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## Homogeneous systems of linear equations in two unknowns with the same solutions are equivalent

Solution to Linear Algebra Hoffman & Kunze Chapter 1.2 Exercise 1.2.6

Solution: Write the two systems as follows:
$$\begin{array}{c} a_{11}x+a_{12}y=0\\ a_{21}x+a_{22}y=0\\ \vdots\\ a_{m1}x+a_{m2}y=0 \end{array} \quad\quad \begin{array}{c} b_{11}x+b_{12}y=0\\ b_{21}x+b_{22}y=0\\ \vdots\\ b_{m1}x+b_{m2}y=0 \end{array}$$  Each system consists of a set of lines through the origin $(0,0)$ in the $x$-$y$ plane. Thus the two systems have the same solutions if and only if they either both have $(0,0)$ as their only solution or if both have a single line $ux+vy-0$ as their common solution. In the latter case all equations are simply multiples of the same line, so clearly the two systems are equivalent. So assume that both systems have $(0,0)$ as their only solution. Assume without loss of generality that the first two equations in the first system give different lines. Then

\frac{a_{11}}{a_{12}}\not=\frac{a_{21}}{a_{22}}
\label{eq1}
We need to show that there’s a $(u,v)$ which solves the following system:
$$\begin{array}{c} a_{11}u+a_{12}v=b_{i1}\\ a_{21}u+a_{22}v=b_{i2} \end{array}$$ Solving for $u$ and $v$ we get
$$u=\frac{a_{22}b_{i1}-a_{12}b_{i2}}{a_{11}a_{22}-a_{12}a_{21}}$$ $$v=\frac{a_{11}b_{i2}-a_{21}b_{i1}}{a_{11}a_{22}-a_{12}a_{12}}$$ By (\ref{eq1}) $a_{11}a_{22}-a_{12}a_{21}\not=0$. Thus both $u$ and $v$ are well defined. So we can write any equation in the second system as a combination of equations in the first. Analogously we can write any equation in the first system in terms of the second.

#### Linearity

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