Hey, I noticed a mistake in the lecture. The lecturer mentions that the system is contradictory, but in reality, it has many solutions.
When you break it down, you’ll see that the equations are consistent with each other. Specifically, you get ( c = 0 ) and ( b = -a ). Since ( a ) can be any value, the system isn’t contradictory at all it actually has infinitely many solutions. So, while it might seem like there’s a contradiction, the system is just more flexible than it appears.
Sorry, but I think the result for the third set looks correct. If you solve the third set of equations, you get:
c = 5
c = 4
So there is no solution. Just subtract the first equation from the second and third to see what I mean.
But I think set #4 is also Contradictory. If you subtract 2 x the first equation from the second equation you get:
0 = -5
Interesting point. Now I’m questioning myself if I really understand the big picture here.
I’d have to go back and listen to the full lecture again to make sure, but just reading what it says on the slide they are making the point that you can tell the system is singular without even looking at the constants on the RHS of the various equations. But just knowing a system is singular does not tell you whether you have no solutions or infinitely many solutions. For that full determination, the RHS matters.