# Having trouble solving system of equations through matrix approach

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?

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.

• The answer I was looking for! Thanks a ton! Oct 29 '15 at 7:51

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.