of multiplying by a fraction and two for one of the new rows you have
to multiply row 1 by something and row 2 by something THEN add them
together. For your math pleasure included below in the row operations
I did to solve the matrix. Follow along and verify that I did my math
right.
The original equations:
X + Y -Z = 6
3X -2Y +Z = -5
X + 3Y -2Z =14
Which produces the augemented matrix
[1 1 -1 : 6]
[3 -2 1 : -5]
[1 3 -2 :14]
[1 0 0 : a]
[0 1 0 : b]
[0 0 1 : c]
So the first thing I need a 1 in the upper left corner. Check. That was easy.
I now need 0s. How do I get zeros? If I used the operations:
R2 = -3r1 +r2 &
R3 = r1 + (-1)r3
I would end up with the matrix
[1 1 -1 : 6]
[0 -5 4 : -23]
[0 -2 1 : -8]
Remember to double check what I am doing. It does you NO good if I
understand how to do this.
This next part I can see where you would be confused where to go next.
I need a zero on the bottom. How do I get it there? What is a
common denominator between 5 and 2? Would 10 work? How do I make
each fisrt number 10?
What if I times row 2 by -2 and row 3 by +5 Then I could add -10 to
+10 and would that = 0?
So try the row operation:
R3 = -2r2 + 5r3
If you do, you should get the matrix
[1 1 -1 : 6]
[0 -5 4 : -23]
[0 0 3 : 6]
Email me if you have any questions about that last step and how I got
it. If you get it we'll continue on.
At this point I can either get ones along the diagonal or go for
zeros. Also at this point if the teacher didn't ask for reduced row
echelon (r.r.e) form I am actually done. Because I know that
3Z = 6 solve for Z
-5Y + 4Z = -23 (I know Z from above so plug in and solve for Y).
& X + Y -Z (Again from above I now know Y and Z so plug in and get X).
But assuming your teacher was a jerk and actually wanted r.r.e form
we'll continue our dreary task.
Because it is super easy I am going to make the diagonal 1s.
let
R2 = (-1/5)r2 &
R3 = (-1/3)r3
If we do this our new matrix is
[1 1 -1 : 6]
[0 1 (-4/5) : (23/5)]
[0 0 1 : -2]
Now let R3 = r2+ (4/5)r3. I leave it to you to figure out why. It is
also left up to you to see if there was another way of proceeding if I
hadn't changed the diagonals to ones. I'll give you a hint. It
involves +/- 12 and its common factors.
[ 1 1 -1 : 6]
[ 0 1 0 : 3]
[ 0 0 1 : 2]
I purposely dropped a negative sign in the above matrix, see if you
can find it. Don't go on until you find it. I want to make sure you
are paying attention and not just mindlessly reading my email.
Found it yet? Good!
For my next trick I will let R1 = r1 + r3. Why? To get a zero where I want it.
[1 1 0 : 4]
[0 1 0 : 3]
[0 0 1 : -2]
I am almost there. I only have one more set of row operations. I am
going to let you figure out what they are. But if done right you
should get the matrix
[1 0 0 : 1]
[0 1 0 : 3]
[0 0 1 : -2]
Did you get the same thing I did? Do you understand what I did and
more importantly, why I did it? Is there anything that didn't make
sense that I need to go over one more time. Speak now or forever hold
your peace, well at least until next time. I hope that helped. If
you need anything else you know how to reach me.
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