<p>You may find better help in the Math SE, but let's see if we can help here.</p>

<p>So we can arrange your system as such:</p>

<p>\begin{align}
 s_1 - \frac{1}{3} s_2 &amp; = 2 \\ 
 -\frac{1}{8} s_1 + s_2&amp; = 7 \\
\end{align}</p>

<p>So we form the matrix setup:</p>

<p>$$\begin{bmatrix}
        1 &amp; -1/3  \\
        -1/8 &amp; 1  \\
        \end{bmatrix}
\begin{bmatrix}
        s_1 \\
        s_2 \\
        \end{bmatrix}
=
\begin{bmatrix}
        2 \\
        7 \\
        \end{bmatrix}
$$</p>

<p>You can multiply the matrices out to see that you get the same system.</p>

<p>The first step of solving any system, regardless of method, is to always check for linear dependence of the coefficient matrix.</p>

<p>$$
\det
\begin{bmatrix}
        1 &amp; -1/3  \\
        -1/8 &amp; 1  \\
        \end{bmatrix}
=
(1 \cdot 1) - (-1/3 \cdot -1/8) = \frac{23}{24} \neq 0
$$</p>

<p>So Cramer's rule is indeed useable here. To solve for $s_1$. You replace the column for $s_1$ with the answer matrix and divide the determinant of that by the determinant of the original coefficient matrix. Use a similar process for $s_2$</p>

<p>$$
s_1^* =
\frac
{\det
\begin{bmatrix}
        2 &amp; -1/3  \\
        7 &amp; 1  \\
        \end{bmatrix}}
{\det
\begin{bmatrix}
        1 &amp; -1/3  \\
        -1/8 &amp; 1  \\
        \end{bmatrix}}
= \frac{\frac{13}{3}}{\frac{23}{24}} = \frac{104}{23}
$$</p>

<p>$$
s_2^* =
\frac
{\det
\begin{bmatrix}
        1 &amp; 2  \\
        -1/8 &amp; 7  \\
        \end{bmatrix}}
{\det
\begin{bmatrix}
        1 &amp; -1/3  \\
        -1/8 &amp; 1  \\
        \end{bmatrix}}
= \frac{\frac{29}{4}}{\frac{23}{24}} = \frac{174}{23}
$$</p>

<p>And then we're done. Plug in the answers and make sure they work.</p>

CC BY-SA 3.0
2015-10-29T06:44:44.313Z
2015-10-29T06:44:44.313Z
Kitsune Cavalry
<p>Your system may be rewrite as :
\begin{equation}
\begin{bmatrix}1&amp;-\frac{1}{3}\\-\frac{1}{8}&amp;1\end{bmatrix}\cdot\begin{bmatrix}s_1\\s_2\end{bmatrix}=\begin{bmatrix}2\\7\end{bmatrix}\end{equation}</p>

<p>Then all you need to do is to invert the coefficients matrix (which is invertible because its det$\neq$0), and confirm that by cramer's rule. In case you don't know, cramer's rule is explained in good details <a href="https://en.wikipedia.org/wiki/Cramer&#39;s_rule" rel="nofollow">here</a>.</p>

CC BY-SA 3.0
2015-10-29T06:26:56.897Z
2015-10-29T06:26:56.897Z
Louis. B
<p>So I have this system of reaction functions:</p>

<p>$$s_1 = 2 + \frac{1}{3} s_2$$<br>
$$s_2 = 7 + \frac{1}{8} s_1$$</p>

<p>I have to solve this through inverse matrix and confirm the result through Cramer's rule. The result is I'm getting awkward values:</p>

<p>$$ \begin{bmatrix} s1 \\ s_2 \end{bmatrix} = \begin{bmatrix} 52.6 &amp; 2.6 \\ 54.6 &amp; 15.6 \end{bmatrix}$$</p>

<p>I don't think so this is the correct result since Cramer's rule cannot really be applied to this system. Any suggestions?</p>

CC BY-SA 3.0
2015-10-29T14:55:42.313Z
Editing to include math type setting, but please consider using this reference on math type setting for future questions: http://meta.economics.stackexchange.com/questions/1417/mathjax-basic-tutorial-and-quick-reference
cc7768
<microeconomics>
Having trouble solving system of equations through matrix approach
Shiv_90
# Having trouble solving system of equations through matrix approach

(Asked by: username_0 on 2015-10-29)

So I have this system of reaction functions:

$$s\_1 = 2 + \frac{1}{3} s\_2$$  
$$s\_2 = 7 + \frac{1}{8} s\_1$$

I have to solve this through inverse matrix and confirm the result through Cramer's rule. The result is I'm getting awkward values:

$$ \begin{bmatrix} s1 \\ s\_2 \end{bmatrix} = \begin{bmatrix} 52.6 & 2.6 \\ 54.6 & 15.6 \end{bmatrix}$$

I don't think so this is the correct result since Cramer's rule cannot really be applied to this system. Any suggestions?

**Tags:** microeconomics

> **Comments:**
> **username_1**: Editing to include math type setting, but please consider using this reference on math type setting for future questions: http://meta.economics.stackexchange.com/questions/1417/mathjax-basic-tutorial-and-quick-reference  

---

## 2 answers

### Answer by username_2 on 2015-10-29. Score: 2 | Accepted answer

You may find better help in the Math SE, but let's see if we can help here.

So we can arrange your system as such:

\begin{align}
s\_1 - \frac{1}{3} s\_2 & = 2 \\
-\frac{1}{8} s\_1 + s\_2& = 7 \\
\end{align}

So we form the matrix setup:

$$\begin{bmatrix}
1 & -1/3 \\
-1/8 & 1 \\
\end{bmatrix}
\begin{bmatrix}
s\_1 \\
s\_2 \\
\end{bmatrix}
=
\begin{bmatrix}
2 \\
7 \\
\end{bmatrix}
$$

You can multiply the matrices out to see that you get the same system.

The first step of solving any system, regardless of method, is to always check for linear dependence of the coefficient matrix.

$$
\det
\begin{bmatrix}
1 & -1/3 \\
-1/8 & 1 \\
\end{bmatrix}
=
(1 \cdot 1) - (-1/3 \cdot -1/8) = \frac{23}{24} \neq 0
$$

So Cramer's rule is indeed useable here. To solve for $s\_1$. You replace the column for $s\_1$ with the answer matrix and divide the determinant of that by the determinant of the original coefficient matrix. Use a similar process for $s\_2$

$$
s\_1^\* =
\frac
{\det
\begin{bmatrix}
2 & -1/3 \\
7 & 1 \\
\end{bmatrix}}
{\det
\begin{bmatrix}
1 & -1/3 \\
-1/8 & 1 \\
\end{bmatrix}}
= \frac{\frac{13}{3}}{\frac{23}{24}} = \frac{104}{23}
$$

$$
s\_2^\* =
\frac
{\det
\begin{bmatrix}
1 & 2 \\
-1/8 & 7 \\
\end{bmatrix}}
{\det
\begin{bmatrix}
1 & -1/3 \\
-1/8 & 1 \\
\end{bmatrix}}
= \frac{\frac{29}{4}}{\frac{23}{24}} = \frac{174}{23}
$$

And then we're done. Plug in the answers and make sure they work.


---

### Answer by username_3 on 2015-10-29. Score: 0

Your system may be rewrite as :
\begin{equation}
\begin{bmatrix}1&-\frac{1}{3}\\-\frac{1}{8}&1\end{bmatrix}\cdot\begin{bmatrix}s\_1\\s\_2\end{bmatrix}=\begin{bmatrix}2\\7\end{bmatrix}\end{equation}

Then all you need to do is to invert the coefficients matrix (which is invertible because its det$\neq$0), and confirm that by cramer's rule. In case you don't know, cramer's rule is explained in good details [here](https://en.wikipedia.org/wiki/Cramer's_rule).


---

