
Daniel K. answered 02/24/20
UIUC Grad to Math Tutor and more!
Part (a.) has a finite solution, while Part (b.) was found to have infinite solutions on one variable. The solutions to each is imaged below.
Part (a.): Set up the matrix as:
1.) Multiply 1st row by (-2) and add it to 2nd row
1 1 -5
2 -4 -4
2.) Divide 2nd row by (6)
-2 -2 10
0 -6 6
3.) Add 2nd row to 1st row and multiply 2nd row by (-1)
1 1 -5
0 -1 1
4.) Thus x = -4 and y = -1
1 0 -4
0 1 -1
Part (b.) is solved as below:
1.) Multiply 1st row by (-1) and add that to 2nd row.
2.) Multiply 1st row by (-2) and add that to 3rd row.
1 1 -1 1
1 2 2 1
2 1 -5 2
3.) Add 2nd row to the 3rd row.
1 1 -1 1
0 1 3 0
0 -1 -3 0
4.) Since the 3rd row is all zero’s, this translates to the “z” variable to be any value.
1 1 -1 1
0 1 3 0
0 0 0 0
Thus, these equations have infinite solutions for its “z” value.