**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

\begin{equation}

\frac{a_{11}}{a_{12}}\not=\frac{a_{21}}{a_{22}}

\label{eq1}

\end{equation} 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.