Question

In an economics text,${}^{\mathbf{10}}$we find the following system:

$$\begin{gathered} \mathit{s}\mathit{Y}\mathbf{+}\mathit{a}\mathit{r}\mathbf{=}\mathit{I}\mathbf{°}\mathbf{+}\mathit{G} \\ \mathit{m}\mathit{Y}\mathbf{-}\mathit{h}\mathit{r}\mathbf{=}{\mathit{M}}_{\mathbf{s}}\mathbf{-}\mathit{M} \end{gathered}$$

Solve for Y and r.

Short answer

Therefore,

$\left[\begin{array}{c}Y\\ r\end{array}\right]=\left[\begin{array}{c}\frac{\left(I°+G\right)h+\left(M_s+M_0\right)a}{sh-am}\\ \frac{\left(M_s+M_0\right)s-\left(I°+G\right)m}{sh-am}\end{array}\right]$

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  and

First, we note

$$\begin{gathered} \alpha :=I°+G, \\ \mathrm{\beta }:=M_s+M_0 \end{gathered}$$
Now, using Cramer’s rule, we solve:

$$\begin{gathered} \det \mathrm{A}=\left|\begin{array}{cc}\mathrm{s}& \mathrm{a}\\ \mathrm{m}& -\mathrm{h}\end{array}\right| \\ \det \mathrm{A}=\mathrm{sh}-\mathrm{am} \\ \mathrm{Y}=\frac{\left|\begin{array}{cc}\mathrm{\alpha }& \mathrm{a}\\ \mathrm{\beta }& -\mathrm{h}\end{array}\right|}{\mathrm{sh}-\mathrm{am}} \\ \mathrm{Y}=\frac{-\mathrm{\alpha h}+\mathrm{\beta a}}{\mathrm{sh}-\mathrm{am}} \\ \mathrm{r}=\frac{\left|\begin{array}{cc}\mathrm{s}& \mathrm{\alpha }\\ \mathrm{m}& \mathrm{\beta }\end{array}\right|}{\mathrm{sh}-\mathrm{am}} \\ \mathrm{r}=\frac{\mathrm{\beta s}-\mathrm{\alpha m}}{\mathrm{sh}-\mathrm{am}} \\ \left[\begin{array}{c}\mathrm{Y}\\ \mathrm{r}\end{array}\right]=\left[\begin{array}{c}\frac{-\mathrm{\alpha h}+\mathrm{\beta a}}{\mathrm{ah}-\mathrm{am}}\\ \frac{\mathrm{\beta s}-\mathrm{\alpha m}}{\mathrm{ah}-\mathrm{am}}\end{array}\right] \\ \left[\begin{array}{c}\mathrm{Y}\\ \mathrm{r}\end{array}\right]=\left[\begin{array}{c}\frac{\left(\mathrm{I}°+\mathrm{G}\right)\mathrm{h}+\left({\mathrm{M}}_{\mathrm{s}}+{\mathrm{M}}_0\right)\mathrm{a}}{\mathrm{sh}-\mathrm{am}}\\ \frac{\left({\mathrm{M}}_{\mathrm{s}}+{\mathrm{M}}_0\right)\mathrm{s}\left(\mathrm{I}°+\mathrm{G}\right)\mathrm{m}}{\mathrm{sh}-\mathrm{am}}\end{array}\right] \end{gathered}$$