Answer:
(t, t -1, t)
Step-by-step explanation:
You have three unknowns but only 2 equations, so you can't really SOLVE this...you can get a solution with a variable still in it (I forget what this is called. I think it refers to infinite many solutions). Here's how it works:
Set up your matrix:
[tex]\left[\begin{array}{ccc}1&-2&1\\2&-1&-1\\\end{array}\right] \left[\begin{array}{ccc}2\\1\\\end{array}\right][/tex]
You want to change the number in position 21 (the 2 in the scond row) to a 0 so you have y and z left. Do this by multiplying the top row by -2 then adding it to the second row to get that 2 to become a 0. Multiplying in a -2 to the top row gives you:
[tex]\left[\begin{array}{ccc}-2&4&-2\\2&-1&-1\\\end{array}\right]\left[\begin{array}{ccc}-4\\1\\\end{array}\right][/tex]
Then add, keeping the first row the same and changing the second to reflect the addition:
[tex]\left[\begin{array}{ccc}-2&4&-2\\0&3&-3\\\end{array}\right] \left[\begin{array}{ccc}-4\\-3\\\end{array}\right][/tex]
The second equation is this now:
3y - 3z = -3. Solving for y gives you y = z - 1. Let's let z = t (some random real number that will make the system true. Any number will work. I'll show you at the end. Just bear with me...)
lf z = t, and if y = z - 1, then y = t - 1. So far we have that y = t - 1 and z = t. Now we solve for x:
From the first equation in the original system,
x - 2y + z = 2. Subbing in t - 1 for y and t for z:
x - 2(t - 1) + t = 2. Simplify to get
x - 2t + 2 + t = 2 and x - t = 0, and x = t. So the solution set is (t, t - 1, t). Picking a random value for t of, let's say 2, sub that in and make sure it works. If:
x - 2y + z = 2, then t - 2(t - 1) + t = 2 becomes t - 2t + 2 + t = 2, and with t = 2, 2 - 2(2) + 2 + 2 = 2. Check it: 2 - 4 + 4 = 2 and 2 = 2. You could pick any value for t and it will work.
