I'm confussed on how to solve this because of the constants a for the given value of s.
x1 + 0x2 + x3 = a
5x1 + (s-1)x2 + 3x3 = a
sx1 + 0x2 + 4x3 = 1
Find what value of s does the system have on solution?
For the general case, a makes no difference:
Have , which is ,
so, when exists, have and it's solved. But exists iff .
Thus, the general condition that will make that system solvable has nothing to do with the B. It's decided by entirely by M, via .
So when when
Factoring gives , so when .
Thus this is solveable whenever , no matter what a is.
Now, when , it *might* have solutions, but then the value of will decide that.
Check by doing row reduction on:
When you get:
so (used row 1 to clear column 1)
so (used row 2 to clear column 3)
Thus and .
The last equation puts a restriction on a: implies implies .
Thus when s = 1, it has a solution, but only if a = 1/7.
The s = 1 and a = 1/7 solution is: .
I'll leave it to you to see what happens when s = 4.