## Systems of Linear Equations## IntroductionLinear algebra is a fundamental tool in mathematics and natural sciences. It begins by studying the systems of linear equations. The subject evolves by polishing the techniques for solving the systems and generalizing them into operator theory. Many techniques of linear algebra are familiar to students even before they start studying the subject. Thus it is important to emphasize that in order to properly interiorize the concepts, one may be required to abandon his/her favorite methods of solving the systems. The old methods will be replaced by techniques that allow better generalizations in the future. Let us, for one last time solve a system of two equations in non-linear-algebra-approved ways - using the method of guessing.
## Definitions
A real linear equation in variables \( x_1 \), \( \dots \), \( x_n \) is any equation of the form \[ a_1x_1+a_2x_2+\cdots+a_nx_n=b,\] where
\( a_1 \), \( \dots \), \( a_n \), \( b \) are real numbers. It is called A system of linear equations is several linear equations put together. A general form of the system is: \begin{eqnarray*} a_{11}x_1+a_{12}x_2+\cdots+a_{1n}x_n&=&b_1\\ a_{21}x_1+a_{22}x_2+\cdots+a_{2n}n_x&=&b_2\\ &\vdots&\\ a_{m1}x_1+a_{m2}x_2+\cdots+a_{mn}x_n&=&b_m, \end{eqnarray*} where \( a_{11} \), \( \dots \), \( a_{1n} \), \( \dots \), \( a_{m1} \), \( \dots \), \( a_{mn} \), \( b_1 \), \( \dots \), \( b_m \) are real numbers. A solution to the system is any \( n \)-tuple \( (x_1, \dots, x_n) \) that satisfies each of the \( m \) equations. ## Equivalent systemsTwo systems of equations are called ## Gaussian eliminationThe previous theorem is the cornerstone of majority of methods for solving systems of linear equations. In each step a system is transformed to an equivalent one that is in some way simpler than the original. After several steps one hopes to obtain the trivial system.
Let us first start from explaining the meaning of the expression The method of Gaussian elimination aims to reduce the system in the row-echelon form. The goal is for the last equation to have only one variable, thus making it trivial to solve. Then once we have the variable from the last equation, the preceding equation would ideally have only one more variable allowing us to find it if we know the last one. In order to put the system in the row-echelon form we will first fix one variable in one of the equations, and using the transformations from Theorem 1 obtain an equivalent system in which that variable is removed from every other equation. As an outline of the method we will consider the following system of three equations with three variables: \begin{eqnarray*} 2x-2y+8z&=&6\\ 6x+7y-9z&=&11\\ -4x+11y-3z&=&15. \end{eqnarray*} Let us consider the variable \( x \) in the first equation. We will put the box around it and eliminate it from the other two equations: Here is how the system looks with a box: \begin{eqnarray*} 2\begin{array}{|c|}\hline x\\\hline\end{array}-2y+8z&=&6\\ 6x+7y-9z&=&11\\ -4x+11y-3z&=&15. \end{eqnarray*} We multiply the first equation by \( -3 \), add it to the second, and use the obtained equation to replace the second in the system. This will give us an equivalent system. As a part of the same step, we will multiply the first equation by \( 2 \), add it to the third, and use the obtained equation to replace the third in the system. We arrive to the following system that is equivalent to the original one: \begin{eqnarray*} 2x-2y+8z&=&6\\ 13y-33z&=&-7\\ 7y+13z&=&27. \end{eqnarray*} We now concentrate on the last two equations that are forming a system with two variables. We will not write the first equation until the very end of the problem. A box in the equation also serves as a mark that the equation is going to be a part of any subsequent system. \begin{eqnarray*} 13y-33z&=&-7\\ 7y+13z&=&27. \end{eqnarray*} Let us now put the box around \( y \) variable in the first equation: \begin{eqnarray*} 13\begin{array}{|c|}\hline y\\\hline\end{array}-33z&=&-7\\ 7y+13z&=&27. \end{eqnarray*} We multiply the first equation by \( -\frac7{13} \) and add it to the second. We get: \begin{eqnarray*} \frac{400}{13}z&=&\frac{400}{13}. \end{eqnarray*} We put a box around \( z \): \begin{eqnarray*} \frac{400}{13}\begin{array}{|c|}\hline z\\\hline\end{array}&=&\frac{400}{13}. \end{eqnarray*} Solving for \( z \) yields \( z=1 \). Now we look at the preceding equation with box and solve it for \( y \). We get \( y=2 \). Finally, the first equation with box gives us \( x=1 \). ## Examples
## Matrix of the system
In this section we will introduce matrices as a way to represent systems of equations in a more compact form. We will study the matrices instead of systems, and we will be able to conclude wether systems are solvable just by looking at their matrices. It will turn out that there are We will start by the definition of the matrix of the system.
We will now provide several examples of systems of equations and their corresponding coefficient matrices and augmented matrices.
Recall that we said that two systems of equations are equivalent, if they have the same solutions. We now extend that definition to matrices: Two matrices are said to be ## Row reduction and echelon formsRecall that the system of equation is in a row-echelon form if the last equation contains only \( x_n \) variable; the second-to last only \( x_n \) and \( x_{n-1} \); the third equation from the bottom only \( x_n \), \( x_{n-1} \), and \( x_{n-2} \); etc. Such system is as good as the solved one. The matrix corresponding to such a system is called a matrix in a row-echelon form. The precise definition is this: Row echelon form is also called the Each symbol \( \ast \) above can be replaced by any number, while each occurrence of \( \star \) be replaced by any non-zero number. We were solving the systems of linear equations by reducing the systems to the row echelon form. We will now see that in an analogous way we can reduce the matrix in a row echelon form. ## Reduced echelon formExamples of matrices in reduced echelon forms are given below. \begin{eqnarray*} \left[\begin{array}{ccccc} 1& 0 & \ast &\ast &\ast\\ 0&1&\ast & \ast &\ast\\ 0&0&0&0&0\\ 0&0&0&0&0 \end{array}\right]\,, \quad \left[\begin{array}{cccccc} 0 &1 &0 &\ast &0 &0\\ 0 &0 &1 &\ast &0 &0\\ 0 &0 &0 &0 &1 &0\\ 0 &0 &0 &0 &0 &1 \end{array}\right]\,, \quad \left[\begin{array}{cccccc} 1&\ast & 0 &\ast &0 &0\\ 0&0 &1 &\ast &0 &0\\ 0&0 &0 &0 &1 &0\\ 0&0 &0 &0 &0 &1 \end{array}\right]\,. \end{eqnarray*} |

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