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Found in: Page 307

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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

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.

See the step by step solution

## 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,

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

Now, using Cramer's rule, we solve:

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

## 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.