Hi, can someone check my solution to this question? Not so sure if I'm right or not.
Find the value of s,t for R so that the following system can have no solutions, infinite solutions and unique.
3w + x + (t+1)y + (t+1)z=1
tw + x + 4y + sz = 1
w + ty = 1
2w + x + y + tz = 0
Unique solution: s#0, t#1
Infinite solutions: t=1
No solution: s=0
Thank guys!
 & (t+1) & 1 \\<br />t & 1 & 4 & s & 1\\<br />1 & 0 & t & 0 & 1\\<br />2 & 1 & 1 & t & 0\\<br />\end{bmatrix}<br /><br /> )
I don't know why there is a <br/> thing there, so just ignore it
Anyway, try and row reduce this thing.
When row reducing it, make sure you don't divide by any pronumerals. For instance, if you get a row that says (t+1) (t+1) (t+1)^2 (2(t+1), do NOT divide by t+1 as if t=-1, you've just divided by zero.
See where that leads you.
No solutions are if the leading one in that row is in the last column.
You get infinitely many solutions if the rank of your matrix is less than four. AKA row echelon form has at least one row all zeros in this case.
Now, from what I see (plugged into CAS), you get a unique solution if t^2 - 2t -3 is not 0. If t = -1, the row echelon form has a 'dud' row in which you have 0 0 0 0 1, implying 0w + 0x + 0y + 0z = 1, which doesn't make sense. Thus, no solutions for t = -1
If t = 3, then you have an empty row, so the matrix is rank 3 with no 'dud' rows. Infinitely many solutions here.
The variable s plays no part in this.
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