A system of linear equations consists of several linear equations involving the same unknowns. To find a single, unique solution it is neccessary to have at least as many equations as unknowns. So if there are three unknown quantities (e.g. a, b and c), three equations are needed for solving the system.

#### Example

$\begin{array}{cccccccc}I&a&+&\frac12b&-&10c&=&5\\\mathrm{II}&2a&-&b&\;&\;&=&6\\\mathrm{III}&-2a&+&4b&-&5c&=&-12\end{array}$

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### Solving systems of linear equations

There are five different methods for solving such a system of equations:

### Solution with the help of a matrix

A system of linear equations can always be written as a matrix of coefficients $\left(\mathrm A\left|\mathrm b\right.\right)$.

$\Rightarrow\;\mathrm A\cdot\mathrm x=\mathrm b$

The assistance of the following matrix can make solving a bigger system of equations easier and quicker. (Example Gaussian elimination) $$(A|b)=\left(\begin{array}{ccc}{\mathrm a}_{11}&\cdots&{\mathrm a}_{n1}\\\vdots&&\vdots\\{\mathrm a}_{1m}&\cdots&{\mathrm a}_\mathrm{nm}\end{array}\left|\begin{array}{c}b_1\\\vdots\\b_n\end{array}\right.\right)$$

#### Describing the solution

1.Unique solution: $\mathrm{rg}(\mathrm A)=\mathrm{rg}(\mathrm A\left|\mathrm b\right.)$ with matrix A has full rank

2.No solution: $rg(A)<rg(A|b)$

3.Infinitely many solutions: $rg(A)=rg(A\left|b\right.)$ with matrix A doesn't have full rank

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#### Example:

Determine the solution as a function of a.

$\left|\begin{array}{l}3x_1+4x_2+2x_3=1\\2x_1+ax_2+1x_3=2\\1x_1+2x_2+2x_3=3\end{array}\right.$

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Solution:

$(\mathrm A\left|\mathrm b\right.)=\left(\begin{array}{ccc}3&4&2\\2&\mathrm a&1\\1&2&2\end{array}\left|\begin{array}{c}1\\2\\3\end{array}\right.\right)$

Transforming the matrix using gaussian elimination:

$$\left(\begin{array}{ccc}3&4&2\\2&\mathrm a&1\\1&2&2\end{array}\left|\begin{array}{c}1\\2\\3\end{array}\right.\right)\begin{array}{c}\mathrm I)\;\leftrightarrow\;\mathrm{III})\\\longrightarrow\\\end{array} \left(\begin{array}{ccc}1&2&2\\2&\mathrm a&1\\3&4&2\end{array}\left|\begin{array}{c}3\\2\\1\end{array}\right.\right)\begin{array}{c}\mathrm{II})\;-\;2\mathrm I)\\\longrightarrow\\\mathrm{III})\;-\;3\mathrm I)\end{array} \left(\begin{array}{ccc}1&2&2\\0&\mathrm a-4&-3\\0&-2&-4\end{array}\left|\begin{array}{c}3\\-4\\-8\end{array}\right.\right)\\$$

$$\begin{array}{c}\\\longrightarrow\\III)\;:(-2)\end{array}\left(\begin{array}{ccc}1&2&2\\0&a-4&-3\\0&1&2\end{array} \left|\begin{array}{c}3\\-4\\4\end{array}\right.\right) \begin{array}{c}\\\longrightarrow\\II)\;-\;(a-4)\;III)\end{array}\\$$

$$\left(\begin{array}{ccc}1&2&2\\0&0&-2a+5\\0&1&2\end{array} \left|\begin{array}{c}3\\-4a+12\\4\end{array}\right.\right)\begin{array}{c}\\\longrightarrow\\II)\leftrightarrow III)\end{array} \left(\begin{array}{ccc}1&2&2\\0&1&2\\0&0&5-2a\end{array} \left|\begin{array}{c}3\\4\\12-4a\end{array}\right.\right)$$

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To be able to divide the last row by (5-2a) and thus create "1" in this position the following must apply: $5-2a\neq0$ $$\;$$

• $$\underline {1.Case}: 5-2a=0\;\;\rightarrow a=2,5$$ Look at rg(A) and rg(AIB):

$$\; \; \; \; \; \; \; rg(A) = 2 \ ; \ rg(A|b) = 3$$

$$\Rightarrow\;\;rg(A)<rg(A|b)$$ $\Rightarrow$ No solution $$\;$$

• $$\underline{2.Case}: 5-2a\neq0\;\;\rightarrow\;\;a\neq2,5$$ Look at rg(A) and rg(AIb) and calculate specific values for ${\mathrm x}_1,{\mathrm x}_2$ and ${\mathrm x}_3$:

$\Rightarrow\;\;$ Unique solution since $rg(A)=rg(A\left|b\right.)$ with full rank

$$\Rightarrow\;x_1=\frac{12-4a}{5-2a}$$

$$\Rightarrow\;x_2+2\cdot\frac{12-4a}{5-2a}=4\;\;\Rightarrow\;x_2=\frac4{2a-5}$$

$$\Rightarrow\;\;x_3+2\cdot\frac4{2a-5}+2\cdot\frac{12-4a}{5-2a}=3\;\;\Rightarrow\;\;x_3=\frac{2a-1}{5-2a}$$