Question

In an economics text${}^{\mathbf{11}}$we find the following system:

$\left[\begin{array}{ccc}-R_1& R_1& -\left(1-\alpha \right)\\ \alpha & 1-\alpha & -{\left(1-\alpha \right)}^2\\ R_2& -R_2& \frac{-{\left(1-\alpha \right)}^2}{\alpha }\end{array}\right]\left[\begin{array}{c}d{x}_1\\ d{y}_1\\ dp\end{array}\right]\mathbf{=}\left[\begin{array}{c}0\\ 0\\ -R_{2}d{e}_2\end{array}\right]$.

Solve for $\mathit{d}{\mathit{x}}_{\mathbf{1}}\mathbf{,}\mathit{d}{\mathit{y}}_{\mathbf{1}}$, and dp. In your answer, you may refer to the determinant of the coefficient matrix as D. (You need not compute D.) The quantities${\mathit{R}}_{\mathbf{1}}\mathbf{,}{\mathit{R}}_{\mathbf{2}}$and D are positive, and ais between zero and one. If $\mathit{d}{\mathit{e}}_{\mathbf{2}}$is positive, what can you say about the signs of $\mathit{d}{\mathit{y}}_{\mathbf{1}}$and dp?

Short answer

Therefore,

$\left[\begin{array}{c}d{x}_1\\ d{y}_1\\ dp\end{array}\right]=\left[\begin{array}{c}\frac{R_{2}d{e}_2\left(R_1-1\right){\left(1-\alpha \right)}^{\alpha }}{D}\\ \frac{R_{2}d{e}_2\left(R_1-R_1\alpha +\alpha \right)\left(1-\alpha \right)}{D}\\ \frac{R_1{R}_{2}d{e}_2}{D}\end{array}\right]$

If $R_1{R}_2$and Dare positive, $\alpha$is between 0 and 1, and $d{e}_2$is positive, then $d{y}_2$and are also positive.

Step-by-step solution

Definition. 

In matrices, Cramer's rule expresses the solution in terms of the determinants of the coefficient matrix (i.e., for a square matrix) and of matrices obtained from it by replacing one column by the column vector of right-hand-sides of the equations.

Step 2: To solve dx1,dy1and dp

Let,

$\left[\begin{array}{ccc}-{\mathrm{R}}_1& {\mathrm{R}}_1& -(1-\mathrm{\alpha })\\ \mathrm{\alpha }& 1-\mathrm{\alpha }& -{(1-\mathrm{\alpha })}^2\\ {\mathrm{R}}_2& -{\mathrm{R}}^2& \frac{-{(1-\mathrm{\alpha })}^2}{\mathrm{\alpha }}\end{array}\right]$

Now, using Cramer's rule, we solve:

$$\begin{gathered} d{x}_1=\frac{\left|\begin{array}{ccc}0& R_1& -\left(1-\alpha \right)\\ 0& 1-\alpha & -{\left(1-\alpha \right)}^2\\ -R_{2}d{e}_2& -R^2& {\displaystyle \frac{-{\left(1-\alpha \right)}^2}{\alpha }}\end{array}\right|}{D} \\ d{x}_1=\frac{R_{2}d{e}_2(R_1-1){\left(1-\alpha \right)}^2}{D} \\ d{y}_1=\frac{\left|\begin{array}{ccc}-R_1& 0& -\left(1-\alpha \right)\\ R_2& 0& -{\left(1-\alpha \right)}^2\\ R_2& -R_{2}d{e}_2& {\displaystyle \frac{-{\left(1-\alpha \right)}^2}{\alpha }}\end{array}\right|}{D} \\ d{y}_1=\frac{R_{2}d{e}_2\left(R_1-R_1\alpha +\alpha \right)\left(1-\alpha \right)}{D} \\ dp=\frac{\left|\begin{array}{ccc}-R_1& R_1& 0\\ \alpha & 1-\alpha & 0\\ R_2& -R^2& -R_{2}d{e}_2\end{array}\right|}{D} \\ dp=\frac{R_1{R}_{2}d{e}_2}{D} \end{gathered}$$

$\left[\begin{array}{c}d{x}_1\\ d{y}_1\\ dp\end{array}\right]=\left[\begin{array}{c}\frac{R_{2}d{e}_2\left(R_1-1\right){\left(1-\alpha \right)}^{\alpha }}{D}\\ \frac{R_{2}d{e}_2\left(R_1-R_1\alpha +\alpha \right)\left(1-\alpha \right)}{D}\\ \frac{R_1{R}_{2}d{e}_2}{D}\end{array}\right]$

Step 3: If de2 is positive, what can you say about the signs of dy1 and dp

If $R_1,R_2$and D are positive, $\alpha$is between 0 and 1 and $d{e}_2$is positive, then $d{y}_2$ and dp are also positive.