Question 507122: Let T: R^3 to R^2 be defined by T(a,b,c)=(a+b,2a-c). Determine T^-1(1,11)
Answer by tinbar(133)   (Show Source):
You can put this solution on YOUR website! There are many answers to this.
We know that some T(a,b,c) = (a+b,2a-c) = (1,11). From this it follows that a+b = 1, and 2a-c = 11
or a+b+0c = 1
and 2a+0b-c = 11
which we can make into a matrix:
1 1 0|1
2 0 -1|11
Then doing 2*row1-row2 -> row 2 makes the matrix:
1 1 0|1
0 2 1|-9
divide row 2 by half so the first non-zero entry is 1 (instead of 2)
1 1 0|1
0 1 (1/2)|(-9/2)
Now replace row1 by doing row1-row2:
1 0 (-1/2)|(11/2)
0 1 (1/2)|(-9/2)
So what is this saying now? It's basically saying you choose any value for c, then a = (11/2)+(1/2)c = (11+c)/2
and b = (-9/2)+(-1/2)c = (-9-c)/2.
Let's check then: a + b = (11+c)/2 + (-9-c)/2 = (11-9+c-c)/2 = 2/2 = 1, as required.
Next, 2a-c = 2(11+c)/2 - c = (11+c)-c = 11, as required.
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