Short Answer

In an economics text $^{11}$ we find the following system:
$[\begin{matrix} - R_{1} & R_{1} & - (1 - α) \\ α & 1 - α & - {(1 - α)}^{2} \\ R_{2} & - R_{2} & \frac{- {(1 - α)}^{2}}{α} \end{matrix}] [\begin{matrix} d x_{1} \\ d y_{1} \\ d p \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ - R_{2} d e_{2} \end{matrix}]$ .
Solve for $d x_{1}, d y_{1}$ , and dp. In your answer, you may refer to the determinant of the coefficient matrix as D. (You need not compute D.) The quantities $R_{1}, R_{2}$ and D are positive, and a is between zero and one. If $d e_{2}$ is positive, what can you say about the signs of $d y_{1}$ and dp?

Therefore,

$[\begin{matrix} d x_{1} \\ d y_{1} \\ d p \end{matrix}] = [\begin{matrix} \frac{R_{2} d e_{2} (R_{1} - 1) {(1 - α)}^{α}}{D} \\ \frac{R_{2} d e_{2} (R_{1} - R_{1} α + α) (1 - α)}{D} \\ \frac{R_{1} R_{2} d e_{2}}{D} \end{matrix}]$

If $R_{1} R_{2}$ and Dare positive, $α$ is between 0 and 1, and $d e_{2}$ is positive, then $d y_{2}$ and are also positive.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: 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,

$[\begin{matrix} - R_{1} & R_{1} & - (1 - α) \\ α & 1 - α & - {(1 - α)}^{2} \\ R_{2} & - R^{2} & \frac{- {(1 - α)}^{2}}{α} \end{matrix}]$

Now, using Cramer's rule, we solve:

$d x_{1} = \frac{|\begin{matrix} 0 & R_{1} & - (1 - α) \\ 0 & 1 - α & - {(1 - α)}^{2} \\ - R_{2} d e_{2} & - R^{2} & \frac{- {(1 - α)}^{2}}{α} \end{matrix}|}{D} d x_{1} = \frac{R_{2} d e_{2} (R_{1} - 1) {(1 - α)}^{2}}{D} d y_{1} = \frac{|\begin{matrix} - R_{1} & 0 & - (1 - α) \\ R_{2} & 0 & - {(1 - α)}^{2} \\ R_{2} & - R_{2} d e_{2} & \frac{- {(1 - α)}^{2}}{α} \end{matrix}|}{D} d y_{1} = \frac{R_{2} d e_{2} (R_{1} - R_{1} α + α) (1 - α)}{D} d p = \frac{|\begin{matrix} - R_{1} & R_{1} & 0 \\ α & 1 - α & 0 \\ R_{2} & - R^{2} & - R_{2} d e_{2} \end{matrix}|}{D} d p = \frac{R_{1} R_{2} d e_{2}}{D}$

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, $α$ is between 0 and 1 and $d e_{2}$ is positive, then $d y_{2}$ and dp are also positive.

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Step 1: Definition.

Step 2: To solve dx1,dy1and dp

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

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Step 1: Definition.

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

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