**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.

## Pedro Cardoso

18 Aug 2021Salutations. Doesn't this solution only work if we assume the coefficients ai1, ai2, bi1, bi2 and that x,y are real numbers? Does it also work for complex numbers?