometrics are usually employed for partialling out some covariates from a linear regression. Observe that in general P_X\neq I. Consider X=\begin{bmatrix}1&5\\1&0\\1&1\end{bmatrix}. Then X'X=\begin{bmatrix}3&6\\6&26\end{bmatrix} and (X'X)^{-1}=\begin{bmatrix}26/42&-6/42\\-6/42&3/42\end{... 2 First, note that the growth rate of \mu is defined as \dot\mu = \frac{ d\mu }{ \mu }. Therefore, you will have to take the total differential of the equation; and divide by \mu. For the first step, as in your previous question, simply use the rules of total differentials, specifically the product rule that states that for variables X and Y, the ... 2 For a variable X, let dX denote its total differential. Let k be a constant, and X and Y variables. You'll need the following rules:dk = 0$$(constant rule),$$d(X + Y) = dX + dY$$(sum rule),$$d(XY) = Y \cdot dX + X \cdot dY$$(product rule) and$$d\left(\frac{X}{Y}\right) = \frac{ Y \cdot dX - X \cdot dY }{ Y^2 }$$(quotient rule). ... 1 First of all, I think that 'linear and homogeneous' is a typo of 'linearly homogeneous.' Indeed, it can be shown that if the production function V is linearly homogeneous, Allen Elasticity of Substitution between K and L can be expressed as$$ \sigma = \frac{V_K V_L}{V \cdot V_{KL}} $$by using the fact that V_K and V_L are homogeneous of degree 0, ... 1 I don't really understand your optimization problem as it stands. Just as a few examples, when you write out$$\min \sum_m \sum_v \left( \frac{\partial \vec f_v (\vec Q_v)}{\partial \vec f_m (\vec Q_m)} - \frac{\partial \vec f_m (\vec Q_m)}{\partial \vec f_m (\vec Q_m)} \right)$$I have questions about why you are minimizing the difference between optimal ... 1 I got some outside help for the ending of the proof I was attempting. I'll leave this question if by chance someone else finds it useful. So if we want to show p_1x_1 + \cdots + p_{n−1}x_{n−1} = 0 \implies p_i = 0 \quad \forall i, then assume without loss of generality that p_1 \neq 0. We have$$x_1 = \left(-\frac{p_2}{p_1}\right) x_2 + \cdots + \left(-...

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All right kids. You gave me the answer when Dismalscience wrote "this is still not a problem as long as the value of the good produced differs from the sum of the value of its inputs" ... actually, one of the branches had a complete set of zeros in the input table exept for one input and a zero in terms of production ... and another one had a complete set ...

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I didn't divided each column by the sum of the column. This would make no sence since the goal is to produce a technical coefficient matrix that links each input (row) needed by the industry (column) in order to produce their output. I divided each element of the input-output table by the total output of the branch, according to the method presented in ...

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