**Solution to Linear Algebra Hoffman & Kunze Chapter 1.3 Exercise 1.3.8**

Solution:

(a) In this case the system of equations is

\begin{alignat*}{1}

0\cdot x_1 + 0\cdot x_2 &= 0\\

0\cdot x_1 + 0\cdot x_2 &= 0

\end{alignat*}Clearly any $(x_1,x_2)$ satisfies this system since $0\cdot x=0$ $\forall$ $x\in F$.

(b) Let $(u,v)\in F^2$. Consider the system:

\begin{alignat*}{1}

a\cdot x_1 + b\cdot x_2 &= u\\

c\cdot x_1 + d\cdot x_2 &= v

\end{alignat*}If $ad-bc\not=0$ then we can solve for $x_1$ and $x_2$ explicitly as

$$

x_1=\frac{du-bv}{ad-bc}\quad x_2=\frac{av-cu}{ad-bc}.$$Thus there's a unique solution for all $(u,v)$ and in partucular when $(u,v)=(0,0)$.

(c) Assume WOLOG that $a\not=0$. Then $ad-bc=0$ $\Rightarrow$ $d=\frac{cb}{a}$. Thus if we multiply the first equation by $\frac ca$ we get the second equation. Thus the two equations are redundant and we can just consider the first one $a x_1+bx_2=0$. Then any solution is of the form $(-\frac bay,y)$ for arbitrary $y\in F$. Thus letting $y=1$ we get the solution $(-b/a,1)$ and the arbitrary solution is of the form $y(-b/a,1)$ as desired.