Definition of a linear equation

An equation in unknowns x1,x2,…,x(n) is linear if it only involves constant multiples of the xi and perhaps a constant term\text{An equation in unknowns } x_{1}, x_{2}, \dots, x_(n) \text{ is linear if it only involves constant multiples of the } x_{i} \text{ and perhaps a constant term}

Example: 5x+3y+7z=0,x+y=1\text{Example: } 5x + 3y + 7z = 0, x + y = 1 Nonlinear equation: x2+y2=1\text{Nonlinear equation: } x^{2} + y^{2} = 1

  • Typically want to put all variables on one side
  • Equations with functions (such as sin and cosine) are not considered linear
  • Equations with the form 1 = 1 are also linear

System of linear equations (SLE)

Example system={x+2y+z=0x−y+3z=−2x+0y−z=4\text{Example system} = \begin{cases} x + 2y + z = 0 \\ x - y + 3z = -2 \\ x + 0y - z = 4 \\ \end{cases}
  • Note the 0y; must have all variables present in the system for each equation
  • Acts as a constraint on a set of points
  • Even one equation can count as a system of linear equations; the constraints are just limited to that one equation
  • In two dimensions, linear equations can be thought of as lines (such as x - 3y = -3 or 2x + y = 8)
    • Usually has one solution: the intersecting point
  • Systems of linear equations can also have no solutions (such as when two lines are parallel to each other) or infinite solutions (such as when the lines are the same)
    • Systems with no solutions are unsolvable
    • With these observations, systems of linear equations can either have zero, one, or infinite solutions; no in-betweens
Suppose that (r1,r2) and (s1,s2) both solve {ax+by=αcx+dy=βand t∈R. Therefore, the point (tr1+(1−t)s1,tr2+(1−t)s2) also solves the system of equations. Plugging in this point into the original system gives us:a(tr1+(1−t)s1)+b(tr2+(1−t)s2)=atr1+a(1−t)s1+btr2+b(1−t)s2=t(ar1+br2)+(1−t)(as1+bs2)=tα+(1−t)α=αTherefore, any value of t can transform the two points, meaning that there are infinite solutions.\text{Suppose that } (r_{1}, r_{2}) \text{ and } (s_{1}, s_{2}) \text{ both solve } \\ \begin{cases} ax + by = \alpha \\ cx + dy = \beta \\ \end{cases} \\ \text{and } t\in\mathbb{R} \text{. Therefore, the point } (tr_{1} + (1 - t)s_{1}, tr_{2} + (1 - t)s_{2}) \text{ also solves the system of equations. Plugging in this point into the original system gives us:} \\ a(tr_{1} + (1 - t)s_{1}) + b(tr_{2} + (1 - t)s_{2}) \\ = atr_{1} + a(1 - t)s_{1} + btr_{2} + b(1 - t)s_{2} \\ = t(ar_{1} + br_{2}) + (1-t)(as_{1} + bs_{2}) \\ = t\alpha + (1 - t)\alpha = \alpha\\ \text{Therefore, any value of t can transform the two points, meaning that there are infinite solutions.}
  • Essentially, if there are 2, 3, 4, etc. points, then there are infinite points that satisfy the system