- #1

- 2

- 0

**1. I am having a trouble with this math problem and would like help finding the answer. Here goes: Let (the ~~ indicates where a line should straight line should be going downward and /// are like a space between the numbers)**

1 /// (-1) /// 1 ~~ 2

a /// 0 /// a+1 ~~ 2a+3

2a /// (-2a) /// (a^2)-3 ~~ 5a-3

be the augmented matrix of a linear system. For what values of a does the system have no solution? One solution? Infinitely many solutions?

1 /// (-1) /// 1 ~~ 2

a /// 0 /// a+1 ~~ 2a+3

2a /// (-2a) /// (a^2)-3 ~~ 5a-3

be the augmented matrix of a linear system. For what values of a does the system have no solution? One solution? Infinitely many solutions?

So, if I correctly reduced it, that would mean that

1 /// 0 /// 1+1/a ~~ 2+3/a

0 /// 1 /// 1/a ~~ 3/a

0 /// 0 ///(a-3)(a+1) ~~ a-3

0 solutions when a=-1

infinite solutions when a=3

one solution when a/=/(doest not equal) 1, 3

And also a/=/ 0 because that would be dividing by 0.

Is this the way you are supposed to do this type of problem?

So, if I correctly reduced it, that would mean that

1 /// 0 /// 1+1/a ~~ 2+3/a

0 /// 1 /// 1/a ~~ 3/a

0 /// 0 ///(a-3)(a+1) ~~ a-3

0 solutions when a=-1

infinite solutions when a=3

one solution when a/=/(doest not equal) 1, 3

And also a/=/ 0 because that would be dividing by 0.

Is this the way you are supposed to do this type of problem?