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Linear Algebra

Systems of Linear Equations

linear equation in n variables \( x_1, \ldots , x_n \) is an equation that can be written in the form 

\[ a_1 x_1 + a_2 x_2 + a_3 x_3 + \ldots a_n x_n = b \]

The numbers \( a_1, \ldots , a_n \)  are called the coefficients of the equations, and is usually referred to as "the right-hand side" or the "constant term". The \(x_i \) are the variables to be solved for.

A set of linear equations in the same variables \( x_1, \ldots , x_n \)  is called a system of m linear equations in variables.

A general system of m linear equations in variables written in the form

\begin{align} a_{11} x_1 + a_{12} x_2 + \ldots a_{1n} x_n &= b_1 \\ a_{21} x_1 + a_{22} x_2 + \ldots a_{2n} x_n &= b_2 \\ &\vdots \\ a_{m1} x_1 + a_{m2} x_2 + \ldots a_{mn} x_n &= b_m \end{align}

A vector \( \vec{s} =  \begin{bmatrix} s_1 \\ \vdots \\ s_n \end{bmatrix} \in \mathbb{R}^n \) is called a solution of a system of m linear equations in n variables if all m equations are satisfied when we set \(x_i = s_i \) for \(1 \leq i \leq n \). The set of all solutions of a system of linear equations is called the solution set of the system.
If a system of linear equations has at least one solution, then it is said to be consistent. Otherwise, it is said to be inconsistent.

\begin{align} x_1 + 3x_2 - 4x_3 &= 1 \\ 2x_1 + 6x_2 - 8x_3 &= 2 \end{align}

does not have any solutions since the corresponding planes are parallel. Hence, the system is inconsistent.

Solving Systems of Linear Equations

Let's find all solutions of the following system of linear equations

\begin{align} x_1 + x_2 - 2x_3 &= 4 \\ x_1 + 3x_2 - x_3 &= 7 \\ 2x_1 + x_2 - 5x_3 &= 7 \end{align}

Solution:

Add (-1) times the first equation to the second equation. The first and third equations are unchanged

\begin{align} x_1 + x_2 - 2x_3 &= 4 \\ 2x_2 + x_3 &= 3 \\ 2x_1 + x_2 - 5x_3 &= 7 \end{align}

Add (-2) times the first equation to the third equation. 

\begin{align} x_1 + x_2 - 2x_3 &= 4 \\ 2x_2 + x_3 &= 3 \\ - x_2 - x_3 &= -1 \end{align}

Interchange the second and third equations.

\begin{align} x_1 + x_2 - 2x_3 &= 4 \\ - x_2 - x_3 &= -1 \\ 2x_2 + x_3 &= 3 \end{align}

Multiply the second equation by (-1).

\begin{align} x_1 + x_2 - 2x_3 &= 4 \\ x_2 + x_3 &= 1 \\ 2x_2 + x_3 &= 3 \end{align}

Add (-2) times the second equation to the third equation.

\begin{align} x_1 + x_2 - 2x_3 &= 4 \\ x_2 + x_3 &= 1 \\ - x_3 &= 1 \end{align}

We could continue but it is simpler to complete the solutions through back-substitution, since we know that \(x_3 = -1\).

\begin{align} x_2 &= 1 - x_3 = 1 - (-1) = 2 \\ x_1 &= 4 - x_2 + 2x_3 = 4 - 2 + 2(-1) = 0 \end{align}

Thus, the only solution of this system is 

\[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 2 \\ -1 \end{bmatrix} \]

This solution checks out with the original equations.

\begin{align} 0 + 2 - 2(-1) = 4 \\ 0 + 3(2) - (-1) = 7 \\ 2(0) + 2 -5(-1) = 7 \end{align}

The solution process we just performed is what we call Gaussian Elimination with back-subsitution. It involved the elimination steps:

  • Multiply one equation by a non-zero constant.
  • Interchange two equations.
  • Add a multiple of one equation to another equation.
Creative Commons License
Designed by Matthew Cheung. This work is licensed under a Creative Commons Attribution 4.0 International License.
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