## Elementary Linear Algebra 7th Edition

a) We can choose $a=11$, $b=5$ and $c=1$ b) We can choose $a=11$, $b=0$ and $c=0$ c) We can choose $a=11$, $b=0$ and $c=1$
The first two equations are independent and mutually consistent because we cannot derive one from another and working with them does not lead to any contradictions. a) If we want to have the unique solution the third equation must not be inconsistent with the first two but it has to be INDEPENDENT on them, i. e. it must not be derivable from them because otherwise it won't count as a new condition. Note that if we add together the first two equations we get $$x+x+5y+6y+z+(-z)=0+0$$ which becomes $$2x+11y=0$$ The last equation also starts with $2x$ so if we want it to be independent we can set $a=11$ and $b=5$, $c=1$ for example. We can newer obtain $2x+11y+5z$ just by using the first two equations so this is indeed a new equation, independent on the first two and not in contradiction with them. b) Here we want something different, we want the last equation to be derivable form the first two so that it does not really count as an independent equation. To this we can put $a=11$, $b=0$, $c=0$. In this way the last equation is $$2x+11y=0$$ which is just the first two equations added together so it is derivable from the first two and we will have infinitely many solutions. c) Now we want something different again. We want that the left side of the equation to be derivable from the left sides of the first two equations but to CONTRADICT them. We will achieve this by setting $a=11$, $b=0$ and $c=1$ because the last equation will say $$2x+11y=1$$ but the first two added together give $$2x+11y = 0$$ which implies $0=1$ and this is a contradiction so we have no solutions!