Solve the system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
. . .3x + ..y - . .z = ..0
. . ...x + ..y + 2z = ..6
. . .2x + 2y + 3z = 10
First I created the augmented matrix:
. . .[ 3 1 -1 | ..0 ]
. . .[ 1 1. 2 | ..6 ]
. . .[ 2 2. 3 | 10 ]
Then I switched R1 and R2:
. . .[ 1 1. 2 | ..6 ]
. . .[ 3 1 -1 | ..0 ]
. . .[ 2 2. 3 | 10 ]
To get rid of the "3" in row 2, I multiplied R1 by -3 and added it to R2 (R1(-3)+R2)
. . .[ 1. 1..2 | . . 6 ]
. . .[ 0 -2 -7 | -18 ]
. . .[ 2. 2..3 | ..10 ]
Now I'm stuck! How do I get -2 (row 2 column 2) to 1 without affecting row 1 column 1 or row 2 column 1? I was going to divide R2 by -2 but then I get an ugly fraction for row 2 column 3. Is there a better solution?
Thanks in advance.
. . .3x + ..y - . .z = ..0
. . ...x + ..y + 2z = ..6
. . .2x + 2y + 3z = 10
First I created the augmented matrix:
. . .[ 3 1 -1 | ..0 ]
. . .[ 1 1. 2 | ..6 ]
. . .[ 2 2. 3 | 10 ]
Then I switched R1 and R2:
. . .[ 1 1. 2 | ..6 ]
. . .[ 3 1 -1 | ..0 ]
. . .[ 2 2. 3 | 10 ]
To get rid of the "3" in row 2, I multiplied R1 by -3 and added it to R2 (R1(-3)+R2)
. . .[ 1. 1..2 | . . 6 ]
. . .[ 0 -2 -7 | -18 ]
. . .[ 2. 2..3 | ..10 ]
Now I'm stuck! How do I get -2 (row 2 column 2) to 1 without affecting row 1 column 1 or row 2 column 1? I was going to divide R2 by -2 but then I get an ugly fraction for row 2 column 3. Is there a better solution?
Thanks in advance.