SOLUTION: The question asks:
Solve each system of equations using matrices (row operations).
Problem 49:
{{{x-2y+3z=7}}}
{{{2x+y+z=4}}}
{{{-3x+2y-2z=-10}}}
Here's what I've tried
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Question 138333This question is from textbook College Algebra Enhanced With Graphing Utilities
: The question asks:
Solve each system of equations using matrices (row operations).
Problem 49:
Here's what I've tried:
Original augmented matrix:
[1 -2 3 7]
[2 1 1 4]
[-3 2 -2 -10]
*R[3]=r[1]+r[3]
[1 -2 3 7]
[2 1 1 4]
[-2 0 1 -3]
*R[2]=-2r[1]+r[2]
*R[3]=2r[1]+r[3]
[1 -2 3 7]
[0 5 -5 -10]
[0 -4 7 11]
*R[2]=(1/5)r[2]
*R[3]=-2r[1]+r[3]
[1 -2 3 7]
[0 1 -1 -2]
[0 0 1 -3]
*R[1]=2r[2]+r[1]
[1 0 1 3]
[0 1 -1 -2]
[0 0 1 -3]
*R[1]=-1r[3]+r[1]
[1 0 0 6]
[0 1 -1 -2]
[0 0 1 -3]
*R[2]=r[3]+r[2]
[1 0 0 6]
[0 1 0 -5]
[0 0 1 -3]
Therefore, I assume that the answer is x=6, y=-5, z=-3
BUT it's not. The answer is actually x=2, y=-1, z=1
When I input the matrix into my TI 84 Plus, I get
[1 0 0 2]
[0 1 0 -1]
[0 0 1 1]
which is the correct answer.
So, I know the start, I know the end, but I'm having trouble with the innards. Any suggestions? Thanks in advance for your help!
This question is from textbook College Algebra Enhanced With Graphing Utilities
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Solve each system of equations using matrices (row operations).
Problem 49:
-----------------------------------------
Here's what I've tried:
Original augmented matrix:
[1 -2 3 7]
[2 1 1 4]
[-3 2 -2 -10]
------------------------
*R[3]=r[1]+r[3]
[ 1 -2 3 7]
[ 2 1 1 4]
[-2 0 1 -3]
-------------------------
*R[2]=-2r[1]+r[2]
*R[3]=2r[1]+r[3]
[1 -2 3 7]
[0 5 -5 -10]
[0 -4 7 11]
-----------------------------
*R[2]=(1/5)r[2]
*R[3]=-2r[1]+r[3]
[1 -2 3 7]
[0 1 -1 -2]
[0 0 1 -3]
You made an error in forming R[3}
You should leave r(1)alone and work r(2) against r(3)
I would suggest you not try to do 2 steps in one transformation of the matrix.
====================
Cheers,
Stan H.
*R[1]=2r[2]+r[1]
[1 0 1 3]
[0 1 -1 -2]
[0 0 1 -3]
*R[1]=-1r[3]+r[1]
[1 0 0 6]
[0 1 -1 -2]
[0 0 1 -3]
*R[2]=r[3]+r[2]
[1 0 0 6]
[0 1 0 -5]
[0 0 1 -3]
Therefore, I assume that the answer is x=6, y=-5, z=-3
BUT it's not. The answer is actually x=2, y=-1, z=1
When I input the matrix into my TI 84 Plus, I get
[1 0 0 2]
[0 1 0 -1]
[0 0 1 1]
which is the correct answer.
So, I know the start, I know the end, but I'm having trouble with the innards. Any suggestions? Thanks in advance for your help!
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