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Parameters: \[ \begin{array}{l} \left\{\alpha, \beta, \phi, \theta, \beta_{c}...

Feb 20, 2024

Parameters:

\begin{array}{l} \left\{\alpha, \beta, \phi, \theta, \beta_{c}, \beta_{h}, \delta_{h}, \delta, \tau\right\} \\ \left\{A, \mu, j, \Omega, A_{h}\right\} \end{array}

Variables:

\left\{Y, K_{y}, l, K_{g}, L_{e}, R_{k}, w, q_{e}, H, L_{h}, K_{h}, p_{h}, q_{h}, h, C, R, \lambda^{1}, \lambda^{2}, D, B, S, C_{h}, K\right\}

Steady states

\begin{array}{l} Y=A K_{y}^{\alpha} l^{1-\alpha-\beta} K_{g}^{\phi} L_{e}^{\beta} \\ \alpha \frac{Y}{K_{y}}=R_{k} \\ (1-\alpha-\beta) \frac{Y}{l}=w \\ \beta \frac{Y}{L_{e}}=q_{e} \\ H=A_{h} K_{h}^{\theta} L_{h}^{1-\theta} \\ \theta p_{h} \frac{H}{K_{h}}=R_{k} \\ (1-\theta) \beta_{h} p_{h} \frac{H}{L_{h}}-\left(1-\left(\beta_{c}-\beta_{h}\right) \mu\right) q_{h}=R_{k} \\ \frac{j}{h}+\beta_{c} C^{-\sigma} p_{h}\left(1-\delta_{h}\right)=C^{-\sigma} p_{h} \\ \beta_{c}\left[(1-\tau) R_{k}+1-\delta\right]=1 \\ a_{n} l^{\varphi}=C^{-\sigma}(1-\tau) w \\ \frac{1}{R}=\beta_{c} \\ A\left(\frac{\alpha}{\beta}\right)^{\alpha}\left(\frac{1-\alpha-\beta}{\beta}\right)^{1-\alpha-\beta}\left(\frac{q_{e}}{R_{k}}\right)^{\alpha}\left(\frac{q_{e}}{w}\right)^{1-\alpha-\beta} K_{g}^{\phi}+\lambda^{1} q_{e}-\lambda^{2} \Omega q_{h}=0 \\ \lambda^{1} q_{h}-\lambda^{2} \Omega q_{h}+\beta_{g} p_{h} A_{h}\left(\frac{\left(1-\left(\beta_{c}-\beta_{h}\right) \mu\right) \theta}{(1-\theta) \beta_{h}}\right)^{\theta}\left(\frac{q_{h}}{R_{k}}\right)^{\theta}=0 \\ \lambda^{1}=\beta_{g}\left[A \phi\left(\frac{\alpha}{\beta}\right)^{\alpha}\left(\frac{1-\alpha-\beta}{\beta}\right)^{1-\alpha-\beta}\left(\frac{q_{e}}{R_{k}}\right)^{1-\beta} L_{e} K_{g}^{\phi-1}+\lambda^{1}\left(1-\delta_{g}\right)\right] \\ \lambda^{1}\left(1-\frac{\beta_{g}}{\beta_{c}}\right)=\lambda^{2} \frac{1}{\beta_{c}} \\ R D=\mu q_{h} L_{h} \\ R B=\Omega q_{h}\left(L-L_{h}-L_{e}\right) \\ \delta_{g} K_{g}+(R-1) B=\tau\left(w l+R_{k} K\right)+q_{h} L_{h}+q_{e} L_{e} \\ C+p_{h} \delta_{h} h+\delta K=(1-\tau)\left(w l+R_{k} K\right)+(R-1) S \\ C_{h}=p_{h} H+(1-R) D-R_{k} K_{h}-q_{h} L_{h} \\ D+B=S \\ H=\delta_{h} h \\ K_{y}+K_{h}=K \\ \end{array}

Solution by Steps

step 1

First, we need to transpose the vectors and matrices involved in the calculation

step 2

The transpose of a row vector is a column vector, so we transpose

\{0,1,0\}

to get

\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}

step 3

Next, we transpose the matrix

\begin{pmatrix} 1 &amp; 2 &amp; 3 \\ 4 &amp; 5 &amp; 6 \end{pmatrix}

to get

\begin{pmatrix} 1 &amp; 4 \\ 2 &amp; 5 \\ 3 &amp; 6 \end{pmatrix}

step 4

We also transpose the matrix

\begin{pmatrix} 1 &amp; 0 &amp; 1 \\ 0 &amp; 1 &amp; 0 \end{pmatrix}

to get

\begin{pmatrix} 1 &amp; 0 \\ 0 &amp; 1 \\ 1 &amp; 0 \end{pmatrix}

step 5

Now we subtract the transposed matrices:

\begin{pmatrix} 1 &amp; 4 \\ 2 &amp; 5 \\ 3 &amp; 6 \end{pmatrix} - \begin{pmatrix} 1 &amp; 0 \\ 0 &amp; 1 \\ 1 &amp; 0 \end{pmatrix}

to get

\begin{pmatrix} 0 &amp; 4 \\ 2 &amp; 4 \\ 2 &amp; 6 \end{pmatrix}

step 6

We multiply the resulting matrix by the column vector

\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}

to get

\begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix}

step 7

Finally, we multiply the resulting vector by the matrix

\begin{pmatrix} 1 &amp; 2 \\ 0 &amp; 6 \end{pmatrix}

to get the final result

step 8

The multiplication is

\begin{pmatrix} 1 &amp; 2 \\ 0 &amp; 6 \end{pmatrix} \begin{pmatrix} 0 \\ 2 \end{pmatrix}

which equals

\begin{pmatrix} 4 \\ 12 \end{pmatrix}

Answer

\begin{pmatrix} 4 \\ 12 \end{pmatrix}

Key Concept

Matrix and vector transposition and multiplication

Explanation

The solution involves transposing vectors and matrices, then performing matrix subtraction followed by matrix-vector multiplication to arrive at the final result.