Daniel K. answered • 02/24/20

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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 1^{st} row by (-2) and add it to 2nd row

1 1 -5

2 -4 -4

2.) Divide 2^{nd} row by (6)

-2 -2 10

0 -6 6

3.) Add 2^{nd} row to 1^{st} row and multiply 2^{nd} 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 1^{st} row by (-1) and add that to 2^{nd} row.

2.) Multiply 1^{st} row by (-2) and add that to 3^{rd} row.

1 1 -1 1

1 2 2 1

2 1 -5 2

3.) Add 2^{nd} row to the 3^{rd} row.

1 1 -1 1

0 1 3 0

0 -1 -3 0

4.) Since the 3^{rd} 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.